let us c solutions ( by chetan )

Let us C solutions ( By Chetan )

http://letuscalllessons.blogspot.com/[07-Apr-14 8:37:39 PM]

will be uploaded soon ( solutions of Let us C++ and Let us JAVA ), Let me know if you need help in any program here.


Monday, 3 February 2014

Drawing a hut using graphics

NOTE: please change the "initgraph" path of BGI directory according to your turbo C++ IDE's directory.

#include<graphics.h>void main() {

int gd=DETECT,gm;

initgraph(&gd,&gm,"c:\\turboc3\\bgi");

line(200,100,170,200);line(200,100,250,200);line(200,100,400,100);line(400,100,450,200);line(450,200,250,200);line(178,176,178,277);line(245,190,245,300);line(440,200,440,300);line(440,300,245,300);line(245,300,178,277);line(210,288,210,215);line(190,280,190,215);line(190,215,210,215);

rectangle(300,210,380,260);

setfillstyle(5,RED);floodfill(399,101,WHITE);

setfillstyle(SOLID_FILL,WHITE^23);floodfill(299,209,WHITE);

Chetan Raikwar

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▼ 2014 (16)

► January (14)

▼ February (2)

Dynamic calender animated without graphics library...


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Posted by Chetan Raikwar at 11:11 No comments:

floodfill(179,275,WHITE);

setfillstyle(SOLID_FILL,BLACK^9);floodfill(191,216,WHITE);


setcolor(BLACK);settextstyle(8,HORIZ_DIR,3);outtextxy(200,200,"Chetan Raikwar");

setcolor(BLUE);setfillstyle(CLOSE_DOT_FILL,BLUE);floodfill(0,0,WHITE);getch();closegraph();restorecrtmode();

}

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Dynamic calender animated without graphics library

#include<stdio.h>#include<conio.h>#include<dos.h>

/*********************************//* function to tackle arrow keys *//*********************************/

getkey() {

union REGS i,o;

while(!kbhit());

i.h.ah=0;int86 (22,&i,&o);

return (o.h.ah);

}

void main() {

int x,y,i,lastday,key;int month=2,year=2014,a;



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void box();

clrscr();

/***********************************************//* starting the program with a condition *//***********************************************/

if(month<=12 && month>=1 && year>=1900 && year<=2045) {

do {

/* if up arrow key is hit */

if(key==72) {

sound(1400);delay(20);sound(750);delay(10);nosound();

if(year+1 > 2045) {}else {year=year+1; /* increment of year */}

}

/* if down arrow key is hit */

if(key==80) {


if(year-1 < 1900) {}else {year=year-1; /* decrement of year */}

}

/* if left arrow key is hit */

if(key==75) {


if(month-1 < 1){}else {month=month-1; /* decrement of month */}

}

/* if right arrow key is hit */

if(key==77) {

sound(1600);



delay(20);sound(1250);delay(10);

nosound();

if(month+1 > 12){}

else {month=month+1; /* increment of month */}

}

x=49,y=9,i=1; /* calender printing objects */

x = dayfinder(month,year); /* calculating first day of the month */

lastday = totaldays(month,year); /* calculating total days of the month*/

clrscr();

box(month,year); /* drawing boxes and headings of calender */

/*************************//* printing the calender *//*************************/

while(i<=lastday) {

gotoxy(x,y);

if(x==22)textcolor(RED);

elsetextcolor(YELLOW);

cprintf("%2d",i);

i++;x+=5;

if(x>52) { /* if the position of 7 days is covered, again print frombeginning from a new line */

x=22;y+=2;

}}

gotoxy(1,1); /* moving cursor away from calender */

key=getkey(); /* taking the arrow key input */

} while(key==72 || key==75 || key==77 || key==80);

}

elseprintf("Error! invalid input\n");

getch();

}/*********************** main ends ************************/

/**********************************************************//* function to find first day of the given month and year */



/**********************************************************/

int dayfinder(int month, int year)

{

int a,day=1;

/* this is a general purpose formula to calculate first day */

a=(14-month)/12;year=year-a;month=month+12*a-2;

day=(day+year+(year/4)-(year/100)+(year/400)+((31*month)/12)) % 7;

/* determining the position to print the first day in the calender */

if(day==0)day=22;

else if(day==1)day=27;






return (day); /* return the position */

}

/********************************************************//* function to draw the boxes, headings of the calender *//********************************************************/

void box(int m,int y) {

int i,j,k,l;

/*************//* inner box *//*************/

textcolor(BLUE);/* corners of inner box */

gotoxy(20,3);cprintf("%c",218);




/* boundries of inner box */



for(j=4;j<=20;j++) {

gotoxy(20,j);cprintf("%c",179);


}

for(i=21;i<=54;i++) {

gotoxy(i,3);cprintf("%c",196);


}

/*************//* outer box *//*************/

/* corners of outer box */





/* boundries of outer box */

for(k=2;k<=22;k++) {

gotoxy(17,k);cprintf("%c",179);


}

for(l=18;l<=57;l++) {

gotoxy(l,1);cprintf("%c",196);


}

textcolor(GREEN+BLINK);gotoxy(24,22);cprintf("%c%c%c%c %c%c %c%c%c%c%c%c%c%c%c%c%c%c%c",77,65,68,69,66,89,67,72,69,84,65,78,82,65,73,75,87,65,82);

/********************************************//* writing heading on appropriate positions *//********************************************/



textcolor(RED);

gotoxy(22,6);cprintf("Sun");

textcolor(YELLOW);

gotoxy(27,6);cprintf("Mon");

gotoxy(32,6);cprintf("Tue");

gotoxy(37,6);cprintf("Wed");

gotoxy(42,6);cprintf("Thu");

gotoxy(47,6);cprintf("Fri");

gotoxy(52,6);cprintf("Sat");

textcolor(LIGHTGREEN);

gotoxy(32,4);

if(m==1)cprintf("January ");

if(m==2)cprintf("February ");

if(m==3)cprintf("March ");

if(m==4)cprintf("April ");

if(m==5)cprintf("May ");

if(m==6)cprintf("June ");

if(m==7)cprintf("July ");

if(m==8)cprintf("August ");

if(m==9)cprintf("September ");

if(m==10)cprintf("October ");

if(m==11)cprintf("November ");

if(m==12)cprintf("December ");

textcolor(CYAN);cprintf("%d",y);/*************************//* printing instructions *//*************************/



textcolor(BROWN);

gotoxy(60,16);cprintf("%c : Next year",30);

gotoxy(60,18);cprintf("%c : Previous year",31);

gotoxy(60,20);cprintf("%c : Next month",16);

gotoxy(60,22);cprintf("%c : Previous month",17);

}

/***************************************************//* function to determine total days of given month *//***************************************************/

int totaldays(int m,int y) {

int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;

/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */




if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}


Thursday, 30 January 2014

Let Us C / Chapter 14 (Operations on Bits)

(a) The information about colors is to be stored in bits of a char variable called color. The bit number 0 to 6, each represent 7 colors of a rainbow, i.e. bit 0 represents violet, 1 representsindigo, and so on. Write a program that asks the user to enter a number and based on this number it reports which colors in the rainbow does the number represents.

#include<stdio.h> #include<conio.h> void main() {

char color; int num;

clrscr();

printf("Please enter the number(0-6): "); scanf("%d",&num);

color=1<<num;

printf("\n\n");

if(num==0 && color==1) printf("Violet");

else if(num==1 && color==2) printf("Indigo");

else if(num==2 && color==4) printf("Blue");

Exercise [A]

Solution:










else if(num==3 && color==8) printf("Green");

else if(num==4 && color==16) printf("Yellow");

else if(num==5 && color==32) printf("Orange");

else if(num==6 && color==64) printf("Red");

else printf("Wrong color number!");

getch();

}

----------------------------------------------------------------------------------------------------------------

(b) A company planning to launch a new newspaper in market conducts a survey. The various parameters considered in the survey were, the economic status (upper, middle, and lower class) the languages readers prefer (English, Hindi, Regional language) and category of paper (daily, supplement, tabloid). Write a program, which reads data of 10 respondents through keyboard, and stores the information in an array of integers. The bit-wise information to be stored in an integer is given below:Bit Number Information0 Upper class1 Middle class2 Lower class3 English4 Hindi5 Regional Language6 Daily7 Supplement8 TabloidAt the end give the statistical data for number of persons who read English daily, number of upper class people who read tabloid and number of regional language readers.


int arr[10][3],i,j; unsigned int infr; int eng=0,utab=0,rgl=0;

for(i=0;i<10;i++) {

clrscr();

gotoxy(20,2); printf("Enter data of respondent %d:\n\n",i+1);

for(j=0;j<3;j++) {

Solution:



if(j==0){ printf("Economic Status:\n\n"); printf("0: Upper class\t1: Middle class\t2: Lower class\n\n");

scanf("%d",&arr[i][j]); }

if(j==1){ printf("\n\nLanguage Preferred:\n\n"); printf("3: English\t4: Hindi\t5:Regional Language\n\n");


if(j==2){

printf("\n\nType of Paper:\n\n"); printf("6: Daily\t7: Supplement\t8: Tabloid\n\n");


} }

/***********************************************/ /* converting the whole array using left shift */ /***********************************************/

for(i=0;i<10;i++) {

for(j=0;j<3;j++) {

arr[i][j]= 1 << arr[i][j]; /* conversion */

} }

for(i=0;i<10;i++) {

if(arr[i][1]==8) /* english readers */ eng++;

if(arr[i][0]==1 && arr[i][2]==256) /* upper class,tabloid readers */ utab++;

if(arr[i][1]==32) /* regional language readers */ rgl++; }

clrscr();

gotoxy(20,2); printf("Reader's statistics:\n\n\n\n"); printf("\tEnglish Reader: %d\n",eng); printf("\tUpper class Tabloid Readers: %d\n",utab); printf("\tRegional Language readers: %d\n",rgl); getch();

}

----------------------------------------------------------------------------------------------------------------

(c) In an inter-college competition, various sports and games are played between different colleges like cricket, basketball, football,



hockey, lawn tennis, table tennis, carom and chess. The information regarding the games won by a particular college is stored in bit numbers 0, 1, 2, 3, 4, 5, 6, 7 and 8 respectively of an integer variable called game. The college that wins in 5 or more than 5 games is awarded the Champion of Champions trophy. If a number is entered through the keyboard, then write a program to find out whether the college won the Champion of the Champions trophy or not, along with the names of the games won by the college.


int game; int i,won=0,name;

clrscr();

printf("Enter number: ");

scanf("%d",&name);

printf("\n\n");

for(i=1;i<=8;i++) {

game=i&name;

if(game==1) { printf("Cricket\n");

won++; }

if(game==2) { printf("Basketball\n");

won++; }

if(game==3) { printf("Football\n");

won++; }

if(game==4) { printf("Hockey\n");

won++; }

if(game==5) { printf("Lawn-tennis\n");

won++; }

if(game==6) { printf("Table-tennis\n");

won++; }

Solution:



if(game==7) { printf("Carrom\n");

won++; }

if(game==8) { printf("Chess\n");

won++; }

}

if(won>=5) {

printf("\n\n\nCollege %d has won the champions trophy\n",name);

}

else {

printf("\n\n\nCollege %d didn't win the champions trophy\n",name);

}

getch(); }

-----------------------------------------------------------------------------------------------------------------

(d) An animal could be either a canine (dog, wolf, fox, etc.), a feline (cat, lynx, jaguar, etc.), a cetacean (whale, narwhal, etc.) or a marsupial (koala, wombat, etc.). The information whether a particular animal is canine, feline, cetacean, or marsupial is stored in bit number 0, 1, 2 and 3 respectively of a integer variable called type. Bit number 4 of the variable type stores the information about whether the animal is Carnivore or Herbivore.For the following animal, complete the program to determine whether the animal is a herbivore or a carnivore. Also determine whether the animal is a canine, feline, cetacean or a marsupial.struct animal{char name[30] ;int type ;}struct animal a = { "OCELOT", 18 } ;


struct animal { char name[30]; int type; };

struct animal a={"OCELOT",18};

Solution:



int t1,t2,typ1,typ2;

clrscr();

t1= a.type >> 3;

typ1= t1 & 2; /* checking the bit after shifting */

t2= t1 << 3;

typ2= t2 & 16; /* checking another bit after shifting */

/**********************************************************/ /* checking if ocelot is canine/feline/cetacean/marsupial */ /**********************************************************/

if(typ1==1) printf("\n\nOCELOT is Canine\n");

if(typ1==2) printf("\n\nOCELOT is Feline\n");

if(typ1==4) printf("\n\nOCELOT is Cetacean\n");

if(typ1==8) printf("\n\nOCELOT is Marsupial\n");

/************************************************/ /* checking if ocelot is carnivore or herbivore */ /************************************************/

if(typ2!=0) printf("\n\nOCELOT is Carnivore\n");

if(typ2==0) printf("\n\nOCELOT is Herbivore\n");

getch();

}

-----------------------------------------------------------------------------------------------------------------

(e)The time field in the directory entry is 2 bytes long. Distribution of different bits which account for hours, minutes and seconds is given below. Write a function which would receive the two-byte time entry and return to the calling function, the hours, minutes and seconds. 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 H H H H H M M M M M M S S S S S


unsigned hr,mn,sc,input;

void timer();

Solution:



clrscr();

printf("Enter the time entry: ");

scanf("%d",&input);

timer(input,&hr,&mn,&sc);

printf("\n\nTIME: %u : %u : %u\n",*(&hr),*(&mn),*(&sc));

getch();

}

void timer(unsigned time, unsigned *hr, unsigned *mn, unsigned *sc) {

*hr=time >> 11;

*mn=(time<<5) >> 10;

*sc=(time<<11) >> 11;

}

-----------------------------------------------------------------------------------------------------------------

(f) In order to save disk space information about student is stored in an integer variable. If bit number 0 is on then it indicates Ist year student, bit number 1 to 3 stores IInd year, IIIrd year and IVth year student respectively. The bit number 4 to 7 stores stream Mechanical, Chemical, Electronics and IT. Rest of the bits store room number. Based on the given data, write a program that asks for the room number and displays the information about the student, if its data exists in the array. The contents of array are,int data[ ] = { 273, 548, 786, 1096 } ;

#include<stdio.h> #include<conio.h>

void main() {

unsigned int yr,br,data_yr,data_br; int i,j,k,rn,a,b,flag=0; int data[]={ 273,548,786,1096};

clrscr();

printf("Enter room number: "); scanf("%d",&rn);

printf("\n\n\n Year: ");

/*********************/ /* checking for year */ /*********************/

for(i=1;i<=8;i=i*2) {

yr= rn & i;

Solution:



if(yr==1) { printf("First year"); break; }

if(yr==2) { printf("Second year"); break; }

if(yr==4) { printf("Third year"); break; }

if(yr==8) { printf("Fourth year"); break; }

}

/***********************/ /* Checking for branch */ /***********************/

printf("\n\nBranch: ");

for(i=16;i<=128;i=i*2) {

br= rn & i;

if(br==16) { printf("Mechanical"); break; }

if(br==32) { printf("Chemical"); break; }

if(br==64) { printf("Electronics"); break; }

if(br==128) { printf("I.T."); break; }

}

/***********************************************/ /* checking if data matches with that of array */ /***********************************************/

for(i=1,j=16;i<=8,j<=128;i=i*2,j=j*2) {

yr=rn&i;

br=rn&j;

for(k=0,a=1,b=16;k<4,a<=8,b<=128;k++,a=a*2,b=b*2) {

data_yr=data[k] & a;




data_br=data[k] & b;

if(yr==data_yr && br==data_br) {

flag+=1;

break;

}

}

}

if(flag==0) printf("\n\n\ndata doesn't matche with the array.\n");

else printf("\n\n\ndata matches with the array.\n");

getch(); }______________________________________________________________________

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Let Us C / Chapter 13 (More Issues in Input Output)

(a) Write a program to carry out the following:(a) Read a text file provided at command prompt(b) Print each word in reverse orderFor example if the file containsINDIA IS MY COUNTRYOutput should beAIDNI SI YM YRTNUOC

#include"stdio.h"<+>#include<conio.h>#include<string.h>

void main(int argc, char *argv[]) {

FILE *fs;char ch,s[80];void rev();

if(argc!=2) {

printf("File name not supplied\n");

exit();

}

fs=fopen(argv[1],"r");

Exercise [B]

Solution:










if(fs==NULL) {

printf("File not found!");

exit();}

while(fgets(s,79,fs)!=NULL)rev(s);

fclose(fs);

}

void rev(char s1[80]) {

char s2[80];int i=0,j=0;

while(s1[i]!='\0') {

s2[j]=s1[i];

if(s1[i]==' ' || s1[i]=='\0') {

s2[j]='\0';

strrev(s2);

printf("%s ",s2);

j=-1;

}

i++;j++;}

s2[j]='\0';

printf("%s",strrev(s2));

}

-----------------------------------------------------------------------------------------------------------------

(b) Write a program using command line arguments to search for a word in a file and replace it with the specified word. The usage of the program is shown below.C> change <old word> <new word> <filename>

#include<stdio.h>#include<conio.h>#include<string.h>

void replace();

void main ( int argc, char *argv[] ) {

FILE *fp,*ft;char str[80],str1[30],str2[30],nwstr[100];

clrscr();

if(argc!=4) {

Solution:



puts("Improper argements passed!");exit();}

strcpy(str1,argv[1]);strcpy(str2,argv[2]);

fp=fopen(argv[3],"r");

if(fp==NULL) {

puts("cannot open source file!");exit();}

ft=fopen("NEW.TXT","w+");

if(ft==NULL) {

puts("cannot open target file!");exit();}

while(fgets(str,79,fp)!=NULL) {

replace(str,str1,str2,&nwstr);

fputs(nwstr,ft);}

fclose(fp);fclose(ft);

remove("FILE.TXT");rename("NEW.TXT","FILE.TXT");

printf("\n\nTask completed!\n\n");

getch();

}

void replace(char *s, char s1[80], char s2[80], char *n) {

int i=0,j=0,k=0;char temp[100],temp2[100],main[100],*t=temp,*m=main;

/*******************************//* copying to temporary string *//*******************************/

while(*s!='\0') {

*t=*s;

t++;s++;

}

*t='\0';

/**********************//* checking each word *//**********************/

while(temp[i]!='\0') {



temp2[j]=temp[i];

if(temp[i]==' ' || temp[i]==',' || temp[i]=='.') {

temp2[j]='\0';

if(strcmpi(temp2,s1)==0) {

strcpy(temp2,s2);

}

j=0;

while(temp2[j]!='\0') {

main[k]=temp2[j];

k++;j++;}

main[k]=' '; /* adding space after each word is copied */

k++; /* increment so that the next word won't replace the space */

j=-1;

}

i++;j++;}

temp2[j]='\0'; /* last word terminated */

if(strcmpi(temp2,s1)==0){ /* checking last word too */

strcpy(temp2,s2);

}

/***************************//* last word of the string *//***************************/

j=0;


main[k]=temp2[j];

k++;j++;}

main[k]='\0'; /* new string is completely ready */

/******************************************************************//* preparing new string to return to main and printing it on file *//******************************************************************/

while(*m!='\0') {

*n=*m;

n++;m++;}



*n='\0';

}

----------------------------------------------------------------------------------------------------------------

(c) Write a program that can be used at command prompt as a calculating utility. The usage of the program is shown below.C> calc <switch> <n> <m>Where, n and m are two integer operands. switch can be any one of the arithmetic or comparison operators. If arithmetic operator is supplied, the output should be the result of the operation. If comparison operator is supplied then the output should be True or False.



int n,m,calc;

clrscr();

if(argc!=4)printf("Improper number of arguments");

/* converting both operands to integers */

n=atoi(argv[2]);

m=atoi(argv[3]);

printf("\n\nOutput: ");

/* comparing the operators passed at command prompt */

if(strcmp("+",argv[1])==0) {

printf("%d\n\n",n+m);

}

if(strcmp("-",argv[1])==0) {

printf("%d\n\n",n-m);

}

if(strcmp("*",argv[1])==0) {

printf("%d\n\n",n*m);

}

if(strcmp("/",argv[1])==0) {

printf("%d\n\n",n/m);

Solution:




}

if(strcmp("%",argv[1])==0) {

printf("%d\n\n",n%m);

} /* IMPORTANT */

/* I have used dedicated variable for '<' and '>' comparison operators *//* because at command prompt, these are redirecting operators and cannot *//* be used simply for comparison. */

/* 1: "<" = "less" */ /* 2: ">" = "greater" */ /* type "less" for "<" and "greater" for ">" respectively if you want to perform comparison */

if(strcmp("greater",argv[1])==0) {

if(n>m)printf("True\n\n");

if(n<m)printf("False\n\n");

}

if(strcmp("less",argv[1])==0) {

if(n<m)printf("True\n\n");

if(n>m)printf("False\n\n");

}

if(strcmp("=",argv[1])==0) {

if(n==m)printf("True\n\n");

elseprintf("False\n\n");

}

}

_______________________________________________________________________


Let Us C / Chapter 11 (Console Input Output) Exercise [D]










(a) Write down two functions xgets( ) and xputs( ) which work similar to the standard library functions gets( ) and puts( ).

#include<stdio.h>#include<conio.h>

void xgets();void xputs();

void main() {

char str[80];clrscr();

xgets(str);

printf("\n");

xputs(str);

getch();

}

void xgets( char *s) {

int i=0;char ch;

for(i=0;i<=79;i++) {

ch=getche();

if(ch=='\r') {

*s='\0';break;}

if(ch=='\b') {

printf("\b");

i-=2;s-=2;}

else {

*s=ch;s++;

}

}

}

void xputs( char *s) {

while(*s!='\0') {

putch(*s);

Solution:




s++;

}

}

---------------------------------------------------------------------------------------------------------------

(b) Write down a function getint( ), which would receive a numeric string from the keyboard, convert it to an integer number and return the integer to the calling function. A sample usage of getint( ) is shown below:main( ){int a ;a = getint( ) ;printf ( "you entered %d", a )}

#include<stdio.h>#include<conio.h> void main() {

int a; char s[80];

printf("Enter any numeric string: "); gets(s);

a=getint(s); /* converting and receiving string as integer */

printf("\n\nYou entered = %d\n",a);

getch();

}

int getint(char s1[80]) {

int digit;

digit=atoi(s1); /* converting string to integer */

return digit;

}

______________________________________________________________________

Solution:


Let Us C / Chapter 10 (Structures)

(a) Create a structure to specify data on students given below:Roll number, Name, Department, Course, Year of joining

Exercise [D]










Assume that there are not more than 450 students in the collage.(a) Write a function to print names of all students who joined in a particular year.(b) Write a function to print the data of a student whose roll number is given.

/* NOTE: since number of students to be assumed is too much ( about 450 )I have alloted full size (about 450) to it but array has been kept empty,if you have time then you can fill up all 450 names and can search throughthem.program has been tested by me and it works accurately. you can reducethe size of array of structure conveniently by changing the value of N */

#include<stdio.h>#include<conio.h>#define N 450

struct students { int rlnm; char name[25]; char dept[25]; /* structure defined outside of main(); */ char course[25]; int year; };

void main() { /* main() */

struct students s[N];int i,ch;

clrscr();

/* taking input of 450 students in an array of structure */

for(i=0;i<N;i++) {

printf(" Enter data of student %d\t\t\t\ttotal students: %d\n",i+1,N);printf("****************************\n\n");

printf("enter rollnumber: ");scanf("%d",&s[i].rlnm);

printf("\n\nenter name: ");scanf(" %s",&s[i].name);

printf("\n\nenter department: ");scanf("%s",&s[i].dept);

printf("\n\nenter course: ");scanf("%s",&s[i].course);

printf("\n\nenter year of joining: ");scanf("%d",&s[i].year);

clrscr();

}

/* displaying a menu */

printf("\n\tenter your choice: \n");printf("\t**********************\n\n");

printf("1: enter year to search all students who took admission in that:\n\n");printf("2: enter roll number to see details of that student\n\n\n");

printf("your choice: "); /* taking input of your choice */

Solution:



scanf("%d",&ch);

clrscr();

switch(ch) {

case 1: clrscr();

dispyr(&s); /* function call to display names of students who joined in\ a particular year */

break;

case 2: clrscr();

disprl(&s); /* function call to display information of a student \ whose roll number is given */

break;

default:

printf("\n\nerror! wrong choice");

}

getch();

}/******************* main() ends **************/

dispyr(struct students *a) { /* function for displaying names of students\ who took admission in a particular year */

int j,yr;

printf("\nenter year: ");scanf("%d",&yr);

printf("\n\nthese students joined in %d\n\n",yr);

for(j=0;j<N;j++) {

if(a[j].year==yr) {

printf("\n%s\n",a[j].name);}

}return 0;}

disprl(struct students *a) { /* function to print information of a\ student whose roll number has been \ given. */

int k,rl;

printf("\nenter roll number: ");scanf("%d",&rl);

for(k=0;k<N;k++) {

if(a[k].rlnm==rl) {

printf("\n\n\t Details of roll number: %d\n",a[k].rlnm);printf("\t***************************\n\n");printf(" Name: %s\n",a[k].name);



printf(" Department: %s\n",a[k].dept);printf(" Course: %s\n",a[k].course);printf("Year of joining: %d",a[k].year);

break;}

else {printf("\nRoll number you entered does not exist\n\n");

break;}

}

return 0;}

--------------------------------------------------------------------------------------------------------------

(b) Create a structure to specify data of customers in a bank. The data to be stored is: Account number, Name, Balance in account. Assume maximum of 200 customers in the bank.(a) Write a function to print the Account number and name of each customer with balance below Rs. 100.(b) If a customer request for withdrawal or deposit, it is given in the form:Acct. no, amount, code (1 for deposit, 0 for withdrawal)Write a program to give a message, “The balance is insufficient for the specified withdrawal”.

/* NOTE: since number of customers to be assumed is too much ( about 200 ) I have alloted full size (about 200) to it but array has been kept empty, if you have time then you can fill up all 200 names and can search through them.program has been tested by me and it works accurately. you can reducethe size of array of structure conveniently by changing the value of N */

#include<stdio.h> #include<conio.h> #define N 200

struct bank { int acn; char name[20]; int bal; /* defined out of main() */ };

void main() {

struct bank b[N];

int i,ch,lw=100,ch2,ac,am;

clrscr();

for(i=0;i<N;i++) { /* inputting customer data */

printf("\tEnter information of customers \n"); printf("\t******************************\n\n");

Solution:



printf("enter account no.: "); scanf("%d",&b[i].acn);

printf("\n\nenter customer name: "); scanf("%s",&b[i].name);

printf("\n\nenter balance: "); scanf("%d",&b[i].bal);

clrscr();

}

clrscr();

printf("\tEnter your choice\n"); /* further processing of transaction */ printf("\t*****************\n\n");

printf("1: to know whose balance is below 100.\n\n"); printf("2: to process request or withdrawl.\n\n\n");

scanf("%d",&ch);

switch(ch) {

case 1:

clrscr();

disp(&b); /* displaying whose balance is below 100 */

break;

case 2:

clrscr();

printf("enter your account number: "); scanf("%d",&ac);

for(i=0;i<N;i++) {

if((b[i].acn)==ac) {

clrscr();

printf("\tHello %s\n",b[i].name); printf("\n\n");

printf("\n\nenter your choice\n"); printf("\n1: deposite:\n"); printf("\n0: withdrawl:\n\n"); scanf("%d",&ch2);

switch(ch2) {

case 0:

clrscr();

if(b[i].bal<lw) {

printf("\n\nsorry! account balance is too low for withdrawl.\n");

break; }



else {

printf("\n\nenter amount for withdrawl: "); scanf("%d",&am);

}

if(b[i].bal<am) {

printf("\n\nyou don't have enough balance for withdrawl.\n");

}

else {

b[i].bal=b[i].bal+am;

printf("\n\nwithdrawl was successful.\n");

} break;

case 1:

clrscr();

printf("\n\nenter amount to deposite: "); scanf("%d",&am);


printf("\n\ncash deposited successfully.\n\n");

break;

}

} } } getch(); }

disp(struct bank *a) {

int k;

printf("\tCustomers whose balance is below 100:\n"); printf("\t*************************************\n\n");

for(k=0;k<N;k++) {

if((a[k].bal)<100) {

printf("%2d\t%s\n",a[k].acn,a[k].name);

} } return 0;

}

-----------------------------------------------------------------------------------------------------------------

(c) An automobile company has serial number for engine parts starting from AA0 to FF9. The other characteristics of parts to be specified in a structure are: Year of manufacture, material and quantity



manufactured.(a) Specify a structure to store information corresponding to a part.(b) Write a program to retrieve information on parts with serial numbers between BB1 and CC6.

#include<stdio.h> #include<conio.h> struct automob { char name[5]; int year; char mtr[10]; int qty; }; void main() {

struct automob a[54]={{"AA1",2009,"AAA",10}, {"AA2",2009,"AAA",10}, {"AA3",2009,"AAA",10}, {"AA4",2009,"AAA",10}, {"AA5",2009,"AAA",10}, {"AA6",2009,"AAA",10}, {"AA7",2009,"AAA",10}, {"AA8",2009,"AAA",10}, {"AA9",2009,"AAA",10}, {"BB1",2010,"BBB",11}, {"BB2",2010,"BBB",11}, {"BB3",2010,"BBB",11}, {"BB4",2010,"BBB",11}, {"BB5",2010,"BBB",11}, {"BB6",2010,"BBB",11}, {"BB7",2010,"BBB",11}, {"BB8",2010,"BBB",11}, {"BB9",2010,"BBB",11}, {"CC1",2011,"CCC",12}, {"CC2",2011,"CCC",12}, {"CC3",2011,"CCC",12}, {"CC4",2011,"CCC",12}, {"CC5",2011,"CCC",12}, {"CC6",2011,"CCC",12}, {"CC7",2011,"CCC",12}, {"CC8",2011,"CCC",12}, {"CC9",2011,"CCC",12}, {"DD1",2012,"DDD",13}, {"DD2",2012,"DDD",13}, {"DD3",2012,"DDD",13}, {"DD4",2012,"DDD",13}, {"DD5",2012,"DDD",13}, {"DD6",2012,"DDD",13}, {"DD7",2012,"DDD",13}, {"DD8",2012,"DDD",13}, {"DD9",2012,"DDD",13}, {"EE1",2013,"EEE",14}, {"EE2",2013,"EEE",14}, {"EE3",2013,"EEE",14}, {"EE4",2013,"EEE",14}, {"EE5",2013,"EEE",14}, {"EE6",2013,"EEE",14}, {"EE7",2013,"EEE",14}, {"EE8",2013,"EEE",14}, {"EE9",2013,"EEE",14}, {"FF1",2014,"FFF",15}, {"FF2",2014,"FFF",15}, {"FF3",2014,"FFF",15}, {"FF4",2014,"FFF",15},

Solution:



{"FF5",2014,"FFF",15}, {"FF6",2014,"FFF",15}, {"FF7",2014,"FFF",15}, {"FF8",2014,"FFF",15}, {"FF9",2014,"FFF",15}};

int i,j,k;

clrscr();

printf("Information of parts between BB1 to CC6:\n\n");

printf("\n\nName\t\tyear(manu.)\t\tmaterial\tquantity\n\n\n");

for(i=0;i<54;i++) {

if(i>=9 && i<=23) { printf("%3s\t\t%4d\t\t\t%5s\t\t%4d\n",a[i].name,a[i].year,a[i].mtr,a[i].qty);

} }

getch(); }

-----------------------------------------------------------------------------------------------------------------

(d) A record contains name of cricketer, his age, number of test matches that he has played and the average runs that he has scored in each test match. Create an array of structure to hold records of 20 such cricketer and then write a program to read these records and arrange them in ascending order by average runs. Use the qusort( ) standard library function.

#include<stdio.h> #include<conio.h> #include<string.h>

void main() {

struct cricketer { char name[50]; int age; int match; float avrn; }; struct cricketer c[20],temp; int i,j; clrscr();

for(i=0;i<20;i++) {

printf("Enter data of cricketer %d\n\n\n",i+1);

fflush(stdin); printf("Name: "); gets(c[i].name);

Solution:



printf("\n\nAge: "); scanf("%d",&c[i].age);

printf("\n\nMatches: "); scanf("%d",&c[i].match);

printf("\n\nAverage runs: "); scanf("%f",&c[i].avrn);

clrscr();

}

/*******************/ /* sorting records */ /*******************/

for(i=0;i<20;i++) {

for(j=i+1;j<20;j++) {

if(c[i].avrn > c[j].avrn) {

temp=c[i]; c[i]=c[j]; c[j]=temp; } } }

printf("Sorted records:\n\n");

for(i=0;i<20;i++) {

printf("%d\t%s\t%d\t%d\t%f\n\n\n",i+1,c[i].name,c[i].age,c[i].match,c[i].avrn);

}

getch(); }

linkfloat() {

float a=0,*b;

b=&a; a=*b;

return 0; }

-----------------------------------------------------------------------------------------------------------------

(e) There is a structure called employee that holds information like employee code, name, date of joining. Write a program to create an array of the structure and enter some data into it. Then ask the user to enter current date. Display the names of those employees whose tenure is 3 or more than 3 years according to the given current date.

#include<stdio.h>#include<conio.h>#define N 3void main() {

Solution:



struct employee { int code; char name[20]; int dd,mm,yy; } e[N];

int d,m,y,i;clrscr();

for(i=0;i<N;i++) {

printf("\tEnter employee data:\n");printf("\t*********************\n");

printf("\nEnter employee code: ");scanf("%d",&e[i].code);

printf("\n\nEnter employee name: ");scanf("%s",&e[i].name);

printf("\n\nEnter date of joining (dd mm yy): ");scanf("%d%d%d",&e[i].dd,&e[i].mm,&e[i].yy);

clrscr();

}

clrscr();

printf("\nenter current date(dd mm yy): \n\n");scanf("%d%d%d",&d,&m,&y);

clrscr();

for(i=0;i<N;i++) {

if((y-(e[i].yy))>=3) {

printf("%s\n\N",e[i].name);

}

}

getch();}

-----------------------------------------------------------------------------------------------------------------

(f) Write a menu driven program that depicts the working of a library. The menu options should be:1. Add book information2. Display book information3. List all books of given author4. List the title of specified book5. List the count of books in the library6. List the books in the order of accession number7. ExitCreate a structure called library to hold accession number, title of the book, author name, price of the book, and flag indicating whether book is issued or not.

Solution:



#include<stdio.h> #include<conio.h> #include<dos.h>

void main() {

struct library { int acn; char title[25]; char auth[25]; float price; int flag; };

struct library a[5]={ {2,"AAA","AAA",154.8,1}, {1,"BBB","BBB",124.6,0}, {5,"EEE","EEE",234.3,0}, {3,"CCC","CCC",232.3,1}, {4,"DDD","DDD",121.3,0} };

struct library temp; /* temporary structure to overwrite information / and for assistance in sorting the whole structure */

int i,ch,k,flag,j=0; char author[10];

clrscr();

while(1) { /* initialized an indefinite loop */

clrscr(); printf("\t1: Add book information\n"); printf("\t2: Display book information\n"); printf("\t3: List all books of given author\n"); printf("\t4: List the title of specified book\n"); printf("\t5: List the count of books in the library\n"); printf("\t6: List the books in order of accesion number\n"); printf("\t7: Exit\n\n\n");

printf("Choice: "); scanf("%d",&ch);

switch(ch) {

case 1: /* adding book information over current information */

clrscr();

printf("\t\t1: Add book information:\n\n\n");

printf("enter book information:\n\n");

printf("Accesion number of book to be modified: "); scanf("%d",&j); fflush(stdin);

for(i=0;i<5;i++) {

if(j==a[i].acn) {

printf("\n\nenter new information:\n\n"); printf("accesion no.: "); scanf("%d",&temp.acn); fflush(stdin);



printf("\n\ntitle: "); scanf("%s",&temp.title); fflush(stdin); printf("\n\nauthor: "); scanf("%s",&temp.auth); fflush(stdin); printf("\n\nprice: "); scanf("%f",&temp.auth); fflush(stdin); printf("\n\nflag(1/0): "); scanf("%d",&temp.flag); fflush(stdin);

} }

a[j]=temp; /* overwriting current information with new one */

clrscr();

printf("\n\nInformation added succesfully!\n\n");

delay(2000);

break;

case 2: /* displaying current book information */

clrscr();

printf("\t\t2: Display book information:\n\n\n");

printf("Enter accesion number: "); scanf("%d",&k);

for(i=0;i<5;i++) {

if(k==a[i].acn) { clrscr(); printf("Book information: \n\n"); printf("\n\nAccesion Number: %d",a[i].acn); printf("\nTitle: %s",a[i].title); printf("\nAuthor: %s",a[i].auth); printf("\nPrice: %f",a[i].price);

if(a[i].flag==0) { printf("\nFlag: Not issued\n\n"); } else{ printf("\nFlag: Issued\n\n"); } } } delay(1000);

break;

case 3: /* listing all books of given author */

clrscr();

printf("\t\t3: List all books of given author:\n\n\n");

printf("Enter author name: "); scanf("%s",&author); fflush(stdin);

printf("\n\nbooks of %s\n\n",author); for(i=0;i<5;i++) {

k=strcmp(author,a[i].auth);



if(k==0) { j=j+1; printf("%d.\t%s\n",j,a[i].title);

} } delay(2000); break;

case 4: /* displaying title of given book */

clrscr(); printf("\t\t4: Title of the specified book:\n\n\n");

printf("Enter the accesion number: "); scanf("%d",&k);

printf("\n\n");

for(i=0;i<5;i++) {

if(k==a[i].acn) {

printf("Title: %s\n",a[i].title);


case 5: /* counting total books in library */

clrscr(); printf("\t\t5: List the count of books in the library: \n\n\n");

for(i=0;i<5;i++) { j=j+1; } printf("\tTotal books: %2d",j);

delay(2000); break;

case 6: /* sorting the books by their accesion numbers */

clrscr(); printf("6: List the books in order of accesion number: \n\n\n");

for(i=0;i<5;i++) {

for(j=i+1;j<4;j++) {

if(a[i].acn>a[j].acn) { temp=a[i]; a[i]=a[j]; a[j]=temp; } } }

printf("Access no.\tTitle\tAuthor\tFlag\n\n\n"); for(i=0;i<5;i++) {

printf("%4d\t\t%5s\t%5s\t",a[i].acn,a[i].title,a[i].auth);



if(a[i].flag==0) printf("not issued\n\n");

else printf("issued\n\n"); }

delay(2000); break;

case 7: /* condition for going out */

exit();

} }

}

linkfloat() { /* special function to solve compilar error */ float a=0,*b; /* not used withing the program but still defined */ b=&a; a=*b; return 0; }

------------------------------------------------------------------------------------------------------------------

(g) Write a program that compares two given dates. To store date use structure say date that contains three members namely date, month and year. If the dates are equal then display message as "Equal" otherwise "Unequal".


void main() {

struct date { int date; int month; int year; }d1;


printf("enter the date to save in structure: \n\n");

printf("Enter the date: ");scanf("%d",&d1.date);

printf("\n\nEnter the month: ");scanf("%d",&d1.month);

printf("\n\nEnter the year: ");scanf("%d",&d1.year);

clrscr();

printf("\nenter the date to compare with structure:\n\n");

printf("Enter the date: ");scanf("%d",&d);

Solution:



printf("\n\nEnter the month: ");scanf("%d",&m);

printf("\n\nEnter the year: ");scanf("%d",&y);

if(d==d1.date && m==d1.month && y==d1.year) {

printf("\n\ndates are equal\n\n");

}

else {

printf("\n\ndates are unequal\n\n");

}

getch();

}

-----------------------------------------------------------------------------------------------------------------

(h) Linked list is a very common data structure often used to store similar data in memory. While the elements of an array occupy contiguous memory locations, those of a linked list are not constrained to be stored in adjacent location. The individual elements are stored “somewhere” in memory, rather like a family dispersed, but still bound together. The order of the elements is maintained by explicit links between them. Thus, a linked list is a collection of elements called nodes, each of which stores two item of information—an element of the list, and a link, i.e., a pointer or an address that indicates explicitly the location of the node containing the successor of this list element.Write a program to build a linked list by adding new nodes at the beginning, at the end or in the middle of the linked list. Also write a function display( ) which display all the nodes present in the linked list.

#include<stdio.h>#include<conio.h>#include<alloc.h>#include<ctype.h>

struct linklist { int data; linklist *link; };

void main() {

struct linklist *first, *temp;

char choice='y';

int count=0;

clrscr();

/* taking the input of first element */

printf("Enter element %d: ",++count);

Solution:



first=(linklist *)malloc(sizeof(linklist));

scanf("%d",&first->data);

first->link=NULL;

printf("\n\nEnter another element(Y/N): ");

choice=getche();

tolower(choice); /* converting to lowercase to match the condition of switch */

switch(choice) {

case 'y':

clrscr();

while(choice=='y' || choice=='Y') {

printf("\nEnter element %d: ",++count);

temp=(linklist *)malloc(sizeof(linklist));

scanf("%d",&temp->data);

temp->link=first;

first=temp;

printf("\n\n\nEnter another elements(Y/N): ");choice=getche();

clrscr();

}

break;

case 'n':

break;

}

printf("\nTotal elements = %d\n\n",count);

/* printing all elements */

for(temp=first; temp!=NULL; temp=temp->link) {

printf("%d ",temp->data);

}

getch();

}

-----------------------------------------------------------------------------------------------------------------



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(i) A stack is a data structure in which addition of new element or deletion of existing element always takes place at the same end. This end is often known as ‘top’ of stack. This situation can be compared to a stack of plates in a cafeteria where every new plate taken off the stack is also from the ‘top’ of the stack. There are several application where stack can be put to use. For example, recursion, keeping track of function calls, evaluation of expressions, etc. Write a program to implement a stack using a linked list.

NOTE: Topic not discussed in the book. I am learning from other resources.

-----------------------------------------------------------------------------------------------------------------

(j) Unlike a stack, in a queue the addition of new element takes place at the end (called ‘rear’ of queue) whereas deletion takes place at the other end (called ‘front’ of queue). Write a program to implement a queue using a linked list

NOTE: Topic not discussed in the book. I am learning from other resources._______________________________________________________________________

Solution:

Solution:


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Let us C solutions ( By Chetan ): Drawing a hut using graphics

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int gd=DETECT,gm;



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}





getkey() {

union REGS i,o;

while(!kbhit());

i.h.ah=0;int86 (22,&i,&o);

return (o.h.ah);

}

void main() {













void box();

clrscr();



do {


if(key==72) {



}


if(key==80) {



}


if(key==75) {



}


if(key==77) {

sound(1600);




nosound();

if(month+1 > 12){}


}




clrscr();



while(i<=lastday) {

gotoxy(x,y);



cprintf("%2d",i);

i++;x+=5;


x=22;y+=2;

}}




}


getch();

}/*********************** main ends ************************/




/**********************************************************/


{

int a,day=1;





if(day==0)day=22;








}



int i,j,k,l;

/*************//* inner box *//*************/









for(j=4;j<=20;j++) {



}

for(i=21;i<=54;i++) {



}

/*************//* outer box *//*************/







for(k=2;k<=22;k++) {



}

for(l=18;l<=57;l++) {



}





textcolor(RED);


textcolor(YELLOW);








gotoxy(32,4);
















textcolor(BROWN);





}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;

/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */




if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}







clrscr();


color=1<<num;

printf("\n\n");




Exercise [A]

Solution:













else if(num==6 && color==64) printf("Red");


getch();

}

----------------------------------------------------------------------------------------------------------------




for(i=0;i<10;i++) {

clrscr();


for(j=0;j<3;j++) {

Solution:







if(j==2){



} }


for(i=0;i<10;i++) {

for(j=0;j<3;j++) {


} }

for(i=0;i<10;i++) {




clrscr();


}

----------------------------------------------------------------------------------------------------------------

(c) In an inter-college competition, various sports and games are played between different colleges like cricket, basketball, football,



hockey, lawn tennis, table tennis, carom and chess. The information regarding the games won by a particular college is stored in bit numbers 0, 1, 2, 3, 4, 5, 6, 7 and 8 respectively of an integer variable called game. The college that wins in 5 or more than 5 games is awarded the Champion of Champions trophy. If a number is entered through the keyboard, then write a program to find out whether the college won the Champion of the Champions trophy or not, along with the names of the games won by the college.



clrscr();


scanf("%d",&name);

printf("\n\n");

for(i=1;i<=8;i++) {

game=i&name;


won++; }


won++; }


won++; }


won++; }


won++; }


won++; }

Solution:




won++; }


won++; }

}

if(won>=5) {


}

else {


}

getch(); }

-----------------------------------------------------------------------------------------------------------------





Solution:




clrscr();

t1= a.type >> 3;


t2= t1 << 3;










getch();

}

-----------------------------------------------------------------------------------------------------------------




void timer();

Solution:



clrscr();


scanf("%d",&input);



getch();

}


*hr=time >> 11;

*mn=(time<<5) >> 10;

*sc=(time<<11) >> 11;

}

-----------------------------------------------------------------------------------------------------------------



void main() {


clrscr();




for(i=1;i<=8;i=i*2) {

yr= rn & i;

Solution:





if(yr==4) { printf("Third year"); break; }


}



for(i=16;i<=128;i=i*2) {

br= rn & i;





}


for(i=1,j=16;i<=8,j<=128;i=i*2,j=j*2) {

yr=rn&i;

br=rn&j;

for(k=0,a=1,b=16;k<4,a<=8,b<=128;k++,a=a*2,b=b*2) {







flag+=1;

break;

}

}

}



getch(); }______________________________________________________________________







if(argc!=2) {


exit();

}


Exercise [B]

Solution:










if(fs==NULL) {


exit();}


fclose(fs);

}



while(s1[i]!='\0') {

s2[j]=s1[i];

if(s1[i]==' ' || s1[i]=='\0') {

s2[j]='\0';

strrev(s2);

printf("%s ",s2);

j=-1;

}

i++;j++;}

s2[j]='\0';


}

-----------------------------------------------------------------------------------------------------------------



void replace();



clrscr();

if(argc!=4) {

Solution:






if(fp==NULL) {



if(ft==NULL) {




fputs(nwstr,ft);}




getch();

}




while(*s!='\0') {

*t=*s;

t++;s++;

}

*t='\0';





temp2[j]=temp[i];


temp2[j]='\0';


strcpy(temp2,s2);

}

j=0;


main[k]=temp2[j];

k++;j++;}



j=-1;

}

i++;j++;}



strcpy(temp2,s2);

}


j=0;


main[k]=temp2[j];

k++;j++;}



while(*m!='\0') {

*n=*m;

n++;m++;}



*n='\0';

}

----------------------------------------------------------------------------------------------------------------

(c) Write a program that can be used at command prompt as a calculating utility. The usage of the program is shown below.C> calc <switch> <n> <m>Where, n and m are two integer operands. switch can be any one of the arithmetic or comparison operators. If arithmetic operator is supplied, the output should be the result of the operation. If comparison operator is supplied then the output should be True or False.



int n,m,calc;

clrscr();



n=atoi(argv[2]);

m=atoi(argv[3]);





}



}



}



Solution:




}



} /* IMPORTANT */






}




}




}

}

_______________________________________________________________________


Let Us C / Chapter 11 (Console Input Output) Exercise [D]













void main() {


xgets(str);

printf("\n");

xputs(str);

getch();

}


int i=0;char ch;

for(i=0;i<=79;i++) {

ch=getche();

if(ch=='\r') {

*s='\0';break;}

if(ch=='\b') {

printf("\b");

i-=2;s-=2;}

else {

*s=ch;s++;

}

}

}


while(*s!='\0') {

putch(*s);

Solution:




s++;

}

}

---------------------------------------------------------------------------------------------------------------



int a; char s[80];




getch();

}


int digit;


return digit;

}

______________________________________________________________________

Solution:



(a) Create a structure to specify data on students given below:Roll number, Name, Department, Course, Year of joining

Exercise [D]










Assume that there are not more than 450 students in the collage.(a) Write a function to print names of all students who joined in a particular year.(b) Write a function to print the data of a student whose roll number is given.






clrscr();


for(i=0;i<N;i++) {







clrscr();

}




printf("your choice: "); /* taking input of your choice */

Solution:



scanf("%d",&ch);

clrscr();

switch(ch) {

case 1: clrscr();


break;

case 2: clrscr();


break;

default:


}

getch();

}/******************* main() ends **************/


int j,yr;



for(j=0;j<N;j++) {

if(a[j].year==yr) {


}return 0;}


int k,rl;


for(k=0;k<N;k++) {

if(a[k].rlnm==rl) {

printf("\n\n\t Details of roll number: %d\n",a[k].rlnm);printf("\t***************************\n\n");printf(" Name: %s\n",a[k].name);



printf(" Department: %s\n",a[k].dept);printf(" Course: %s\n",a[k].course);printf("Year of joining: %d",a[k].year);

break;}


break;}

}

return 0;}

--------------------------------------------------------------------------------------------------------------





void main() {

struct bank b[N];


clrscr();



Solution:






clrscr();

}

clrscr();



scanf("%d",&ch);

switch(ch) {

case 1:

clrscr();


break;

case 2:

clrscr();


for(i=0;i<N;i++) {


clrscr();



switch(ch2) {

case 0:

clrscr();

if(b[i].bal<lw) {


break; }



else {


}

if(b[i].bal<am) {


}

else {



} break;

case 1:

clrscr();




break;

}

} } } getch(); }


int k;


for(k=0;k<N;k++) {

if((a[k].bal)<100) {


} } return 0;

}

-----------------------------------------------------------------------------------------------------------------

(c) An automobile company has serial number for engine parts starting from AA0 to FF9. The other characteristics of parts to be specified in a structure are: Year of manufacture, material and quantity



manufactured.(a) Specify a structure to store information corresponding to a part.(b) Write a program to retrieve information on parts with serial numbers between BB1 and CC6.


struct automob a[54]={{"AA1",2009,"AAA",10}, {"AA2",2009,"AAA",10}, {"AA3",2009,"AAA",10}, {"AA4",2009,"AAA",10}, {"AA5",2009,"AAA",10}, {"AA6",2009,"AAA",10}, {"AA7",2009,"AAA",10}, {"AA8",2009,"AAA",10}, {"AA9",2009,"AAA",10}, {"BB1",2010,"BBB",11}, {"BB2",2010,"BBB",11}, {"BB3",2010,"BBB",11}, {"BB4",2010,"BBB",11}, {"BB5",2010,"BBB",11}, {"BB6",2010,"BBB",11}, {"BB7",2010,"BBB",11}, {"BB8",2010,"BBB",11}, {"BB9",2010,"BBB",11}, {"CC1",2011,"CCC",12}, {"CC2",2011,"CCC",12}, {"CC3",2011,"CCC",12}, {"CC4",2011,"CCC",12}, {"CC5",2011,"CCC",12}, {"CC6",2011,"CCC",12}, {"CC7",2011,"CCC",12}, {"CC8",2011,"CCC",12}, {"CC9",2011,"CCC",12}, {"DD1",2012,"DDD",13}, {"DD2",2012,"DDD",13}, {"DD3",2012,"DDD",13}, {"DD4",2012,"DDD",13}, {"DD5",2012,"DDD",13}, {"DD6",2012,"DDD",13}, {"DD7",2012,"DDD",13}, {"DD8",2012,"DDD",13}, {"DD9",2012,"DDD",13}, {"EE1",2013,"EEE",14}, {"EE2",2013,"EEE",14}, {"EE3",2013,"EEE",14}, {"EE4",2013,"EEE",14}, {"EE5",2013,"EEE",14}, {"EE6",2013,"EEE",14}, {"EE7",2013,"EEE",14}, {"EE8",2013,"EEE",14}, {"EE9",2013,"EEE",14}, {"FF1",2014,"FFF",15}, {"FF2",2014,"FFF",15}, {"FF3",2014,"FFF",15}, {"FF4",2014,"FFF",15},

Solution:



{"FF5",2014,"FFF",15}, {"FF6",2014,"FFF",15}, {"FF7",2014,"FFF",15}, {"FF8",2014,"FFF",15}, {"FF9",2014,"FFF",15}};

int i,j,k;

clrscr();



for(i=0;i<54;i++) {


} }

getch(); }

-----------------------------------------------------------------------------------------------------------------



void main() {


for(i=0;i<20;i++) {



Solution:






clrscr();

}

/*******************/ /* sorting records */ /*******************/

for(i=0;i<20;i++) {

for(j=i+1;j<20;j++) {




for(i=0;i<20;i++) {


}

getch(); }

linkfloat() {

float a=0,*b;

b=&a; a=*b;

return 0; }

-----------------------------------------------------------------------------------------------------------------



Solution:





for(i=0;i<N;i++) {





clrscr();

}

clrscr();


clrscr();

for(i=0;i<N;i++) {

if((y-(e[i].yy))>=3) {


}

}

getch();}

-----------------------------------------------------------------------------------------------------------------


Solution:




void main() {





clrscr();




switch(ch) {


clrscr();




for(i=0;i<5;i++) {

if(j==a[i].acn) {

printf("\n\nenter new information:\n\n"); printf("accesion no.: "); scanf("%d",&temp.acn); fflush(stdin);



printf("\n\ntitle: "); scanf("%s",&temp.title); fflush(stdin); printf("\n\nauthor: "); scanf("%s",&temp.auth); fflush(stdin); printf("\n\nprice: "); scanf("%f",&temp.auth); fflush(stdin); printf("\n\nflag(1/0): "); scanf("%d",&temp.flag); fflush(stdin);

} }


clrscr();


delay(2000);

break;


clrscr();



for(i=0;i<5;i++) {



break;


clrscr();












printf("\n\n");

for(i=0;i<5;i++) {

if(k==a[i].acn) {






delay(2000); break;



for(i=0;i<5;i++) {

for(j=i+1;j<4;j++) {








delay(2000); break;


exit();

} }

}


------------------------------------------------------------------------------------------------------------------



void main() {







clrscr();



Solution:







}

else {


}

getch();

}

-----------------------------------------------------------------------------------------------------------------




void main() {


char choice='y';

int count=0;

clrscr();



Solution:





first->link=NULL;


choice=getche();


switch(choice) {

case 'y':

clrscr();





temp->link=first;

first=temp;


clrscr();

}

break;

case 'n':

break;

}





}

getch();

}

-----------------------------------------------------------------------------------------------------------------



Newer Posts Older PostsHome





-----------------------------------------------------------------------------------------------------------------



Solution:

Solution:












Let us C solutions ( By Chetan ): January 2014

http://letuscalllessons.blogspot.com/2014_01_01_archive.html[07-Apr-14 8:40:59 PM]








clrscr();


color=1<<num;

printf("\n\n");







else if(num==6 && color==64)

Exercise [A]

Solution:

Chetan Raikwar


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About Me

▼ 2014 (16)

▼ January (14)

Let Us C / Chapter 2 (The Decision Control Structu...

Let Us C / Chapter 12 (File Input Output)

Let Us C / Chapter 1 (Getting started)

Let Us C / Chapter 3 (The Loop Control Structure)

Let Us C / Chapter 4 (The Case Control Structure)

Let Us C / Chapter 5 (Functions & Pointers)

Let Us C / Chapter 6 (Data Types Revisited)

Let Us C / Chapter 7 (The C Pre-processor)

Let Us C / Chapter 8 (Arrays)

Let Us C / Chapter 9 (Puppeting on Strings)


Let Us C / Chapter 11 (Console Input Output)

Let Us C / Chapter 13 (More Issues in Input Output...


► February (2)

Blog Archive

















printf("Red");


getch();

}

----------------------------------------------------------------------------------------------------------------




for(i=0;i<10;i++) {

clrscr();


for(j=0;j<3;j++) {




Solution:




if(j==2){



} }


for(i=0;i<10;i++) {

for(j=0;j<3;j++) {


} }

for(i=0;i<10;i++) {




clrscr();


}

----------------------------------------------------------------------------------------------------------------

(c) In an inter-college competition, various sports and games are played between different colleges like cricket, basketball, football, hockey, lawn tennis, table tennis, carom and chess. The information regarding the games won by a particular college is stored in bit numbers 0, 1, 2, 3, 4, 5, 6, 7 and 8 respectively of an integer variable called game. The college that wins in 5 or more than 5 games is awarded the Champion of Champions trophy. If a number is entered through the keyboard, then write a program to find out whether the college won the Champion of the Champions trophy or not, along with the names of the games won by the college.





clrscr();


scanf("%d",&name);

printf("\n\n");

for(i=1;i<=8;i++) {

game=i&name;


won++; }


won++; }


won++; }


won++; }


won++; }


won++; }


won++; }


won++;

Solution:



}

}

if(won>=5) {


}

else {


}

getch(); }

-----------------------------------------------------------------------------------------------------------------






clrscr();

t1= a.type >> 3;


t2= t1 << 3;


Solution:











getch();

}

-----------------------------------------------------------------------------------------------------------------




void timer();

clrscr();


scanf("%d",&input);



getch();

Solution:



}


*hr=time >> 11;

*mn=(time<<5) >> 10;

*sc=(time<<11) >> 11;

}

-----------------------------------------------------------------------------------------------------------------



void main() {


clrscr();




for(i=1;i<=8;i=i*2) {

yr= rn & i;



if(yr==4) { printf("Third year");

Solution:



break; }


}



for(i=16;i<=128;i=i*2) {

br= rn & i;





}


for(i=1,j=16;i<=8,j<=128;i=i*2,j=j*2) {

yr=rn&i;

br=rn&j;

for(k=0,a=1,b=16;k<4,a<=8,b<=128;k++,a=a*2,b=b*2) {




flag+=1;

break;

}

}




}



getch(); }______________________________________________________________________







if(argc!=2) {


exit();

}


if(fs==NULL) {


exit();}


fclose(fs);

Exercise [B]

Solution:










}



while(s1[i]!='\0') {

s2[j]=s1[i];

if(s1[i]==' ' || s1[i]=='\0') {

s2[j]='\0';

strrev(s2);

printf("%s ",s2);

j=-1;

}

i++;j++;}

s2[j]='\0';


}

-----------------------------------------------------------------------------------------------------------------



void replace();



clrscr();

if(argc!=4) {




if(fp==NULL) {

Solution:





if(ft==NULL) {




fputs(nwstr,ft);}




getch();

}




while(*s!='\0') {

*t=*s;

t++;s++;

}

*t='\0';



temp2[j]=temp[i];


temp2[j]='\0';


strcpy(temp2,s2);

}



j=0;


main[k]=temp2[j];

k++;j++;}



j=-1;

}

i++;j++;}



strcpy(temp2,s2);

}


j=0;


main[k]=temp2[j];

k++;j++;}



while(*m!='\0') {

*n=*m;

n++;m++;}

*n='\0';

}

----------------------------------------------------------------------------------------------------------------

(c) Write a program that can be used at command prompt as a calculating utility. The usage of the program is shown below.C> calc <switch> <n> <m>



Where, n and m are two integer operands. switch can be any one of the arithmetic or comparison operators. If arithmetic operator is supplied, the output should be the result of the operation. If comparison operator is supplied then the output should be True or False.



int n,m,calc;

clrscr();



n=atoi(argv[2]);

m=atoi(argv[3]);





}



}



}



}



} /* IMPORTANT */

Solution:









}




}




}

}

_______________________________________________________________________





Exercise [D]

Solution:











void main() {


xgets(str);

printf("\n");

xputs(str);

getch();

}


int i=0;char ch;

for(i=0;i<=79;i++) {

ch=getche();

if(ch=='\r') {

*s='\0';break;}

if(ch=='\b') {

printf("\b");

i-=2;s-=2;}

else {

*s=ch;s++;

}

}

}


while(*s!='\0') {

putch(*s);

s++;

}

}

---------------------------------------------------------------------------------------------------------------

(b) Write down a function getint( ), which would receive a numeric string from the keyboard, convert it to an integer number and return




the integer to the calling function. A sample usage of getint( ) is shown below:main( ){int a ;a = getint( ) ;printf ( "you entered %d", a )}


int a; char s[80];




getch();

}


int digit;


return digit;

}

______________________________________________________________________

Solution:



(a) Create a structure to specify data on students given below:Roll number, Name, Department, Course, Year of joiningAssume that there are not more than 450 students in the collage.(a) Write a function to print names of all students who joined in a particular year.(b) Write a function to print the data of a student whose roll number is given.

/* NOTE: since number of students to be assumed is too much ( about 450 )I have alloted full size (about 450) to it but array has been kept empty,

Exercise [D]

Solution:










if you have time then you can fill up all 450 names and can search throughthem.program has been tested by me and it works accurately. you can reducethe size of array of structure conveniently by changing the value of N */





clrscr();


for(i=0;i<N;i++) {







clrscr();

}




printf("your choice: "); /* taking input of your choice */scanf("%d",&ch);

clrscr();

switch(ch) {

case 1: clrscr();




break;

case 2: clrscr();


break;

default:


}

getch();

}/******************* main() ends **************/


int j,yr;



for(j=0;j<N;j++) {

if(a[j].year==yr) {


}return 0;}


int k,rl;


for(k=0;k<N;k++) {

if(a[k].rlnm==rl) {

printf("\n\n\t Details of roll number: %d\n",a[k].rlnm);printf("\t***************************\n\n");printf(" Name: %s\n",a[k].name);printf(" Department: %s\n",a[k].dept);printf(" Course: %s\n",a[k].course);printf("Year of joining: %d",a[k].year);

break;}


break;}



}

return 0;}

--------------------------------------------------------------------------------------------------------------





void main() {

struct bank b[N];


clrscr();






clrscr();

}

Solution:



clrscr();



scanf("%d",&ch);

switch(ch) {

case 1:

clrscr();


break;

case 2:

clrscr();


for(i=0;i<N;i++) {


clrscr();



switch(ch2) {

case 0:

clrscr();

if(b[i].bal<lw) {


break; }

else {


}

if(b[i].bal<am) {




}

else {



} break;

case 1:

clrscr();




break;

}

} } } getch(); }


int k;


for(k=0;k<N;k++) {

if((a[k].bal)<100) {


} } return 0;

}

-----------------------------------------------------------------------------------------------------------------

(c) An automobile company has serial number for engine parts starting from AA0 to FF9. The other characteristics of parts to be specified in a structure are: Year of manufacture, material and quantity manufactured.(a) Specify a structure to store information corresponding to a part.(b) Write a program to retrieve information on parts with serial numbers between BB1 and CC6.


Solution:



struct automob { char name[5]; int year; char mtr[10]; int qty; }; void main() {

struct automob a[54]={{"AA1",2009,"AAA",10}, {"AA2",2009,"AAA",10}, {"AA3",2009,"AAA",10}, {"AA4",2009,"AAA",10}, {"AA5",2009,"AAA",10}, {"AA6",2009,"AAA",10}, {"AA7",2009,"AAA",10}, {"AA8",2009,"AAA",10}, {"AA9",2009,"AAA",10}, {"BB1",2010,"BBB",11}, {"BB2",2010,"BBB",11}, {"BB3",2010,"BBB",11}, {"BB4",2010,"BBB",11}, {"BB5",2010,"BBB",11}, {"BB6",2010,"BBB",11}, {"BB7",2010,"BBB",11}, {"BB8",2010,"BBB",11}, {"BB9",2010,"BBB",11}, {"CC1",2011,"CCC",12}, {"CC2",2011,"CCC",12}, {"CC3",2011,"CCC",12}, {"CC4",2011,"CCC",12}, {"CC5",2011,"CCC",12}, {"CC6",2011,"CCC",12}, {"CC7",2011,"CCC",12}, {"CC8",2011,"CCC",12}, {"CC9",2011,"CCC",12}, {"DD1",2012,"DDD",13}, {"DD2",2012,"DDD",13}, {"DD3",2012,"DDD",13}, {"DD4",2012,"DDD",13}, {"DD5",2012,"DDD",13}, {"DD6",2012,"DDD",13}, {"DD7",2012,"DDD",13}, {"DD8",2012,"DDD",13}, {"DD9",2012,"DDD",13}, {"EE1",2013,"EEE",14}, {"EE2",2013,"EEE",14}, {"EE3",2013,"EEE",14}, {"EE4",2013,"EEE",14}, {"EE5",2013,"EEE",14}, {"EE6",2013,"EEE",14}, {"EE7",2013,"EEE",14}, {"EE8",2013,"EEE",14}, {"EE9",2013,"EEE",14}, {"FF1",2014,"FFF",15}, {"FF2",2014,"FFF",15}, {"FF3",2014,"FFF",15}, {"FF4",2014,"FFF",15}, {"FF5",2014,"FFF",15}, {"FF6",2014,"FFF",15}, {"FF7",2014,"FFF",15}, {"FF8",2014,"FFF",15}, {"FF9",2014,"FFF",15}};

int i,j,k;

clrscr();





for(i=0;i<54;i++) {


} }

getch(); }

-----------------------------------------------------------------------------------------------------------------



void main() {


for(i=0;i<20;i++) {






clrscr();

Solution:



}

/*******************/ /* sorting records */ /*******************/

for(i=0;i<20;i++) {

for(j=i+1;j<20;j++) {




for(i=0;i<20;i++) {


}

getch(); }

linkfloat() {

float a=0,*b;

b=&a; a=*b;

return 0; }

-----------------------------------------------------------------------------------------------------------------





for(i=0;i<N;i++) {

printf("\tEnter employee data:\n");

Solution:



printf("\t*********************\n");




clrscr();

}

clrscr();


clrscr();

for(i=0;i<N;i++) {

if((y-(e[i].yy))>=3) {


}

}

getch();}

-----------------------------------------------------------------------------------------------------------------



void main() {

struct library { int acn;

Solution:



char title[25]; char auth[25]; float price; int flag; };




clrscr();




switch(ch) {


clrscr();




for(i=0;i<5;i++) {

if(j==a[i].acn) {

printf("\n\nenter new information:\n\n"); printf("accesion no.: "); scanf("%d",&temp.acn); fflush(stdin); printf("\n\ntitle: "); scanf("%s",&temp.title); fflush(stdin); printf("\n\nauthor: "); scanf("%s",&temp.auth); fflush(stdin); printf("\n\nprice: "); scanf("%f",&temp.auth); fflush(stdin); printf("\n\nflag(1/0): "); scanf("%d",&temp.flag); fflush(stdin);

} }




clrscr();


delay(2000);

break;


clrscr();



for(i=0;i<5;i++) {



break;


clrscr();












printf("\n\n");

for(i=0;i<5;i++) {

if(k==a[i].acn) {






delay(2000); break;



for(i=0;i<5;i++) {

for(j=i+1;j<4;j++) {






delay(2000); break;




exit();

} }

}


------------------------------------------------------------------------------------------------------------------



void main() {







clrscr();







}

else {

Solution:




}

getch();

}

-----------------------------------------------------------------------------------------------------------------




void main() {


char choice='y';

int count=0;

clrscr();





first->link=NULL;


choice=getche();


Solution:



switch(choice) {

case 'y':

clrscr();





temp->link=first;

first=temp;


clrscr();

}

break;

case 'n':

break;

}





}

getch();

}

-----------------------------------------------------------------------------------------------------------------








-----------------------------------------------------------------------------------------------------------------



Solution:

Solution:












Let us C solutions ( By Chetan ): February 2014








int gd=DETECT,gm;



rectangle(300,210,380,260);



Chetan Raikwar


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► January (14)

▼ February (2)



Blog Archive
























}





getkey() {

union REGS i,o;

while(!kbhit());

i.h.ah=0;int86 (22,&i,&o);

return (o.h.ah);

}

void main() {













void box();

clrscr();



do {


if(key==72) {



}


if(key==80) {



}


if(key==75) {



}


if(key==77) {

sound(1600);




nosound();

if(month+1 > 12){}


}




clrscr();



while(i<=lastday) {

gotoxy(x,y);



cprintf("%2d",i);

i++;x+=5;


x=22;y+=2;

}}




}


getch();

}/*********************** main ends ************************/




/**********************************************************/


{

int a,day=1;





if(day==0)day=22;








}



int i,j,k,l;

/*************//* inner box *//*************/









for(j=4;j<=20;j++) {



}

for(i=21;i<=54;i++) {



}

/*************//* outer box *//*************/







for(k=2;k<=22;k++) {



}

for(l=18;l<=57;l++) {



}





textcolor(RED);


textcolor(YELLOW);








gotoxy(32,4);
















textcolor(BROWN);





}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;

/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */






if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}












Let us C solutions ( By Chetan ): Dynamic calender animated without graphics library

http://letuscalllessons.blogspot.com/2014/02/dynamic-calender-animated-without.html[07-Apr-14 8:42:57 PM]







getkey() {

union REGS i,o;

while(!kbhit());

i.h.ah=0;int86 (22,&i,&o);

return (o.h.ah);

}

void main() {

int x,y,i,lastday,key;int month=2,year=2014,a;void box();

clrscr();



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do {


if(key==72) {



}


if(key==80) {



}


if(key==75) {



}


if(key==77) {

sound(1600);delay(20);sound(1250);delay(10);

nosound();

if(month+1 > 12){}




}




clrscr();



while(i<=lastday) {

gotoxy(x,y);



cprintf("%2d",i);

i++;x+=5;


x=22;y+=2;

}}




}


getch();

}/*********************** main ends ************************/

/**********************************************************//* function to find first day of the given month and year *//**********************************************************/


{

int a,day=1;







if(day==0)day=22;








}



int i,j,k,l;

/*************//* inner box *//*************/







for(j=4;j<=20;j++) {





}

for(i=21;i<=54;i++) {



}

/*************//* outer box *//*************/







for(k=2;k<=22;k++) {



}

for(l=18;l<=57;l++) {



}



textcolor(RED);


textcolor(YELLOW);










gotoxy(32,4);














textcolor(BROWN);



gotoxy(60,20);



cprintf("%c : Next month",16);


}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;

/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */

if(m==9)days=30;

/* for october */

if(m==10)days=31;



Newer Post Older PostHome



/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}


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Let us C solutions ( By Chetan ): Let Us C / Chapter 14 (Operations on Bits)

http://letuscalllessons.blogspot.com/2014/01/chapter-14-operations-on-bits.html[07-Apr-14 8:43:44 PM]








clrscr();


color=1<<num;

printf("\n\n");








Exercise [A]

Solution:

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printf("Red");


getch();

}

----------------------------------------------------------------------------------------------------------------




for(i=0;i<10;i++) {

clrscr();


for(j=0;j<3;j++) {




Solution:




if(j==2){



} }


for(i=0;i<10;i++) {

for(j=0;j<3;j++) {


} }

for(i=0;i<10;i++) {




clrscr();


}

----------------------------------------------------------------------------------------------------------------






clrscr();


scanf("%d",&name);

printf("\n\n");

for(i=1;i<=8;i++) {

game=i&name;


won++; }


won++; }


won++; }


won++; }


won++; }


won++; }


won++; }


won++;

Solution:



}

}

if(won>=5) {


}

else {


}

getch(); }

-----------------------------------------------------------------------------------------------------------------






clrscr();

t1= a.type >> 3;


t2= t1 << 3;


Solution:











getch();

}

-----------------------------------------------------------------------------------------------------------------




void timer();

clrscr();


scanf("%d",&input);



getch();

Solution:



}


*hr=time >> 11;

*mn=(time<<5) >> 10;

*sc=(time<<11) >> 11;

}

-----------------------------------------------------------------------------------------------------------------



void main() {


clrscr();




for(i=1;i<=8;i=i*2) {

yr= rn & i;




Solution:



break; }


}



for(i=16;i<=128;i=i*2) {

br= rn & i;





}


for(i=1,j=16;i<=8,j<=128;i=i*2,j=j*2) {

yr=rn&i;

br=rn&j;

for(k=0,a=1,b=16;k<4,a<=8,b<=128;k++,a=a*2,b=b*2) {




flag+=1;

break;

}

}






}



getch(); }______________________________________________________________________



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Let us C solutions ( By Chetan ): Let Us C / Chapter 13 (More Issues in Input Output)

http://letuscalllessons.blogspot.com/2014/01/chapter-13-more-issues-in-input-output.html[07-Apr-14 8:44:36 PM]









if(argc!=2) {


exit();

}


if(fs==NULL) {


exit();}


fclose(fs);

}

Exercise [B]

Solution:

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while(s1[i]!='\0') {

s2[j]=s1[i];

if(s1[i]==' ' || s1[i]=='\0') {

s2[j]='\0';

strrev(s2);

printf("%s ",s2);

j=-1;

}

i++;j++;}

s2[j]='\0';


}

-----------------------------------------------------------------------------------------------------------------



void replace();



clrscr();

if(argc!=4) {




if(fp==NULL) {

puts("cannot open source file!");

Solution:



exit();}


if(ft==NULL) {




fputs(nwstr,ft);}




getch();

}




while(*s!='\0') {

*t=*s;

t++;s++;

}

*t='\0';



temp2[j]=temp[i];


temp2[j]='\0';


strcpy(temp2,s2);

}



j=0;


main[k]=temp2[j];

k++;j++;}



j=-1;

}

i++;j++;}



strcpy(temp2,s2);

}


j=0;


main[k]=temp2[j];

k++;j++;}



while(*m!='\0') {

*n=*m;

n++;m++;}

*n='\0';

}

----------------------------------------------------------------------------------------------------------------







int n,m,calc;

clrscr();



n=atoi(argv[2]);

m=atoi(argv[3]);





}



}



}



}



} /* IMPORTANT */

Solution:









}




}




}

}

_______________________________________________________________________


Google Account

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Let us C solutions ( By Chetan ): Let Us C / Chapter 11 (Console Input Output)

http://letuscalllessons.blogspot.com/2014/01/chapter-11-console-input-output.html[07-Apr-14 8:45:28 PM]








void main() {


xgets(str);

printf("\n");

xputs(str);

getch();

}


int i=0;char ch;

for(i=0;i<=79;i++) {

ch=getche();

if(ch=='\r') {

*s='\0';break;}

if(ch=='\b') {

printf("\b");

Exercise [D]

Solution:

Chetan Raikwar


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i-=2;s-=2;}

else {

*s=ch;s++;

}

}

}


while(*s!='\0') {

putch(*s);

s++;

}

}

---------------------------------------------------------------------------------------------------------------



int a; char s[80];




getch();

}


int digit;

Solution:







return digit;

}

______________________________________________________________________



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Let us C solutions ( By Chetan ): Let Us C / Chapter 10 (Structures)

http://letuscalllessons.blogspot.com/2014/01/chapter-10-structures.html[07-Apr-14 8:46:18 PM]











clrscr();


for(i=0;i<N;i++) {



Exercise [D]

Solution:

Chetan Raikwar


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clrscr();

}





clrscr();

switch(ch) {

case 1: clrscr();


break;

case 2: clrscr();


break;

default:


}

getch();

}/******************* main() ends **************/


int j,yr;



for(j=0;j<N;j++) {



if(a[j].year==yr) {


}return 0;}


int k,rl;


for(k=0;k<N;k++) {

if(a[k].rlnm==rl) {


break;}


break;}

}

return 0;}

--------------------------------------------------------------------------------------------------------------


/* NOTE: since number of customers to be assumed is too much ( about 200 ) I have alloted full size (about 200) to it but array has been kept empty, if you have time then you can fill up all 200 names and can search through

Solution:



them.program has been tested by me and it works accurately. you can reducethe size of array of structure conveniently by changing the value of N */



void main() {

struct bank b[N];


clrscr();






clrscr();

}

clrscr();



scanf("%d",&ch);

switch(ch) {

case 1:

clrscr();


break;

case 2:

clrscr();


for(i=0;i<N;i++) {




clrscr();



switch(ch2) {

case 0:

clrscr();

if(b[i].bal<lw) {


break; }

else {


}

if(b[i].bal<am) {


}

else {



} break;

case 1:

clrscr();




break;

}

} } } getch(); }




int k;


for(k=0;k<N;k++) {

if((a[k].bal)<100) {


} } return 0;

}

-----------------------------------------------------------------------------------------------------------------



struct automob a[54]={{"AA1",2009,"AAA",10}, {"AA2",2009,"AAA",10}, {"AA3",2009,"AAA",10}, {"AA4",2009,"AAA",10}, {"AA5",2009,"AAA",10}, {"AA6",2009,"AAA",10}, {"AA7",2009,"AAA",10}, {"AA8",2009,"AAA",10}, {"AA9",2009,"AAA",10}, {"BB1",2010,"BBB",11}, {"BB2",2010,"BBB",11}, {"BB3",2010,"BBB",11}, {"BB4",2010,"BBB",11}, {"BB5",2010,"BBB",11}, {"BB6",2010,"BBB",11}, {"BB7",2010,"BBB",11}, {"BB8",2010,"BBB",11}, {"BB9",2010,"BBB",11}, {"CC1",2011,"CCC",12}, {"CC2",2011,"CCC",12}, {"CC3",2011,"CCC",12}, {"CC4",2011,"CCC",12}, {"CC5",2011,"CCC",12},

Solution:



{"CC6",2011,"CCC",12}, {"CC7",2011,"CCC",12}, {"CC8",2011,"CCC",12}, {"CC9",2011,"CCC",12}, {"DD1",2012,"DDD",13}, {"DD2",2012,"DDD",13}, {"DD3",2012,"DDD",13}, {"DD4",2012,"DDD",13}, {"DD5",2012,"DDD",13}, {"DD6",2012,"DDD",13}, {"DD7",2012,"DDD",13}, {"DD8",2012,"DDD",13}, {"DD9",2012,"DDD",13}, {"EE1",2013,"EEE",14}, {"EE2",2013,"EEE",14}, {"EE3",2013,"EEE",14}, {"EE4",2013,"EEE",14}, {"EE5",2013,"EEE",14}, {"EE6",2013,"EEE",14}, {"EE7",2013,"EEE",14}, {"EE8",2013,"EEE",14}, {"EE9",2013,"EEE",14}, {"FF1",2014,"FFF",15}, {"FF2",2014,"FFF",15}, {"FF3",2014,"FFF",15}, {"FF4",2014,"FFF",15}, {"FF5",2014,"FFF",15}, {"FF6",2014,"FFF",15}, {"FF7",2014,"FFF",15}, {"FF8",2014,"FFF",15}, {"FF9",2014,"FFF",15}};

int i,j,k;

clrscr();



for(i=0;i<54;i++) {


} }

getch(); }

-----------------------------------------------------------------------------------------------------------------


Solution:




void main() {


for(i=0;i<20;i++) {






clrscr();

}

/*******************/ /* sorting records */ /*******************/

for(i=0;i<20;i++) {

for(j=i+1;j<20;j++) {




for(i=0;i<20;i++) {


}

getch(); }

linkfloat() {



float a=0,*b;

b=&a; a=*b;

return 0; }

-----------------------------------------------------------------------------------------------------------------





for(i=0;i<N;i++) {





clrscr();

}

clrscr();


clrscr();

for(i=0;i<N;i++) {

if((y-(e[i].yy))>=3) {


}

}

Solution:



getch();}

-----------------------------------------------------------------------------------------------------------------



void main() {





clrscr();


clrscr(); printf("\t1: Add book information\n"); printf("\t2: Display book information\n"); printf("\t3: List all books of given author\n"); printf("\t4: List the title of specified book\n"); printf("\t5: List the count of books in the library\n");

Solution:



printf("\t6: List the books in order of accesion number\n"); printf("\t7: Exit\n\n\n");


switch(ch) {


clrscr();




for(i=0;i<5;i++) {

if(j==a[i].acn) {


} }


clrscr();


delay(2000);

break;


clrscr();



for(i=0;i<5;i++) {


if(a[i].flag==0) { printf("\nFlag: Not issued\n\n");



} else{ printf("\nFlag: Issued\n\n"); } } } delay(1000);

break;


clrscr();










printf("\n\n");

for(i=0;i<5;i++) {

if(k==a[i].acn) {






delay(2000);



break;



for(i=0;i<5;i++) {

for(j=i+1;j<4;j++) {






delay(2000); break;


exit();

} }

}


------------------------------------------------------------------------------------------------------------------



void main() {

struct date {

Solution:



int date; int month; int year; }d1;






clrscr();







}

else {


}

getch();

}

-----------------------------------------------------------------------------------------------------------------


Solution:





void main() {


char choice='y';

int count=0;

clrscr();





first->link=NULL;


choice=getche();


switch(choice) {

case 'y':

clrscr();





temp->link=first;

first=temp;


clrscr();

}

break;

case 'n':




break;

}





}

getch();

}

-----------------------------------------------------------------------------------------------------------------



-----------------------------------------------------------------------------------------------------------------



Solution:

Solution:


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http://letuscalllessons.blogspot.com/search?updated-max=2014-01-30T07:04:00-08:00&max-results=7[07-Apr-14 8:47:07 PM]





(c) Write a program that converts all lowercase characters in a given string to its equivalent uppercase character.

#include<stdio.h>#include<conio.h>#include<string.h>void main() {

char str[100];

gets(str);

printf("\n\n");

strupr(str); /* 'strupr()' is a function to convert a lowercase puts(str); string into uppercase */

getch();

}

-----------------------------------------------------------------------------------------------------------

(d) Write a program that extracts part of the given string from the specified position. For example, if the sting is "Working with strings is fun", then if from position 4, 4 characters are to be extracted then the program should return string as "king". Moreover, if the position from where the string is to be extracted is given and the number of characters to be extracted is 0 then the program should extract entire string from the specified position.


void main() {

char s[100];int i=0,n,pos;

Exercise [D]

Solution:

Solution:

Chetan Raikwar


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clrscr();

printf("enter the string: \n\n");gets(s);

printf("\n\nenter the position to extract from: ");scanf("%d",&pos);

printf("\nenter the number of characters to extract: ");scanf("%d",&n);

printf("\n\n\nExtracted string: \n\n");

if(n==0) {

while(s[i]!='\0') {

if(i>=pos-1) {

putch(s[i]);

}

i++;}}

else {

while(s[i]!='\0') {

if(i>=pos-1 && i<=pos-1+(n-1)) {printf("%c",s[i]);}

i++;}}getch();}

--------------------------------------------------------------------------------------------------------------

(e) Write a program that converts a string like "124" to an integer 124.


void main() {

char str[100];int i;clrscr();

printf("Enter the string: \n\n");

gets(str);

i=atoi(str); /* 'atoi' is a function to convert a string into an integer */

printf("%d",i);

getch();

}

Solution:



---------------------------------------------------------------------------------------------------------------

(f) Write a program that replaces two or more consecutive blanks in a string by a single blank. For example, if the input isGrim return to the planet of apes!!the output should beGrim return to the planet of apes!!

#include<stdio.h>#include<conio.h>void main() {

char s[80];int i=0;clrscr();

printf("Enter the string:\n\n\n");gets(s);

printf("\n\n\nOutput:\n\n\n");

while(s[i]!='\0') {

if(s[i]==' ' && s[i+1]==' ') {

/* if there are two or more blanks, do nothing */

}

else {

putch(s[i]);

}

i++;

}

getch();}_______________________________________________________________________

(a) Write a program that uses an array of pointers to strings str[ ]. Receive two strings str1 and str2 and check if str1 is embedded in any of the strings in str[ ]. If str1 is found, then replace it with str2.char *str[ ] = {"We will teach you how to...","Move a mountain","Level a building","Erase the past","Make a million","...all through C!"} ;For example if str1 contains "mountain" and str2 contains "car", then the second string in str should get changed to "Move a car".

Solution:

Exercise [F]




void replace();

void main() {

char *str[] = { "We will teach you how to...", "Move a mountain", "Level a building", "Erase the past", "Make a million", "...all through C !" }; char str1[80],str2[80]; int i;

clrscr();

printf("\n\n");

for(i=0;i<6;i++) {

printf("\t%s\n",*(str+i));

}

printf("\n\n");

printf("Enter the word to search: "); gets(str1);

printf("\n\nEnter the word to replace: "); gets(str2);

clrscr();

printf("\nBefore modification:\n\n");

for(i=0;i<6;i++) {


}

/*******************************************/ /* passing all strings to replace function */ /*******************************************/

printf("\nAfter modification:\n\n");

for(i=0;i<6;i++) {

replace(*(str+i),str1,str2);

}

getch();

}

void replace(char *s, char s1[80], char s2[80]) {

Solution:



int i=0,j=0,k=0; char temp[100],temp2[100],main[100],*t=temp;

/* copying to temporary string */

while(*s!='\0') {

*t=*s;

t++; s++;

}

*t='\0';

/**********************/ /* checking each word */ /**********************/


temp2[j]=temp[i];

if(temp[i]==' ') {

temp2[j]='\0';


strcpy(temp2,s2);

}

j=0;


main[k]=temp2[j];

k++; j++; }



j=-1;

}

i++; j++; }



strcpy(temp2,s2);

}

/***************************/ /* last word of the string */ /***************************/

j=0;




main[k]=temp2[j];

k++; j++; }


printf("\t%s\n",main); /* printing the new string */

}

------------------------------------------------------------------------------------------------------------

(b) Write a program to sort a set of names stored in an array in alphabetical order.


char a[10][10];char t1[10],t2[10];int i,j;clrscr();

printf("\nUnsorted list: \n\n");

for(i=0;i<10;i++) {

scanf("%s",a[i]);

}

printf("\n\n");

/* sorting list *//****************/

for(i=0;i<10;i++) {

for(j=i+1;j<10;j++) {

if(a[i][0]>a[j][0]) { /* testing only first alphabet of each name */

strcpy(t1,a[i]); /* copying both in two temporary strings */strcpy(t2,a[j]);

strcpy((a[i]),t2); /* replacing both from temporary strings */strcpy(a[j],t1);

}}}

/* sorted list *//***************/

printf("\n\nSorted list: \n\n");

Solution:



for(i=0;i<10;i++) {

printf("\n%s\n",a[i]);

}

getch();

}

--------------------------------------------------------------------------------------------------------------

(c) Write a program to reverse the strings stored in the following array of pointers to strings: char *s[ ] = { "To err is human...", "But to really mess things up...", "One needs to know C!!" } ; Hint: Write a function xstrrev ( string ) which should reverse the contents of one string. Call this function for reversing each string stored in s.


void main() {

char *s[]={"To err is human....", "But to really mess up things up...", "One needs to know C!!" };int i;

clrscr();

printf("REVERSED strings\n\n\n\n");/* reversing and printing all strings */

xstrrev(*s);printf("%s\n\n",*s);

xstrrev(*(s+1));printf("%s\n\n",*(s+1));


getch();

}

xstrrev( char *s) {

int i=0;char target[100],*t=target;

Solution:



/* taking 'i' to the null position */

while(*s!='\0') {

i++;s++;

}

/* reversing in temporary target string */

while(i>=0) {

s--;

*t=*s;

t++;i--;

}

*t='\0';

/* reversing original string */

while(target[i]!='\0') {

*s=target[i];

i++;s++;

}

return *s;}

-----------------------------------------------------------------------------------------------------------------

(d) Develop a program that receives the month and year from the keyboard as integers and prints the calendar in the following format. Note that according to the Gregorian calendar 01/01/1900 was Monday. With this as the base the calendar should be generated.


void main() {

int x=49,y=9,i=1,lastday;int month,year,a;void box();

clrscr();

Solution:



printf("Enter month: ");scanf("%d",&month);

printf("\n\nEnter year: ");scanf("%d",&year);



x = dayfinder ( month, year ); /* finding the first day of month */

lastday=totaldays(month,year); /* finding the total days of month */

clrscr();



while(i<=lastday) {

gotoxy(x,y);

printf("%2d",i);

i++;x+=5;

if(x>52) { /* if the position of 7 days is covered, again print from beginning from a new line */

x=22;y+=2;

}}

}

else /* exit message on wrong input */

printf("\n\nSorry! invalid input...");

gotoxy(1,1); /* moving cursor out of the calender */

getch();

}

/*********************** main ends ************************/



{

int a,day=1;







if(day==0)day=22;








}



int i,j,k,l;

/*************//* inner box *//*************/

/* corners of inner box */

gotoxy(20,3);printf("%c",218);





for(j=4;j<=20;j++) {

gotoxy(20,j);printf("%c",179);




}

for(i=21;i<=54;i++) {

gotoxy(i,3);printf("%c",196);


}

/*************//* outer box *//*************/







for(k=2;k<=22;k++) {

gotoxy(17,k);printf("%c",179);


}

for(l=18;l<=57;l++) {

gotoxy(l,1);printf("%c",196);


}


gotoxy(22,6);printf("Sun");

gotoxy(27,6);printf("Mon");

gotoxy(32,6);printf("Tue");



gotoxy(37,6);printf("Wed");

gotoxy(42,6);printf("Thu");

gotoxy(47,6);printf("Fri");

gotoxy(52,6);printf("Sat");

gotoxy(32,4);

if(m==1)printf("January %d",y);

if(m==2)printf("February %d",y);

if(m==3)printf("March %d",y);

if(m==4)printf("April %d",y);

if(m==5)printf("May %d",y);

if(m==6)printf("June %d",y);

if(m==7)printf("July %d",y);

if(m==8)printf("August %d",y);

if(m==9)printf("September %d",y);

if(m==10)printf("October %d",y);

if(m==11)printf("November %d",y);

if(m==12)printf("December %d",y);

}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)



days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;

/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */

if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}

----------------------------------------------------------------------------------------------------------------

(e) Modify the above program suitably so that once the calendar for a particular month and year has been displayed on the screen, then using arrow keys the user must be able to change the calendar in the



following manner: Up arrow key : Next year, same month Down arrow key : Previous year, same month Right arrow key : Same year, next month Left arrow key : Same year, previous month If the escape key is hit then the procedure should stop. Hint: Use the getkey( ) function discussed in Chapter 8, problem number [L](c).



getkey() {

union REGS i,o;

while(!kbhit());

i.h.ah=0;int86 (22,&i,&o);

return (o.h.ah);

}

void main() {

int x,y,i,lastday,key;int month,year,a;void box();

clrscr();





do {


if(key==72) {

Solution:




}


if(key==80) {


}


if(key==75) {


}


if(key==77) {

if(month+1 > 12){}


}




clrscr();



while(i<=lastday) {

gotoxy(x,y);



printf("%2d",i);

i++;x+=5;


x=22;y+=2;

}}




}


getch();

}/*********************** main ends ************************/



{

int a,day=1;





if(day==0)day=22;










}



int i,j,k,l;

/*************//* inner box *//*************/







for(j=4;j<=20;j++) {



}

for(i=21;i<=54;i++) {



}

/*************//* outer box *//*************/




gotoxy(58,1);



printf("%c",191);



for(k=2;k<=22;k++) {



}

for(l=18;l<=57;l++) {



}









gotoxy(32,4);















/*************************//* printing instructions *//*************************/

gotoxy(60,16);printf("%c : Next year",30);

gotoxy(60,18);printf("%c : Previous year",31);

gotoxy(60,20);printf("%c : Next month",16);

gotoxy(60,22);printf("%c : Previous month",17);

}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;



/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */

if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}

--------------------------------------------------------------------------------------------------------------

(f) A factory has 3 division and stocks 4 categories of products. An inventory table is updated for each division and for each product as they are received. There are three independent suppliers of products to the factory: (a) Design a data format to represent each transaction. (b) Write a program to take a transaction and update the inventory.



(c) If the cost per item is also given write a program to calculate the total inventory values.

-------------------------------------------------------------------------------------------------------------

(g) A dequeue is an ordered set of elements in which elements may be inserted or retrieved from either end. Using an array simulate a dequeue of characters and the operations retrieve left, retrieve right, insert left, insert right. Exceptional conditions such as dequeue full or empty should be indicated. Two pointers (namely, left and right) are needed in this simulation.


---------------------------------------------------------------------------------------------------------------

(h) Write a program to delete all vowels from a sentence. Assume that the sentence is not more than 80 characters long.


char s[80];int i=0;

clrscr();

gets(s);

while(s[i]!='\0') {

if(s[i]=='a' || s[i]=='e' || s[i]=='i' || s[i]=='o' || s[i]=='u') {putch(' ');}

elseputch(s[i]);i++;}

getch();}

---------------------------------------------------------------------------------------------------------------

Solution:

Solution:

Solution:



(i) Write a program that will read a line and delete from it all occurrences of the word ‘the’.


void replace();

void main() {

char str[80],str1[]="the";clrscr();

gets(str);

replace(str,str1);

getch();

}void replace(char *s, char s1[80]) {

int i=0,j=0,k=0;char temp[100],temp2[100],main[100],*t=temp;


while(*s!='\0') {

*t=*s;

t++;s++;

}

*t='\0';



temp2[j]=temp[i];

if(temp[i]==' ') {

temp2[j]='\0';


}

else {j=0;


main[k]=temp2[j];

Solution:



k++;j++;}


k++; /* increment so that the next word won't replace the space */}j=-1;

}

i++;j++;}



}


else {

j=0;


main[k]=temp2[j];

k++;j++;}


}printf("%s\n",main); /* printing the new string */

}

---------------------------------------------------------------------------------------------------------------

(j) Write a program that takes a set of names of individuals and abbreviates the first, middle and other names except the last name by their first letter.


void abbr();

void main() {

char s1[10][80];int i=0;

clrscr();

printf("\tEnter 10 names: \n\n\n");

Solution:



for(i=0;i<10;i++) {

gets(s1[i]); /* saving names in 2d string */

}

clrscr();

printf("\tAbbreviated names: \n\n");

for(i=0;i<10;i++) {

abbr(s1[i]); /* sending each name to function */

printf("\n\n");

}

getch();

}

void abbr( char s1[100]) {


while(s1[i]!='\0') {

s2[j]=s1[i];

if(s1[i]==' ') { /* if space is detected then it's not last name */

printf(" %c",toupper(s2[0])); /* printing first character of name after converting to uppercase */j=-1;

}

j++;i++;} /* printing the last name */s2[j]='\0';printf(" %s",s2);

}

--------------------------------------------------------------------------------------------------------------

(k) Write a program to count the number of occurrences of any two vowels in succession in a line of text. For example, in the sentence “Pleases read this application and give me gratuity” such occurrences are ea, ea, ui.


char a[10000],i=0,suc=0;clrscr();

Solution:






printf("enter the line of text (string): \n\n");

gets(a);

printf("\n\n\n\ntwo vowels in succesion are: \n\n");

while(a[i]!='\0') {

if(a[i]=='a'|| a[i]=='e'|| a[i]=='i'|| a[i]=='o'||a[i]=='u' ) {

if(a[i+1]=='a'|| a[i+1]=='e'|| a[i+1]=='i'|| a[i+1]=='o'|| a[i+1]=='u') {

suc++;printf("%c%c ",a[i],a[i+1]);}}i++;}

printf("\n\n\noccurence of two vowels in succesion = %2d ",suc);

getch();}

______________________________________________________________________













http://letuscalllessons.blogspot.com/search?updated-min=2014-01-01T00:00:00-08:00&updated-max=2014-01-30T07:04:00-08:00&max-results=16&start=6&by-date=false[07-Apr-14 8:48:00 PM]







char str[100];

gets(str);

printf("\n\n");


getch();

}

-----------------------------------------------------------------------------------------------------------



void main() {


Exercise [D]

Solution:

Solution:

Chetan Raikwar


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► February (2)

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clrscr();





if(n==0) {

while(s[i]!='\0') {

if(i>=pos-1) {

putch(s[i]);

}

i++;}}

else {

while(s[i]!='\0') {


i++;}}getch();}

--------------------------------------------------------------------------------------------------------------



void main() {



gets(str);


printf("%d",i);

getch();

}

Solution:



---------------------------------------------------------------------------------------------------------------






while(s[i]!='\0') {

if(s[i]==' ' && s[i+1]==' ') {


}

else {

putch(s[i]);

}

i++;

}

getch();}_______________________________________________________________________


Solution:

Exercise [F]




void replace();

void main() {


clrscr();

printf("\n\n");

for(i=0;i<6;i++) {


}

printf("\n\n");



clrscr();


for(i=0;i<6;i++) {


}



for(i=0;i<6;i++) {


}

getch();

}


Solution:





while(*s!='\0') {

*t=*s;

t++; s++;

}

*t='\0';



temp2[j]=temp[i];

if(temp[i]==' ') {

temp2[j]='\0';


strcpy(temp2,s2);

}

j=0;


main[k]=temp2[j];

k++; j++; }



j=-1;

}

i++; j++; }



strcpy(temp2,s2);

}


j=0;




main[k]=temp2[j];

k++; j++; }



}

------------------------------------------------------------------------------------------------------------





for(i=0;i<10;i++) {

scanf("%s",a[i]);

}

printf("\n\n");

/* sorting list *//****************/

for(i=0;i<10;i++) {

for(j=i+1;j<10;j++) {




}}}

/* sorted list *//***************/


Solution:



for(i=0;i<10;i++) {


}

getch();

}

--------------------------------------------------------------------------------------------------------------



void main() {


clrscr();





getch();

}

xstrrev( char *s) {


Solution:




while(*s!='\0') {

i++;s++;

}


while(i>=0) {

s--;

*t=*s;

t++;i--;

}

*t='\0';



*s=target[i];

i++;s++;

}

return *s;}

-----------------------------------------------------------------------------------------------------------------



void main() {


clrscr();

Solution:









clrscr();



while(i<=lastday) {

gotoxy(x,y);

printf("%2d",i);

i++;x+=5;


x=22;y+=2;

}}

}




getch();

}

/*********************** main ends ************************/



{

int a,day=1;







if(day==0)day=22;








}



int i,j,k,l;

/*************//* inner box *//*************/







for(j=4;j<=20;j++) {





}

for(i=21;i<=54;i++) {



}

/*************//* outer box *//*************/







for(k=2;k<=22;k++) {



}

for(l=18;l<=57;l++) {



}











gotoxy(32,4);













}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)



days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;

/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */

if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}

----------------------------------------------------------------------------------------------------------------







getkey() {

union REGS i,o;

while(!kbhit());

i.h.ah=0;int86 (22,&i,&o);

return (o.h.ah);

}

void main() {


clrscr();





do {


if(key==72) {

Solution:




}


if(key==80) {


}


if(key==75) {


}


if(key==77) {

if(month+1 > 12){}


}




clrscr();



while(i<=lastday) {

gotoxy(x,y);



printf("%2d",i);

i++;x+=5;


x=22;y+=2;

}}




}


getch();

}/*********************** main ends ************************/



{

int a,day=1;





if(day==0)day=22;










}



int i,j,k,l;

/*************//* inner box *//*************/







for(j=4;j<=20;j++) {



}

for(i=21;i<=54;i++) {



}

/*************//* outer box *//*************/




gotoxy(58,1);



printf("%c",191);



for(k=2;k<=22;k++) {



}

for(l=18;l<=57;l++) {



}









gotoxy(32,4);




















}



int days;

/* for january */

if(m==1)days=31;

/* for february */

if(m==2) {

if(y%4==0)days=29;

elsedays=28;

}/* for march */

if(m==3)days=31;



/* for april */

if(m==4)days=30;

/* for may */

if(m==5)days=31;

/* for june */

if(m==6)days=30;

/* for july */

if(m==7)days=31;

/* for august */

if(m==8)days=31;

/* for september */

if(m==9)days=30;

/* for october */

if(m==10)days=31;

/* for november */

if(m==11)days=30;

/* for december */

if(m==12)days=31;

return days;}

--------------------------------------------------------------------------------------------------------------





-------------------------------------------------------------------------------------------------------------



---------------------------------------------------------------------------------------------------------------



char s[80];int i=0;

clrscr();

gets(s);

while(s[i]!='\0') {



getch();}

---------------------------------------------------------------------------------------------------------------

Solution:

Solution:

Solution:





void replace();

void main() {


gets(str);

replace(str,str1);

getch();




while(*s!='\0') {

*t=*s;

t++;s++;

}

*t='\0';



temp2[j]=temp[i];

if(temp[i]==' ') {

temp2[j]='\0';


}

else {j=0;


main[k]=temp2[j];

Solution:



k++;j++;}



}

i++;j++;}



}


else {

j=0;


main[k]=temp2[j];

k++;j++;}



}

---------------------------------------------------------------------------------------------------------------



void abbr();

void main() {

char s1[10][80];int i=0;

clrscr();


Solution:



for(i=0;i<10;i++) {


}

clrscr();


for(i=0;i<10;i++) {


printf("\n\n");

}

getch();

}



while(s1[i]!='\0') {

s2[j]=s1[i];



}


}

--------------------------------------------------------------------------------------------------------------




Solution:







gets(a);


while(a[i]!='\0') {





getch();}

______________________________________________________________________












Let us C solutions ( By Chetan ): Let Us C / Chapter 2 (The Decision Control Structure)

http://letuscalllessons.blogspot.com/2014/01/chapter-2-decision-control-structure.html[07-Apr-14 8:48:59 PM]




Let Us C / Chapter 2 (The Decision Control Structure)

(a) If cost price and selling price of an item is input through the keyboard, write a program to determine whether the seller has made profit or incurred loss. Also determine how much profit he made or loss he incurred.

#include<stdio.h>#include<conio.h>main() {

int cp,sp,rslt;clrscr();

printf("Please enter the cost price of the item: \n");scanf("%d",&cp);

printf("Please enter the selling price of the item: \n");scanf("%d",&sp);

if(cp>sp) {rslt=cp-sp;printf("\nSeller has incurred LOSS of %d rupees.\n",rslt);}

if(cp<sp){rslt=sp-cp;printf("\nSeller has made PROFIT of %d rupees.\n",rslt);}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(b) Any integer is input through the keyboard. Write a program to find out whether it is an odd number or even number.

#include<stdio.h>

Exercise [C]

Solution:

Solution:

Chetan Raikwar


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#include<conio.h>

main() {

int i;clrscr();

printf("Please enter the number: \n");scanf("%d",&i);

if(i%2==0)printf("\nThe number is EVEN.\n");

elseprintf("\nThe number is ODD.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(c) Any year is input through the keyboard. Write a program to determine whether the year is a leap year or not.(Hint: Use the % (modulus) operator)


main() {

int yr;clrscr();

printf("Please enter the year: \n");scanf("%d",&yr);

if(yr%4==0)printf("\nThe year is a LEAP YEAR.\n");

elseprintf("\nThe Year is NOT A LEAP YEAR.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(d) According to the Gregorian calendar, it was Monday on the date 01/01/1900. If any year is input through the keyboard write a program to find out what is the day on 1st January of this year.


main() {

int month=1,year,a,day=1;

clrscr();

printf("Enter year: ");

Solution:

Solution:



scanf("%d",&year);





printf("\n\nFirst day is ");

if(day==0)printf("Sunday");

else if(day==1)printf("Monday");

else if(day==2)printf("Tuesday");

else if(day==3)printf("Wednesday");

else if(day==4)printf("Thursday");

else if(day==5)printf("Friday");

else if(day==6)printf("Saturday");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(e) A five-digit number is entered through the keyboard. Write a program to obtain the reversed number and to determine whether the original and reversed numbers are equal or not.


main() {

long i,d1,d2,d3,d4,d5,a,b,c,d,n_num;clrscr();

printf("Please enter a five digit number: \n");scanf("%ld",&i);

d1=i/10000;a=i%10000;

d2=a/1000;b=a%1000;

d3=b/100;

Solution:



c=b%100;

d4=c/10;d=c%10;

d5=d/1;

n_num=(d5*10000)+(d4*1000)+(d3*100)+(d2*10)+(d1*1);

printf("\nThe REVERSED NUMBER is %ld \n",n_num);

if(n_num==i){printf("\nboth numbers are SAME.\n");}else {printf("\nboth number are NOT SAME.\n");}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(f) If the ages of Ram, Shyam and Ajay are input through the keyboard, write a program to determine the youngest of the three.


main() {

int ram,shyam,ajay;clrscr();

printf("Please enter the age of RAM: ");scanf("%d",&ram);

printf("\nPlease enter the age of SHYAM: ");scanf("%d",&shyam);

printf("\nPlease enter the age or AJAY: ");scanf("%d",&ajay);

if(ram<shyam){if(ram<ajay) {printf("\nRam is the YOUNGEST.\n");} } if(shyam<ram){if(shyam<ajay) {printf("\nShyam is the YOUNGEST.\n");} }

if(ajay<ram) {if(ajay<shyam) {printf("\nAjay is the YOUNGEST.\n");} }

getch();

Solution:



return 0;

}

------------------------------------------------------------------------------------------------------------

(g)Write a program to check whether a triangle is valid or not, when the three angles of the triangle are entered through the keyboard. A triangle is valid if the sum of all the three angles is equal to 180 degrees.


int a1,a2,a3;clrscr();

printf("Please enter the first angle: \n");scanf("%d",&a1);

printf("\nPlease enter the second angle: \n");scanf("%d",&a2);

printf("\nPlease enter the third angle: \n");scanf("%d",&a3);

if(a1+a2+a3==180) {printf("\nThe triangle is VALID.\n");}

else {printf("\nThe triangle is NOT VALID.\n");}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(h) Find the absolute value of a number entered through the keyboard.

#include<stdio.h>#include<conio.h>#include<stdlib.h>main() {

int i,av;clrscr();

printf("Please enter any number: ");scanf("%d",&i);

av=abs(i);

/* abs(); is a standard function to calculate absolute value.\this function has been defined in <stdlib.h> library. */

printf("Absolute value = %d",av);

getch();

Solution:

Solution:



return 0;

}

------------------------------------------------------------------------------------------------------------

(i) Given the length and breadth of a rectangle, write a program to find whether the area of the rectangle is greater than its perimeter. For example, the area of the rectangle with length = 5 and breadth = 4 is greater than its perimeter.


int l,b,area,peri;clrscr();

printf("Please enter the length and bredth of a rectangle: \n");scanf("%d%d",&l,&b);

area=l*b;peri=2*(l+b);

if(area>peri) {printf("\nThe Area of Rectangle is GREATER then it's perimeter.\n");}

else {printf("\nThe area of Rectangle is NOT GREATER then it's perimeter.\n");}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(j) Given three points (x1, y1), (x2, y2) and (x3, y3), write a program to check if all the three points fall on one straight line.


int x1,y1,x2,y2,x3,y3,curve1,curve2;clrscr();

printf("Please enter the values of Three points: ");printf("\n(x1,y1):\n");scanf("%d%d",&x1,&y1);printf("\n(x2,y2):\n");scanf("%d%d",&x2,&y2);printf("\n(x3,y3):\n");scanf("%d%d",&x3,&y3);

curve1=(x2-x1)/(y2-y1);curve2=(x3-x2)/(y3-y2);

if(curve1==curve2)printf("\nAll three points fall on one straight line.\n");else

Solution:

Solution:



printf("\nThese three points do not fall on one straight line.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(k) Given the coordinates (x, y) of a center of a circle and it’s radius, write a program which will determine whether a point lies inside the circle, on the circle or outside the circle.(Hint: Use sqrt( ) and pow( ) functions)

#include<stdio.h>#include<conio.h>#include<math.h>main() {

int x,y,x_,y_,r,point;clrscr();

printf("Please enter the coordinates of center: \n");printf("(x,y): ");scanf("%d%d",&x,&y);printf("\nPlease enter the radius of circle: \n");scanf("%d",&r);printf("\nPlease enter the point to check: \n");printf("(x`,y`): ");scanf("%d%d",&x_,&y_);

point=pow(x_,2)+pow(y_,2)+pow(x,2)+pow(y,2)-(2*x_*x)-(2*y_*y)-pow(r,2);

if(point<0)printf("\nThe point lies inside the circle.\n");

if(point>0)printf("\nThe point lies outside the circle.\n");

if(point==0)printf("\nThe point lies on the circle.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(l) Given a point (x, y), write a program to find out if it lies on the x-axis, y-axis or at the origin, viz. (0, 0).


int x,y;clrscr();

printf("Please enter points\n");printf("(x,y): \n");scanf("%d%d",&x,&y);

Solution:

Solution:



if(x==0&&y!=0) {printf("\nThe point lies on Y AXIS.\n");}

else if(x!=0&&y==0) {printf("\nThe point lies on X AXIS.\n");}

else if(x==0&&y==0) {printf("\nThe point lies at the origin.\n");}

else {printf("\nThe point doesn't lie on any axis or at origin.\n");}

getch();return 0;

}

___________________________________________________________________

(a) Any year is entered through the keyboard, write a program to determine whether the year is leap or not. Use the logical operators && and ||.


int yr;clrscr();

printf("Please enter the year: ");scanf("%d",&yr);

if(yr%4==0 || !((yr%4)>0) )printf("\nThe is a LEAP YEAR.\n");

elseprintf("\nThis is NOT A LEAP YEAR.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(b) Any character is entered through the keyboard, write a program to determine whether the character entered is a capital letter, a small case letter, a digit or a special symbol.The following table shows the range of ASCII values for various characters.

The following table shows the range of ASCII values for various characters.

Exercise [F]

Solution:



Characters

Characters ASCII values

A – Z 65 - 90a – z 97 - 1220 – 9 48 - 57special symbols 0 - 47, 58 - 64, 91 - 96, 123 - 127


char ch;clrscr();

printf("Please enter any character:");scanf("%c",&ch);

if(ch>=65&&ch<=90)printf("\nThe character you entered is a CAPITAL LETTER.\n");

if(ch>=97&&ch<=122)printf("\nThe character you entered is a SMALL LETTER.\n");

if(ch>=48&&ch<=57)printf("\nThe character you entered is a DIGIT.\n");

if(ch>=0&&ch<=47 || ch>=58&&ch<=64 || ch>=91&&ch<=96 || ch>=123&&ch<=127)printf("\nThe character you entered is a SPECIAL SYMBOL.\n");

getch();

return 0;

}

------------------------------------------------------------------------------------------------------------

(c) An Insurance company follows following rules to calculate premium.(1) If a person’s health is excellent and the person is between 25 and 35 years of age and lives in a city and is a male then the premium is Rs. 4 per thousand and his policy amount cannot exceed Rs. 2 lakhs.(2) If a person satisfies all the above conditions except that the sex is female then the premium is Rs. 3 per thousand and her policy amount cannot exceed Rs. 1 lakh.(3) If a person’s health is poor and the person is between 25 and 35 years of age and lives in a village and is a male then the premium is Rs. 6 per thousand and his policy cannot exceed Rs. 10,000.(4) In all other cases the person is not insured.

Write a program to output whether the person should be insured or not, his/her premium rate and maximum amount for which he/she can be insured.

Solution:

Solution:




int age,premium,p_a,hlt,rsdnc,sex;long amount;clrscr();

printf("Please tell the HEALTH of the person\n(1 for excellent or 2 for poor) :\n");scanf("%d",&hlt);

printf("Please tell the AGE of the person: \n");scanf("%d",&age);

printf("Please tell the residence\n(1 for city or 2 for village) : \n");scanf("%d",&rsdnc);

printf("Please tell you sex\n(1 for male or 2 for female) : \n");scanf("%d",&sex);

if(hlt==1 && age>=25&&age<=35 && rsdnc==1 && sex==1) {premium=4;amount=200000;

printf("\nThis person is insured.\n");printf("Premium rate = %d rupees per thousand.\n",premium);printf("Person can be insured for maximum amount of %ld rupees.\n",amount);}

else if(hlt==1||hlt==2 && age>=25&&age<=35 && rsdnc==1||rsdnc==2 && sex==2) {premium=3;amount=100000;


else if(hlt==2 && age>=25&&age<=35 && rsdnc==2 && sex==1) {premium=6;amount=10000;

printf("\nThis person is insured.\n");printf("Premium rate is = %d rupees per thousand.\n",premium);printf("Person can be insured for maximum amount of %ld rupees.\n",amount);}

else {printf("\nThis person cannot be insured.\n");}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(d) A certain grade of steel is graded according to the following conditions:(i) Hardness must be greater than 50(ii) Carbon content must be less than 0.7(iii) Tensile strength must be greater than 5600

The grades are as follows:



Grade is 10 if all three conditions are metGrade is 9 if conditions (i) and (ii) are metGrade is 8 if conditions (ii) and (iii) are metGrade is 7 if conditions (i) and (iii) are metGrade is 6 if only one condition is metGrade is 5 if none of the conditions are met

Write a program, which will require the user to give values of hardness, carbon content and tensile strength of the steel under consideration and output the grade of the steel.


int hdn,tnsl,hardness=50,tensile_strength=5600;

float c,carbon_content=0.7;

clrscr();

printf("Please enter the Hardness: \n");scanf("%d",&hdn);

printf("\nPlease enter the Carbon content: \n");scanf("%f",&c);

printf("\nPlease enter the Tensile Strength: \n");scanf("%d",&tnsl);

if(hdn>hardness && c<carbon_content && tnsl>tensile_strength)printf("\nThe steel has been Graded 10.\n");

else if(hdn>hardness && c<carbon_content)printf("\nThe steel has been Graded 9.\n");

else if(c<carbon_content && tnsl>tensile_strength)printf("\nThe steel has been Graded 8.\n");

else if(hdn>hardness && tnsl>tensile_strength)printf("\nThe steel has been Graded 7.\n");

else if(hdn>hardness || c<carbon_content || tnsl>tensile_strength)printf("\nThe steel has been Graded 6.\n");

elseprintf("\nThe steel has been Graded 5.\n");

getch();

return 0;

}

------------------------------------------------------------------------------------------------------------

(e) A library charges a fine for every book returned late. For first 5 days the fine is 50 paise, for 6-10 days fine is one rupee and above 10 days fine is 5 rupees. If you return the book after 30 days your membership will be cancelled. Write a program to accept the number of days the

Solution:



member is late to return the book and display the fine or the appropriate message.

#include<stdio.h>#include<conio.h>main() {int days;clrscr();

printf("Please enter the days of being late in returning the book:\n");scanf("%d",&days);

if(days<=5)printf("\nYou have been fined for 50 paise.\n");

if(days>=6&&days<=10)printf("\nYou have been fined for 1 rupee.\n");

if(days>10&&days<30)printf("\nYou have been fined for 5 rupee.\n");

if(days>=30)printf("\nYour membership has been cancelled.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(f) If the three sides of a triangle are entered through the keyboard, write a program to check whether the triangle is valid or not. The triangle is valid if the sum of two sides is greater than the largest of the three sides.


int s1,s2,s3,ls,ss1,ss2;clrscr();

printf("Please enter the three sides of a triangle: \n");scanf("%d%d%d",&s1,&s2,&s3);

if(s1>s2&&s1>s3){ls=s1;ss1=s2;ss2=s3;}if(s2>s1&&s2>s3) {ls=s2;ss1=s1;ss2=s3;}if(s3>s1&&s3>s2) {ls=s3;ss1=s1;ss2=s2;}

Solution:

Solution:



if((ss1+ss2)>ls) {printf("\nThe Triangle is VALID.\n");}else {printf("The triangle is NOT VALID.\n");}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(g) If the three sides of a triangle are entered through the keyboard, write a program to check whether the triangle is isosceles, equilateral, scalene or right angled triangle.


main() {

int s1,s2,s3,largest,s_1,s_2;clrscr();

printf("Please enter the three sides of a triangle: \n");scanf("%d %d %d",&s1,&s2,&s3);

if(s1==s2&&s1!=s3||s1==s3&&s1!=s2 || s2==s3&&s2!=s1||s2==s1&&s2!=s3 ||s3==s1&&s3!=s2||s3==s2&&s3!=s1){printf("The Triangle is ISOSCELES.\n");}

if(s1==s2&&s2==s3 || s2==s1&&s2==s3 || s3==s1&&s3==s2){printf("The Triangle is EQUILATERAL.\n");}

if(s1!=s2&&s2!=s3 || s2!=s1&&s1!=s3 || s3!=s1&&s1!=s2){printf("The Triangle is SCALENE.\n");}

if(s1>s2&&s1>s3){largest=s1;s_1=s2;s2=s3;}

if(s2>s1&&s2>s3){largest=s2;s_1=s1;s_2=s3;}


if(pow(largest,2)==pow(s_1,2)+pow(s_2,2))printf("The Triangle is RIGHT ANGLED.\n");

getch();return 0;}

Solution:



------------------------------------------------------------------------------------------------------------

(h) In a company, worker efficiency is determined on the basis of the time required for a worker to complete a particular job. If the time taken by the worker is between 2 – 3 hours, then the worker is said to be highly efficient. If the time required by the worker is between 3 – 4 hours, then the worker is ordered to improve speed. If the time taken is between 4 – 5 hours, the worker is given training to improve his speed, and if the time taken by the worker is more than 5 hours, then the worker has to leave the company. If the time taken by the worker is input through the keyboard, find the efficiency of the worker.


int hrs;clrscr();

printf("Please enter the hours of completing job: ");scanf("%d",&hrs);

if(hrs>=2 && hrs<=3)printf("\nThe worker is HIGHLY EFFICIENT.\n");

if(hrs>=3 && hrs<=4)printf("\nThe worker is ordered to IMPROVE SPEED.\n");

if(hrs>=4 && hrs<= 5)printf("\nThe worker should be given TRAINING TO IMPROVE SPEED.\n");

if(hrs>5)printf("\nThe worker should be asked to leave the company.\n");

elseprintf("The worker is HIGHLY EFFICIENT.\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(i) A university has the following rules for a student to qualify for a degree with A as the main subject and B as the subsidiary subject: (a) He should get 55 percent or more in A and 45 percent or more in B. (b) If he gets than 55 percent in A he should get 55 percent or more in B. However, he should get at least 45 percent in A. (c) If he gets less than 45 percent in B and 65 percent or more in A he is allowed to reappear in an examination in B to qualify. (d) In all other cases he is declared to have failed.

Write a program to receive marks in A and B and Output whether the student has passed, failed or is allowed to reappear in B.

#include<stdio.h>

Solution:

Solution:



#include<conio.h>main() {

int a,b;clrscr();

printf("Please enter the marks in subject A: \n");scanf("%d",&a);

printf("\nPlease enter the marks in subject B: \n");scanf("%d",&b);

if(a>=55 && b>=45) {printf("Student is qualified for the degree.\n"); }

else if(a==55 && b>=55 || a==55 && b>=45) {printf("Student is qualified for the degree.\n"); }

else if(a>=65 && b<45) {printf("Student is allowed to reappear in the exam of B to qualify.\n"); }

else {printf("Student has failed.\n"); }

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(j) The policy followed by a company to process customer orders is given by the following rules:

(a) If a customer order is less than or equal to that in stock and has credit is OK, supply has requirement.(b) If has credit is not OK do not supply. Send him intimation.(c) If has credit is Ok but the item in stock is less than has order, supply what is in stock. Intimate to him data the balance will be shipped.

Write a C program to implement the company policy.


int stock,credit,order;clrscr();

printf("Please enter the stock available: ");scanf("%d",&stock);printf("\nPlease enter the order: ");scanf("%d",&order);printf("\nPlease enter the credit: ");scanf("%d",&credit);

if(credit!=0&&order<=stock) {

Solution:



printf("\nSupply\n");}

else if(credit!=0&&order>stock) {printf("\nAvailable items will be supplied.\n");}

else {printf("\nNo supply.\n");}

getch();return 0;

}______________________________________________________________________

(a) Using conditional operators determine: (1) Whether the character entered through the keyboard is a lower case alphabet or not.(2) Whether a character entered through the keyboard is a special symbol or not.


char main() {

char ch;clrscr();

printf("Please enter a character: \n");scanf("%c",&ch);

if(ch>=97 && ch<=122) {printf("\n[A] This character is a SMALL CASE alphabet.\n");}

else {printf("\n[A] This character is NOT A SMALL CASE alphabet.\n");}

if(ch>=0&&ch<=47||ch>=58&&ch<=64||ch>=91&&ch<=96||ch>=123&&ch<=127){printf("\n[B] This character is a SPECIAL SYMBOL.\n");}

else{printf("\n[B] This character is NOT A SPECIAL SYMBOL.\n");}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

Exercise [J]

Solution:



(b) Write a program using conditional operators to determine whether a year entered through the keyboard is a leap year or not.

#include<stdio.h>#include<conio.h>main () {

int yr;clrscr();


if(yr%4!=0)printf("\nThis is NOT A LEAP YEAR.\n");

elseprintf("\nThis is a LEAP YEAR.\n");

getch();return 0;}

------------------------------------------------------------------------------------------------------------

(c) Write a program to find the greatest of the three numbers entered through the keyboard using conditional operators.


main() {

int n1,n2,n3;clrscr();

printf("\nPlease enter 3 numbers: \n");scanf("%d %d %d",&n1,&n2,&n3);

if(n1>n2 && n1>n3) {printf("\n%d is the greatest number of the three numbers.\n",n1);}

else if(n2>n1 && n2>n3) {printf("\n%d is the greatest number of the three numbers.\n",n2);}


else {printf("\nwrong input!\n");}

getch();return 0;

}

_______________________________________________________________________

Solution:

Solution:



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Let us C solutions ( By Chetan ): Let Us C / Chapter 12 (File Input Output)

http://letuscalllessons.blogspot.com/2014/01/chapter-12-file-input-output.html[07-Apr-14 8:49:48 PM]





(a) Write a program to read a file and display contents with its line numbers.

NOTE: Make a file as "DATA.TXT" in bin directory and write/paste some text in it, then run the program.


FILE *fp;char i;int line=1;

clrscr();

fp=fopen("DATA.TXT","r");

if(fp==NULL) {

printf("\n\nCannot open file!");

delay(1000);exit();}

printf("%2d. ",line); /* we already print 1st for the first line */

while(i!=EOF) {

i=fgetc(fp);

printf("%c",i);

/* if the character is newline,the line number will be\printed after it */

if(i=='\n') {line++;printf("%2d. ",line);}

}

Exercise [C]

Solution:

Chetan Raikwar


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fclose(fp);getch();}

---------------------------------------------------------------------------------------------------------------

(b) Write a program to find the size of a text file without traversing it character by character.



void main() {

FILE *fp;char s[80],ch;int len=0;

clrscr();

fp=fopen("data.txt","r");

while(fgets(s,79,fp)!=NULL)

len=len+strlen(s); /* length of each string */ /* spaces and newlines are also counted */

fclose(fp);

printf("length of file = %d",len);

getch();}

----------------------------------------------------------------------------------------------------------------

(c) Write a program to add the contents of one file at the end of another.

NOTE: Make two file as "FILE1.TXT" and "FILE2.TXT" in bin directory and then run the program. The contents of FILE1.TXT will be appended to FILE2.TXT.


FILE *f1,*f2;char ch;

clrscr();

f1=fopen("FILE1.TXT","r"); /* file to append */f2=fopen("FILE2.TXT","a+"); /* file to be appended */

Solution:

Solution:



if(f1==NULL || f2==NULL) {printf("\ncannot open one of files!");

exit();}

while(1) {

ch=fgetc(f1);

if(ch==EOF) {break;}

fputc(ch,f2);

}

fclose(f1);fclose(f2);

printf("\ntask completed successfully!");

getch();

}

---------------------------------------------------------------------------------------------------------------

(d) Suppose a file contains student’s records with each record containing name and age of a student. Write a program to read these records and display them in sorted order by name.

#include<stdio.h>#include<conio.h>#define N 100struct student { char name[30]; int age; };

void main() {

struct student std;struct student s[N]; /* size of array of structure defined globally for convenience */

FILE *fp;int flag=0,ch,i=0,count=0;long recsize;char another='y';void srt_print(); /* funtion to sort and print */

clrscr();

recsize=sizeof(std);

fp=fopen("STUDENT.DAT","rb+");if(fp==NULL) {fp=fopen("STUDENT.DAT","wb+");if(fp==NULL)exit();}

while(1) {

Solution:



clrscr();

printf("\t\t\tStudent database\n");printf("\t\t\t****************\n\n\n");printf("\t\t\n1: Add student data\n");printf("\t\t\n2: List student data\n");printf("\t\t\n0: Exit");

gotoxy(2,24);printf("Your choice: ");scanf("%d",&ch);

switch(ch) {

case 1:

clrscr();

while(another=='y' || another=='Y') {

clrscr();

printf("\t\tAdd student data\n");printf("\t\t****************\n\n");printf("\nEnter student name: ");scanf("%s",&std.name);printf("\n\naEnter student age: ");scanf("%d",&std.age);

fseek(fp,0,SEEK_END);fwrite(&std,recsize,1,fp);

gotoxy(2,24);printf("Add another information(Y/N): ");fflush(stdin);another=getche();

}break;

case 2:

clrscr();

printf("\t\tList student data\n");printf("\t\t*****************\n\n");rewind(fp);while(fread(&std,recsize,1,fp)==1) {s[i]=std;flag=1;i++;count++;}

srt_print(&s,count); /* function to print names */

if(flag==0) {

printf("\n\n\nNo data found!\n");

}printf("\n\n\npress any key to return...");getch();

break;

case 0:

fclose(fp);exit();



default:

printf("wrong input!\n");exit();

}

}

}/******** main ends ************/

/**** sorting and printing function ****/void srt_print(struct student *ss, int n) {

struct student temp;int i,j;

for(i=0;i<=n;i++) {

for(j=i+1;j<n;j++) {

/* checking first alphabets of both names */

if(ss[i].name[0] > ss[j].name[0]) {

temp=ss[i];ss[i]=ss[j];ss[j]=temp;}

/* if first alphabets are same */

else if(ss[i].name[0]==ss[j].name[0] && ss[i].name[1] > ss[j].name[1]) {

temp=ss[i];ss[i]=ss[j];ss[j]=temp;

}

/* if first 2 alphabets are same */

else if(ss[i].name[0]==ss[j].name[0] && ss[i].name[1]==ss[j].name[1]) {


temp=ss[i];ss[i]=ss[j];ss[j]=temp;}}



if(ss[i].name[2]==ss[i].name[2]) {




temp=ss[i];ss[i]=ss[j];ss[j]=ss[i];}}}

/* if first four 4 alphabets are same */


if(ss[i].name[2]==ss[j].name[2] && ss[i].name[2]==ss[j].name[2]) {

if(ss[i].name[3]==ss[j].name[3]) {


temp=ss[i];ss[i]=ss[j];ss[j]=ss[i];

} } } }

}}

/* printing sorted list */

for(i=0;i<n;i++) {

printf("\n%10s\t\t%2d\n",ss[i].name,ss[i].age);

}

}

--------------------------------------------------------------------------------------------------------------

(e) Write a program to copy one file to another. While doing so replace all lowercase characters to their equivalent uppercase characters.

NOTE: Make a file as "FILE.TXT" in bin directory and write/paste some text in it. then run the program and you will receive another file as "RESULT.TXT" with uppercase letters.


void main() {

FILE *fr,*fw;char a[1000];char ch,upr;clrscr();

fr=fopen("SOURCE.TXT","r");

if(fr==NULL) {printf("cannot open source file!\n");

}

Solution:



fw=fopen("RESULT.TXT","w");

if(fw==NULL) {printf("cannot open target file!\n");

}

while(1) {

ch=fgetc(fr);

if(ch==EOF)break;

else {

if(ch>=97 && ch<=122) {

ch=ch-32;

}

fputc(ch,fw);}

}fclose(fr);fclose(fw);

printf("Task completed!");getch();}

----------------------------------------------------------------------------------------------------------------

(f) Write a program that merges lines alternately from two files and writes the results to new file. If one file has less number of lines than the other, the remaining lines from the larger file should be simply copied into the target file.

----------------------------------------------------------------------------------------------------------------

(g) Write a program to display the contents of a text file on the screen. Make following provisions:Display the contents inside a box drawn with opposite cornerco-ordinates being ( 0, 1 ) and ( 79, 23 ). Display the name of the file whose contents are being displayed, and the page numbers in the zeroth row. The moment one screenful of file has been displayed, flash a message ‘Press any key...’ in 24th row. When a key is hit, the next page’s contents should be displayed, and so on till the end of file.

#include<stdio.h>#include<conio.h>#include<string.h>#include<types.h>#include<stat.h>#include<fcntl.h>

Solution:

Solution:



void box();void print();

void main() {

int inhandle,bytes,pg=1;FILE *fp;char source[80],buffer[1400];

clrscr();

printf("Enter file name: ");gets(source);

inhandle=open(source, O_RDONLY);

if(inhandle==-1) {

printf("cannot open file!");exit();

}

clrscr();

while(1) {

bytes=read(inhandle,buffer,1387);

if(bytes>0) {

gotoxy(32,1); /* showing filename */printf("%s",strupr(source));

gotoxy(70,1);printf("Pg: %3d",pg); /* showing page number */

box(); /* passing the heading and page number to the function */

print(buffer); /* passing the buffer to the function */

}

else {

gotoxy(70,1);printf("Pg: %3d",pg);break;}

++pg;}

close(inhandle);

getch();}

/********************************//* function to print the buffer *//********************************/

void print(char s[1400]) {

int x=4,y=3,i=0;



while(s[i]!=EOF) {

gotoxy(x,y);printf("%c",s[i]);

if(x>74) {

x=4;y+=1;

}

if(y>21) {

gotoxy(2,24);printf("press any key to go to next page...");

x=4;y=3;

getch();

clrscr();box();}

x++;i++;

}

}

/****************************//* function to draw the box *//****************************/

void box() {

int i,j;

for(i=2;i<=77;i++) {



}

for(j=3;j<=22;j++) {



}







}

----------------------------------------------------------------------------------------------------------------

(h) Write a program to encrypt/decrypt a file using:(1) An offset cipher: In an offset cipher each character from the source file is offset with a fixed value and then written to the target file.For example, if character read from the source file is ‘A’, then convert this into a new character by offsetting ‘A’ by a fixed value, say 128, and then writing the new character to the target file.(2) A substitution cipher: In this each character read from the source file is substituted by a corresponding predetermined character and this character is written to the target file.For example, if character ‘A’ is read from the source file, and if we have decided that every ‘A’ is to be substituted by ‘!’, then a ‘!’ would be written to the target file in place of every ‘A’ Similarly, every ‘B’ would be substituted by ‘5’ and so on.

Offset cipher Encryption:


void main() { /* offset cipher encryption */

/* every character has been added to 128 and the new value has beenwritten */

FILE *fp,*ft;char ch;

clrscr();

fp=fopen("data.txt","r");ft=fopen("temp.txt","w");

if(fp==NULL) {printf("cannot open one of files!");exit();}

while(1) {

ch=fgetc(fp);

if(ch==EOF)break;

ch=ch+128; /* offset cipher encryption */

fputc(ch,ft);

}


Solution:



remove("data.txt");rename("temp.txt","data.txt");

printf("task completed!");getch();}

Offset cipher Decryption:

#include<stdio.h>#include<conio.h>void main() { /* offset cipher decryption */



if(fp==NULL) {printf("cannot open one of files!");

exit();}

while(1) {

ch=fgetc(fp);

if(ch==EOF)break;

ch=ch-128; /* decryption */

fputc(ch,ft);

}



printf("task completed!");getch();

}

Substitution cipher encryption:


void main() { /* substitution cipher encryption */

/* all vowels have been exchanged with first five ascii characters and every space has been converted to 6th ascii character */




clrscr();



while(1) {

ch=fgetc(fp);

if(ch==EOF)break;

encrypt(&ch); /* function for encryption */

fputc(ch,ft);

}



printf("task completed!");getch();}encrypt(char *c) {

if(*c=='a') {*c='!';}

if(*c=='e') {*c='@';}

if(*c=='i') {*c='#';}

if(*c=='o') {*c='$';}

if(*c=='u') {*c='%';}

if(*c==' ') {*c='^';}

return *c;}

Substitution cipher Decryption:


/* substitution cipher's decryption */






exit();}

while(1) {

ch=fgetc(fp);

if(ch==EOF)break;

decrypt(&ch); /* function for decryption */

fputc(ch,ft);

}




}

decrypt(char *c) {

if(*c=='!')*c='a';

if(*c=='@')*c='e';

if(*c=='#')*c='i';

if(*c=='$')*c='o';

if(*c=='%')*c='u';

if(*c=='^')*c=' ';

return *c;

}----------------------------------------------------------------------------------------------------------------

(i) In the file ‘CUSTOMER.DAT’ there are 100 records with the following structure:struct customer{



int accno ;char name[30] ;float balance ;} ;In another file ‘TRANSACTIONS.DAT’ there are several records with the following structure:struct trans{int accno ,char trans_type ;float amount ;} ;The parameter trans_type contains D/W indicating deposit or withdrawal of amount. Write a program to update ‘CUSTOMER.DAT’ file, i.e. if the trans_type is ‘D’ then update the balance of ‘CUSTOMER.DAT’ by adding amount to balance for the corresponding accno. Similarly, if trans_type is ‘W’ then subtract the amount from balance. However, while subtracting the amount make sure that the amount should not get overdrawn, i.e. at least 100 Rs. Should remain in the account.

NOTE: If on withdrawal, the balance subtracts to less than 100. The withdrawal will not be performed.


struct customer { int accno; char name[30]; float balance; }cust;struct trans { int accno; char trans_type; float amount; }tra;FILE *fp,*ft,*ftemp;int flag=0;long recsize,retsize;char another,ch;clrscr();

fp=fopen("CUSTOMER.DAT","rb+");

if(fp==NULL) {

fp=fopen("CUSTOMER.DAT","wb+");

if(fp==NULL)

printf("cannot open customer data file!\n");

exit();

}

ft=fopen("TRANSACTIONS.DAT","rb+");

if(ft==NULL) {

Solution:



ft=fopen("TRANSACTIONS.DAT","wb+");

if(ft==NULL)

printf("cannot open transactions file!\n");

exit();

}

recsize=sizeof(cust);retsize=sizeof(tra);

while(1) {

clrscr();

printf("\t\tCutomer Transactions:\n");printf("\t\t*********************\n\n\n");printf("\t1: Add customer information:\n\n");printf("\t2: Add transaction information:\n\n");printf("\t3: List customer information:\n\n");printf("\t4: List transaction information:\n\n");printf("\t5: Perform transaction:\n\n");printf("\t0: Exit:\n\n\n");

gotoxy(2,24);printf("Your choice: ");fflush(stdin);ch=getche();

switch(ch) {

case '1':

clrscr();

fseek(fp,0,SEEK_END);

another='y';


printf("\t\tAdd customer information:\n");printf("\t\t*************************\n\n");printf("\nEnter account number: ");scanf("%d",&cust.accno);printf("\n\nEnter name: ");scanf("%s",cust.name);printf("\n\nEnter balance: ");fflush(stdin);scanf("%f",&cust.balance);

fwrite(&cust,recsize,1,fp);

gotoxy(2,24);printf("Add another customer information(Y/N): ");another=getche();

clrscr();

}break;

case '2':

clrscr();

fseek(ft,0,SEEK_END);



another='y';


printf("\t\tAdd transaction information:\n");printf("\t\t****************************\n\n\n");printf("Enter existing customer account number: ");scanf("%d",&tra.accno);printf("\n\nEnter transaction type(D/W): ");fflush(stdin);scanf("%c",&tra.trans_type);printf("\n\nEnter amount for transaction: ");fflush(stdin);scanf("%f",&tra.amount);

fwrite(&tra,retsize,1,ft);

gotoxy(2,24);printf("Enter another information(Y/N): ");another=getche();

clrscr();}break;

case '3':

clrscr();

printf("\t\tList customer information:\n");printf("\t\t**************************\n\n");

rewind(fp);while(fread(&cust,recsize,1,fp)==1) {printf("\n%5d\t%-8s\t%5.2f\n",cust.accno,cust.name,cust.balance);flag=1;}

if(flag==0) {gotoxy(2,12);printf("No customer information found!\n");}printf("\n\npress any key to go back...");getch();

break;

case '4':

clrscr();

printf("\t\tList transaction information:\n");printf("\t\t*****************************\n\n");

rewind(ft);while(fread(&tra,retsize,1,ft)==1) {printf("\n%5d\t%c\t%6.2f\n",tra.accno,tra.trans_type,tra.amount);flag=1;}

if(flag==0) {gotoxy(2,12);printf("No transaction information found!\n");}

printf("\n\npress any key to go back...");getch();

break;



case '5':

clrscr();

printf("\t\tPerform transactions\n");printf("\t\t********************\n\n");

rewind(fp);

while(fread(&cust,recsize,1,fp)==1) {

rewind(ft);

while(fread(&tra,retsize,1,ft)==1) {

if(cust.accno==tra.accno) {

flag=1;

if(tra.trans_type=='D' || tra.trans_type=='d') {

cust.balance+=tra.amount;

fseek(fp,-recsize,SEEK_CUR);fwrite(&cust,recsize,1,fp);

}

else if(tra.trans_type=='W' || tra.trans_type=='w') {

if((cust.balance-tra.amount)>=100) {

cust.balance-=tra.amount;


}

}

}}}

fclose(ft);

ftemp=fopen("TEMP.DAT","rb+");

if(ftemp==NULL) {

ftemp=fopen("TEMP.DAT","wb+");

}

remove("TRANSACTIONS.DAT");rename("TEMP.DAT","TRANSACTIONS.DAT");


if(ft==NULL) {


}

if(flag==0) {

gotoxy(2,12);printf("No active transactions\n");



}

else if(flag>0) {

gotoxy(2,12);printf("Transactions performed seccussfully!\n");gotoxy(2,14);printf("NOTE: withdrawl for low balance accounts has not been performed\n");

}gotoxy(2,24);printf("press any key to return...");getch();

break;

case '0':

fclose(fp);fclose(ft);exit();

}}

}

---------------------------------------------------------------------------------------------------------------

(j) There are 100 records present in a file with the following structure:struct date{int d, m, y ;} ;struct employee{int empcode[6] ;char empname[20] ;struct date join_date ;float salary ;} ;Write a program to read these records, arrange them in ascending order of join_date and write them in to a target file.

----------------------------------------------------------------------------------------------------------------

(k) A hospital keeps a file of blood donors in which each record has the format:Name: 20 ColumnsAddress: 40 ColumnAge: 2 ColumnsBlood Type: 1 Column (Type 1, 2, 3 or 4)Write a program to read the file and print a list of all blood donors whose age is below 25 and blood is type 2.

Solution:

Solution:Program to creat record file of blood donors



/* This program will make a file of blood donors and save information in it */

/* Writing program */


FILE *fp;char another='y';

struct blood { char name[50]; char adr[50]; int age; int bld; } b;

clrscr();

fp=fopen("BLOODBANK.DAT","wb");

if(fp==NULL) {printf("cannot open target file!\n");exit();}

while(another=='Y' || another=='y') {

clrscr();printf("\t\tInformation of Blood donor\n");printf("\t\t**************************\n\n\n");printf("Enter the name: ");scanf("%s",b.name);printf("\n\nenter the address: ");scanf("%s",b.adr);printf("\n\nenter the age: ");scanf("%d",&b.age);printf("\n\nenter the blood group(1/2/3/4): ");scanf("%d",&b.bld);

fprintf(fp,"%s\t%s\t%d\t%d",b.name,b.adr,b.age,b.bld);

printf("\n\n\nenter more information(Y/N): ");fflush(stdin);

another=getch();

}

fclose(fp);

}

/* This program will read the information from the file made by writing program */

/* Reading Program */


FILE *fp;

Program to read record file for specifications



char ch;

struct blood { char name[50]; char adr[50]; int age; int bld; }b;clrscr();

fp=fopen("BLOODBANK.DAT","rb");

if(fp==NULL) {printf("cannot open source file!\n\n");exit();}

while(fscanf(fp,"%s\t%s\t%d\t%d",&b.name,&b.adr,&b.age,&b.bld)!=EOF)if(b.age<25 && b.bld==2) {printf("\n%s\t %s\t%2d\t %d",b.name,b.adr,b.age,b.bld);}fclose(fp);

getch();}

----------------------------------------------------------------------------------------------------------------

(l) Given a list of names of students in a class, write a program to store the names in a file on disk. Make a provision to display the nth name in the list (n is data to be read) and to display all names starting with S.


struct name { int sn; char name[30]; }s;int i,num,flag=0;long recsize;char another,ch;FILE *fp;clrscr();

fp=fopen("NAMES.DAT","rb+");

if(fp==NULL) {

fp=fopen("NAMES.DAT","wb+");

if(fp==NULL)

printf("cannot open file! \n");exit();}

recsize=sizeof(s);

while(1) {

clrscr();printf("\t\tStudent Names:\n");printf("\t\t**************\n\n\n");printf("\t1: Add names of students:\n\n");printf("\t2: Search a student name:\n\n");printf("\t3: List all student names:\n\n");

Solution:



printf("\t4: List all names starting with 'S':\n\n");printf("\t0: Exit\n\n");

gotoxy(2,24);printf("Your choice: ");fflush(stdin);

ch=getche();

switch(ch) {

case '1':

clrscr();


another='y';


printf("\t\tAdd names of students:\n");printf("\t\t**********************\n\n");printf("Enter student number: ");scanf("%d",&s.sn);printf("\n\nEnter student name: ");scanf("%s",s.name);

fwrite(&s,recsize,1,fp);

gotoxy(2,24);printf("Enter another name(Y/N): ");fflush(stdin);another=getche();

clrscr();}break;

case '2':

clrscr();

printf("\t\tSearch a student name:\n");printf("\t\t**********************\n\n");printf("Enter student number: ");scanf("%d",&num);

rewind(fp);

while(fread(&s,recsize,1,fp)==1) {

if(s.sn==num) {

printf("\n\nStudent Number: %5d\n\n",s.sn);printf("Student name: %s\n\n",s.name);

flag=1;break;} }

if(flag==0) {printf("\n\n\nNo such name found!\n");}gotoxy(2,24);printf("press any key to return...\n");getch();



break;

case '3':

clrscr();

printf("\t\tList all student names\n");printf("\t\t**********************\n\n\n");

rewind(fp);

while(fread(&s,recsize,1,fp)==1) {printf("\n%5d\t%-10s\n",s.sn,s.name);flag=1;}

if(flag==0) {printf("\n\n\nNo name found!\n");}

printf("\n\n\npress any key to return...\n");getch();break;

case '4':

clrscr();

printf("\t\tAll name starting with 'S':\n");printf("\t\t***************************\n\n\n");

rewind(fp);while(fread(&s,recsize,1,fp)==1) {

if(strncmp('s',s.name[0])==0) { /* comparing only first character of \ name if it is "s" */printf("\n%5d\t%-10s\n",s.sn,s.name);flag=1;}}

if(flag==0) {printf("\n\n\nNo name starting with \'S\' found!\n");}


case '0':

fclose(fp);exit();

} }

}

----------------------------------------------------------------------------------------------------------------

(m) Assume that a Master file contains two fields, Roll no. and name of the student. At the end of the year, a set of students join the class and another set leaves. A Transaction file contains the roll numbers and an appropriate code to add or delete a student.Write a program to create another file that contains the updated list of names and roll numbers. Assume that the Master file and the



Transaction file are arranged in ascending order by roll numbers. The updated file should also be in ascending order by roll numbers.

Note:- This program ( all the 3 ) are ideally for single. After you have ran it once, check the results and delete the updated list file before running it again. Assign roll number in ascending orders by roll numbers. And process data according to ascending nature.

1- first program will let you save data in masterfile.2- second program will let you add or delete data and will generate an updated list in text mode.3- obtain the updated list and check it ( assuming you have processed data as instructed).

1. program to creat master file.


void main() {

FILE *fp;struct student { int rl; char name[50]; }s;char ch,another;clrscr();

fp=fopen("MASTERFILE.DAT","rb+");

if(fp==NULL) {

fp=fopen("MASTERFILE.DAT","wb+");

if(fp==NULL)puts("cannot open master file!");exit();}

while(1) {

clrscr();

gotoxy(30,2);printf("Masterfile\n");gotoxy(30,3);printf("**********\n\n\n");gotoxy(20,6);printf("1: Enter student data: ");gotoxy(20,8);printf("2: Read student data: ");gotoxy(20,10);printf("0: Exit: ");


ch=getche();

switch(ch) {

case '1':

Solution:



clrscr();


another='y';


clrscr();

gotoxy(2,24);printf("NOTE: assign roll numbers in ascending order");gotoxy(20,5);printf("Enter roll number: ");scanf("%d",&s.rl);

gotoxy(20,7);printf("Enter name: ");fflush(stdin);gets(s.name);

fwrite(&s,sizeof(s),1,fp);

gotoxy(20,10);printf("Add more data(Y/N): ");fflush(stdin);

another=getche();

}

break;

case '2':

clrscr();

gotoxy(30,2);printf("Student data");gotoxy(30,3);printf("************\n\n");

rewind(fp);

while(fread(&s,sizeof(s),1,fp)==1) {

printf("\n%d\t%s\n",s.rl,s.name);

}

printf("\n\npress any key to return...");getch();

break;

case '0':

fclose(fp);exit();

}}

}

2. program to creat transaction file and generate



updated list

#include<stdio.h>#include<conio.h>#include<string.h>#define N 100

void main() {

FILE *ft,*fu,*fm,*ftemp;

struct student { int rl; char name[50]; }s,ss[N],temp;

struct transaction { char stats; int rl; char name[50]; }t;

int flag=0;char ch,another;clrscr();

fm=fopen("MASTERFILE.DAT","rb+");

if(fm==NULL) {

printf("Masterfile doesn't exist!");

}

ft=fopen("TRANSACTION.DAT","rb+");

if(ft==NULL) {

ft=fopen("TRANSACTION.DAT","wb+");

if(ft==NULL)puts("cannot open transactions file!");exit();}

fu=fopen("UPDATELIST.TXT","w+");

if(fu==NULL) {


while(1) {

clrscr();

gotoxy(30,2);printf("Transaction-File\n");gotoxy(30,3);printf("****************\n\n\n");gotoxy(20,6);printf("1: ADD/DELETE student data from master list: ");gotoxy(20,8);printf("2: Read transaction data: ");gotoxy(20,10);printf("3: Creat updated list: ");gotoxy(20,12);



printf("0: Exit:");


ch=getche();

switch(ch) {

case '1':

clrscr();


another='y';


clrscr();

gotoxy(2,23);printf("NOTE: data to be deleted should match master list");gotoxy(2,24);printf("NOTE: data to be added should follow the ascending nature of master list");gotoxy(20,5);printf("ADD/DELETE student(A/D): ");scanf("%c",&t.stats);gotoxy(20,7);printf("Enter roll number: ");scanf("%d",&t.rl);gotoxy(20,9);printf("Enter name: ");fflush(stdin);gets(t.name);

fwrite(&t,sizeof(t),1,ft);


another=getche();

}

break;

case '2':

clrscr();


rewind(ft);

while(fread(&t,sizeof(t),1,ft)==1) {

printf("\n");

if(t.stats=='a' || t.stats=='A') {printf("ADD");}

else {printf("DELETE");}



printf("\t%d\t%s\n",t.rl,t.name);

}


break;

case '3':

clrscr();

gotoxy(30,2);printf("make updated list");gotoxy(30,3);printf("*****************\n\n");

rewind(fm);

while(fread(&s,sizeof(s),1,fm)==1) {

flag=0;

rewind(ft);


if(t.rl==s.rl ) {

if(t.stats=='d' || t.stats=='D') {flag=1;}

}

}

if(flag==0)

fprintf(fu,"\n%d\t%-s\n",s.rl,s.name);

}

rewind(ft);


if(t.stats=='a' || t.stats=='A') {

fprintf(fu,"\n%d\t%-s\n",t.rl,t.name);

}

}


fclose(ftemp);fclose(ft);

remove("TRANSACTION.DAT");rename("TEMP.DAT","TRANSACTION.DAT");


gotoxy(2,24);printf("press any key to continue...");



getch();

break;case '0':

fclose(fm);fclose(ft);fclose(fu);exit();

}}

}

-----------------------------------------------------------------------------------------------------------------

(n) In a small firm employee numbers are given in serial numerical order, that is 1, 2, 3, etc.− Create a file of employee data with following information: employee number, name, sex, gross salary.− If more employees join, append their data to the file.− If an employee with serial number 25 (say) leaves, delete the record by making gross salary 0.− If some employee’s gross salary increases, retrieve the record and update the salary.Write a program to implement the above operations.


struct emp { int empno; char name[30]; char sex; float gs; } e;FILE *fp,*ft;int long recsize;int empn,flag=0;float new_sal;char another,ch;clrscr();

recsize=sizeof(e);

fp=fopen("EMP.DAT","rb+");

if(fp==NULL) {

fp=fopen("EMP.DAT","wb+");if(fp==NULL)

exit();

}

while(1) {

clrscr();

printf("\t\tEmployee database management\n");

Solution:



printf("\t\t****************************\n");printf("\n\n\t1: Add another employee: ");printf("\n\n\t2: Add salary information of employee: ");printf("\n\n\t3: List all records: ");printf("\n\n\t4: Delete employee with 0 salary: ");printf("\n\n\t0: Exit:");

gotoxy(2,24);printf("your choice: ");fflush(stdin);ch=getche();

switch(ch) {

case '1':

clrscr();

fseek(fp,0,SEEK_END); /* seeking cursor to reach at the end of file */

another='Y';


printf("\n\tEnter new employee information\n");printf("\t******************************\n\n\n");

printf("\nEnter employee number: ");scanf("%d",&e.empno);

printf("\n\nEnter employee name: ");scanf("%s",e.name);

printf("\n\nEnter sex(M/F/O): ");scanf(" %c",&e.sex);

printf("\n\nEnter gross salary: ");scanf("%f",&e.gs);

/* writing new information at the end of file */

fwrite(&e,recsize,1,fp);

printf("\n\n\n\nAdd another employee(Y/N): ");fflush(stdin);another=getche();

clrscr();

}

break;

case '2':

clrscr();

another='Y';


printf("\n\tEnter salary information\n");printf("\t************************\n\n");

gotoxy(2,23); /* showing message at the bottom of the screen */

printf("NOTE: to delete an employee, mark his/her salary 0\n");printf(" then use option 4 from main menu.");

gotoxy(3,5); /* returning cursor back from the bottom */



printf("Enter employee number: ");scanf("%d",&empn); /* asking for employee number to search */

rewind(fp);

while(fread(&e,recsize,1,fp)==1) {

if(e.empno-empn==0) { /* if employee number matches with structure */

flag=1; /* condition indicator for printing further messages */

printf("\n\nEnter new salary for employee: ");

scanf("%f",&e.gs);

e.empno=e.empno; /* rest information should be same except only\ salary */e.sex=e.sex;

e.name[30]=e.name[30];

fseek(fp,-recsize,SEEK_CUR); /* seeking the correct location of data within\ structure in the file */fwrite(&e,recsize,1,fp); /* writing data at correct position */

break;

}}

if(flag==0) /* conditional indicator used above */printf("\n\n\n\tinformation does not exist!\n");

printf("\n\nenter another information(Y/N): ");another=getche();

clrscr();

}

break;

case '4':

clrscr();

printf("\n\n\tDelete employee\n");printf("\t***************\n\n");

ft=fopen("TEMP.DAT","w"); /* opening new temporary file */

rewind(fp); /* taking cursor back to the very beginning of file */

while(fread(&e,recsize,1,fp)==1) { /* matching each salary */

if(e.gs!=0.0 || e.gs!=0) { /* if salary is not 0 then data will be written to new file */

flag=1;

fwrite(&e,recsize,1,ft);

}}fclose(fp);fclose(ft);

remove("EMP.DAT"); /* removing original file with 0 salary and renaming\ temporary without 0 salary as the original file */



rename("TEMP.DAT","EMP.DAT");

fp=fopen("EMP.DAT","rt+"); /* opening the new file, it opens because it has not been opened before */ /* a file cannot be opened twice during execution as you know */

if(flag>0) {printf("\n\n\nall records with 0 gross salary have been deleted. \n");}

gotoxy(2,24);printf("\n\n\npress any key to return...");getch();

break;

case '0':

fclose(fp);exit();

case '3':

clrscr();

printf("\t\tList all employees\n");printf("\t\t******************\n\n\n");

rewind(fp);


flag=1;

printf("%2d\t%6s\t%c\t%.2f\n\n",e.empno,e.name,e.sex,e.gs);

}

if(flag==0)

printf("\n\n\tNo records exist!\n\n");

printf("\n\npress any key to return... ");

getch(); /* this is very important place, if we don't stop screen here \ after reading and printing the list,we won't be able to see it and it will disappear and will return to main menu because of "break" statement. */break;

} }

}linkfloat() { /* function to avoid possible errors because of floats */float a=0,*b;b=&a;a=*b;return 0;}

-----------------------------------------------------------------------------------------------------------------

(o) Given a text file, write a program to create another text file deleting the words “a”, “the”, “an” and replacing each one of them with a blank space.



NOTE: Make a file "FILE.TXT" in bin directory and write/paste something in it, then run the program, you will get another file as "NEW.TXT" as result of the program.


void replace();

void main() {

FILE *fp,*ft; char str[80],target[80]; clrscr();

fp=fopen("FILE.TXT","r");

if(fp==NULL) {

puts("cannot open source file!"); exit(); }

ft=fopen("NEW.TXT","w");

if(ft==NULL) {

puts("cannot open target file!"); exit(); }


replace(str,&target);

fputs(target,ft);

}

fclose(fp); fclose(ft);

printf("\nTask completed!"); getch();

}

void replace(char *s, char *s1) {

int i=0,j=0,k=0; char temp[100],temp2[100],main[100],*t=temp,*m=main;


while(*s!='\0') {

*t=*s;

t++; s++;

Solution:



}

*t='\0';



temp2[j]=temp[i];

if(temp[i]==' ') {

temp2[j]='\0';

if(strcmpi(temp2,"the")==0) {

strcpy(temp2," "); }

else if(strcmpi(temp2,"an")==0) {

strcpy(temp2," ");

}

else if(strcmpi(temp2,"a")==0) {

strcpy(temp2," ");

}

j=0;


main[k]=temp2[j];

k++; j++;

}



j=-1;

}

i++; j++; }


if(strcmpi(temp2,"the")==0){ /* checking last word too */









else {

j=0;


main[k]=temp2[j];

k++; j++; }


}

while(*m!='\0') {

*s1=*m;

s1++; m++;

}

*s1='\0';

}

-----------------------------------------------------------------------------------------------------------------

(p) You are given a data file EMPLOYEE.DAT with the following record structure:struct employee {int empno ;char name[30] ;int basic, grade ;} ;Every employee has a unique empno and there are supposed to be no gaps between employee numbers. Records are entered into the data file in ascending order of employee number, empno. It is intended to check whether there are missing employee numbers. Write a program segment to read the data file records sequentially and display the list of missing employee numbers.

NOTE: assign employee numbers in ascending order only.


struct employee {

Solution:



int empno; char name[30]; int basic,grade; }e; FILE *fp; int num=0; long recsize; char another,ch; clrscr();

fp=fopen("EMPLOYEE.DAT","rb+"); if(fp==NULL) { fp=fopen("EMPLOYEE.DAT","wb+");

if(fp==NULL) exit(); }

recsize=sizeof(e);

while(1) {

clrscr();

printf("\t\tEmployee number database:\n"); printf("\t\t*************************\n\n"); printf("\n\t1: Add employee information:\n"); printf("\n\t2: List employee information:\n"); printf("\n\t3: Check missing employee numbers:\n"); printf("\n\t0: Exit:\n\n");

gotoxy(2,24); printf("your choice: "); fflush(stdin); ch=getche();

switch(ch) {

case '1':

clrscr();


another='y';


printf("\t\tAdd employee information:\n"); printf("\t\t*************************\n\n");

printf("Note: employee numbers should be given in ascending order\n\n");

printf("\nEnter employee number: "); scanf("%d",&e.empno); printf("\n\nEnter employee name: "); scanf("%s",e.name); printf("\n\nEnter employee basic salary: "); scanf("%d",&e.basic); printf("\n\nEnter employee grade(1/2/3): "); scanf("%d",&e.grade);


gotoxy(2,24); printf("Add another employee information(Y/N): "); fflush(stdin); another=getche();

clrscr();



}

break;

case '2':

clrscr();

printf("\t\tList employee information:\n"); printf("\t\t**************************\n\n\n");

rewind(fp); while(fread(&e,recsize,1,fp)==1) {

printf("\n%3d\t%8s\t%5d\t%d\n",e.empno,e.name,e.basic,e.grade);

}

printf("\n\npress any key to return...\n"); getch();

break;

case '3':

clrscr();

printf("\t\tMissing employee numbers:\n"); printf("\t\t*************************\n\n");

rewind(fp); while(fread(&e,recsize,1,fp)==1) { num=e.empno; /* assigning the value of first employee number */ break; }

rewind(fp); /* again rewinding the file to read from beginning */


if(num!=e.empno) { /* if assigned number is smaller than employee number we will print all the number between them */ while(num<e.empno) {

printf("%4d ",num);

num++; } num=e.empno+1;

}

/* we will assign greater value than employee number to make sure that both don't match until another greater employee number is found */

else

num=e.empno+1;

}

printf("\n\n press any key to return..."); getch();

break;

case '0':

fclose(fp);



exit();

} } }

-----------------------------------------------------------------------------------------------------------------

(q) Write a program to carry out the following:− To read a text file “TRIAL.TXT” consisting of a maximum of 50 lines of text, each line with a maximum of 80 characters.− Count and display the number of words contained in the file.− Display the total number of four letter words in the text file.Assume that the end of a word may be a space, comma or a full-stop followed by one or more spaces or a newline character.

NOTE: Make a file "TRIAL.TXT" in bin directory and write something in it ,then run the program.


FILE *fp; char s[80]; int twd,fwd,tw=0,fw=0;void word(); clrscr();

fp=fopen("TRIAL.TXT","r"); if(fp==NULL) {

exit(); }

while(fgets(s,79,fp)!=NULL) { word(s,&twd,&fwd); tw=tw+twd; fw=fw+fwd; }

fclose(fp);

printf("\nTotal number of words in text file = %d\n",tw); printf("\n\nTotal number of 4 letter words = %d\n",fw);

getch(); }

void word(char ss[80],int *tw, int *fw) {

int i=0,tot_wd=0,tot_4_wd=0;

while(ss[i]!='\0') {

/************************/ /* to cound total words */ /************************/

if(ss[i]>=65 && ss[i]<=90 || ss[i]>=97 && ss[i]<=122) {

Solution:



if(ss[i+1]==' ' || ss[i+1]=='.' || ss[i+1]==',' || ss[i+1]=='\n' ) {

tot_wd++; } }

/*********************************/ /* to count total 4 letter words */ /*********************************/

if(ss[i]==' ' || ss[i]==',' || ss[i]=='.') {

if(ss[i+1]>=65 && ss[i+1]<=90 || ss[i+1]>=97 && ss[i+1]<=122) {

if(ss[i+2]>=65 && ss[i+2]<=90 || ss[i+2]>=97 && ss[i+2]<=122) {

if(ss[i+3]>=65 && ss[i+3]<=90 || ss[i+3]>=97 && ss[i+3]<=122) {

if(ss[i+4]>=65 && ss[i+4]<=90 || ss[i+4]>=97 && ss[i+4]<=122) {

if(ss[i+5]==' ' || ss[i+5]==',' || ss[i+5]=='.' || ss[i+5]=='\n') {

tot_4_wd++; } } } } } }

if(ss[i]>=65 && ss[i]<=90 || ss[i]>=97 && ss[i]<=122) {

if(ss[i+1]>=65 && ss[i+1]<=90 || ss[i+1]>=97 && ss[i+1]<=122) {

if(ss[i+2]>=65 && ss[i+2]<=90 || ss[i+2]>=97 && ss[i+2]<=122) {

if(ss[i+3]>=65 && ss[i+3]<=90 || ss[i+3]>=97 && ss[i+3]<=122) {


} } } } }

i++;

}

*tw=tot_wd; *fw=tot_4_wd;

}

-----------------------------------------------------------------------------------------------------------------

(r) Write a program to read a list of words, sort the words in alphabetical order and display them one word per line. Also give the total number of words in the list. Output format should be:Total Number of words in the list is _______Alphabetical listing of words is:-----------



Assume the end of the list is indicated by ZZZZZZ and there are maximum in 25 words in the Text file.

NOTE: Make a file as "trial.txt" in bin directory and write some words in it. Words should be in the form of a list. One below another,then run the program.


/* make a list of words, every words should be written under previous one to make sure it's a list of words. */

/* upper case letters will be arranged before small case letter, so it is recommended to use the list of either Capital letter or small case letters but not both together. */struct word { char wrd [30]; };

void main() {

struct word w[N];

FILE *fp;char s1[30];int i=0,count=0;void srt_wrd(); /* function to sort and print list */

clrscr();

fp=fopen("TRIAL.TXT","rb+");

while(fgets(s1,30,fp)!=NULL) {strcpy(w[i].wrd,s1); /* copying each word in array of structure */i++;count++; /* count of all the words */}

printf("Total words if file = %3d\n",count);

printf("\t\tList in alphabetical order:\n");srt_wrd(&w,count); /* function for sorting and printing list */

fclose(fp);getch();

}

void srt_wrd( struct word *w, int n) {

int i,j,k=0;struct word temp;

/***************************************//* sorting words in alphabetical order *//***************************************/

for(i=0;i<=n;i++) {


/* testing the first alphabets of two words */

Solution:



if(w[i].wrd[0] > w[j].wrd[0]) {

temp=w[i];w[i]=w[j];w[j]=temp;}

/* testing the first two alphabets of two words */

else if(w[i].wrd[0]==w[j].wrd[0] && w[i].wrd[1] > w[j].wrd[1]) {


/* testing first three alphabets of two words */

else if(w[i].wrd[0]==w[j].wrd[0] && w[i].wrd[1]==w[j].wrd[1]) {

if(w[i].wrd[2]>w[j].wrd[2]) {

temp=w[i];w[i]=w[j];w[j]=temp;}}

/* testing first four alphabets of two words */


if(w[i].wrd[2]==w[j].wrd[2] && w[i].wrd[3]>w[j].wrd[2]) {


}}

/*****************************//* printing the sorted words *//*****************************/

for(i=0;i<n;i++) {

printf("%-s\n",w[i].wrd);

k++;

if(k==10) {

printf("\n\npress any key to go to next page...");getch();

k=1;clrscr();}}

}

--------------------------------------------------------------------------------------------------------------



(s) Write a program to carry out the following:(a) Read a text file ‘INPUT.TXT’(b) Print each word in reverse orderExample,Input: INDIA IS MY COUNTRYOutput: AIDNI SI YM YRTNUOCAssume that each word length is maximum of 10 characters and each word is separated by newline/blank characters.

NOTE: Make a file "INPUT.TXT" in bin directory and write something in it, then run the program.


void main() {

FILE *fs;char s[80];void rev();

clrscr();

fs=fopen("INPUT.TXT","r");

if(fs==NULL) {

printf("cannot open file!");exit();}


fclose(fs);

getch();

}



while(s1[i]!='\0') {

s2[j]=s1[i];

if(s1[i]==' ' || s1[i]=='\0') {

s2[j]='\0';

strrev(s2);

printf("%s ",s2);

j=-1;

}

i++;j++;

Solution:



}

s2[j]='\0';


}

-----------------------------------------------------------------------------------------------------------------

(s) Write a C program to read a large text file ‘NOTES.TXT’ and print it on the printer in cut-sheets, introducing page breaks at the end of every 50 lines and a pause message on the screen at the end of every page for the user to change the paper.

NOTE: Make a file as "NOTES.TXT" in bin directory and write/paste some text in it, then run the program.


void print();

void main() {

FILE *fp;

char s[80]; int x=4,y=4,c=0,pg=0; clrscr();

fp=fopen("NOTES.TXT","r");

if(fp==NULL) {

puts("cannot open file!"); exit(); }

while(fgets(s,74,fp)!=NULL) {

gotoxy(30,1); /* printing page number */ printf("Page No: %3d",pg);

print(s,x,y,c); /* function to print */

c++; y++;

if(c>51) { /* checking for page end */

pg++;

c=0; gotoxy(2,24); printf("press any key to change paper..."); getch();

clrscr();

Solution:



}

if(y>22) { /* checking for total lines */

gotoxy(2,24); printf("press any key to go to next page..."); getch(); y=5; clrscr(); }

}

fclose(fp); }

void print(char *s,int x, int y, int c) {

/* page border */

int i,bdr,bdr2;

gotoxy(1,2); printf("%c",218); for(bdr=3;bdr<23;bdr++) {

gotoxy(1,bdr); printf("%c",179);


}

gotoxy(79,2); printf("%c",191);



for(bdr2=2;bdr2<=78;bdr2++) {

gotoxy(bdr2,2); printf("%c",196);

}

gotoxy(x,y);

puts(s);

if(c>50) {

for(i=2;i<79;i+=2) {

gotoxy(i,23);

printf("-"); } } }

____________________________________________________








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Let us C solutions ( By Chetan ): Let Us C / Chapter 1 (Getting started)

http://letuscalllessons.blogspot.com/2014/01/chapter-1-getting-started.html[07-Apr-14 8:50:41 PM]





(a) Ramesh’s basic salary is input through the keyboard. His dearness allowance is 40% of basic salary, and house rent allowance is 20% of basic salary. Write a program to calculate his gross salary.

Solution:

(b) The distance between two cities (in km.) is input through the keyboard. Write a program to convert and print this distance in meters, feet, inches and centimeters.

#include<stdio.h>#include<conio.h>main() {long km,m,ft,inc,cm;clrscr();printf("Please enter the distance(in km):\n");scanf("%ld",&km);m=km*1000;ft=m*3;inc=ft*12;cm=ft*30;printf("\nDistance in kilometer = %ld\n",km);printf("\nDistance in meter = %ld\n",m);printf("\nDistance in feet = %ld\n",ft);printf("\nDistance in inches = %ld\n",inc);printf("\nDistance in centimeters = %ld\n",cm);getch();

Exercise [H]

#include<stdio.h>#include<conio.h>main() {long bs,da,hra,gs;clrscr();printf("Please enter Ramesh's Basic Salary: \n");scanf("%ld",&bs);gs=(40*bs/100)+(20*bs/100)+bs;printf("Ramesh's GROSS SALARY = %ld\n",gs);getch();return 0;} ------------------------------------------------------------------------------------------------------------

Solution:

Chetan Raikwar


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return 0;}

------------------------------------------------------------------------------------------------------------

(c) If the marks obtained by a student in five different subjects are input through the keyboard, find out the aggregate marks and percentage marks obtained by the student. Assume that the maximum marks that can be obtained by a student in each subject is 100.

------------------------------------------------------------------------------------------------------------

(d) Temperature of a city in Fahrenheit degrees is input through the keyboard. Write a program to convert this temperature into Centigrade degrees.

#include<stdio.h>#include<conio.h>main() {float fh,cn;clrscr();printf("Please enter the temperature in fahrenheit:\n");scanf("%f",&fh);cn=5*(fh-32)/9;printf("Centigrade Temperature = %f \n",cn);getch();return 0;}

------------------------------------------------------------------------------------------------------------

(e) The length & breadth of a rectangle and radius of a circle are input through the keyboard. Write a program to calculate the area & perimeter of the rectangle, and the area & circumference of the circle.

Solution:

#include<stdio.h>#include<conio.h>main() {int m1,m2,m3,m4,m5,ttl;float prcnt;clrscr();printf("Please enter the marks of the student:\n");scanf("%d%d%d%d%d",&m1,&m2,&m3,&m4,&m5);ttl=m1+m2+m3+m4+m5;prcnt=ttl/5;printf("AGGREGATE MARKS = %d\nPERCENTAGE MARKS = %f%\n",ttl,prcnt);getch();return 0;}

Solution:

Solution:

#include<stdio.h>#include<conio.h>main() {float l,b,area,peri,ca,cr,r;clrscr();printf("Please enter the length of rectangle:");scanf("%f",&l);printf("\nPlease enter the breadth of rectangle:");scanf("%f",&b);area=l*b;peri=2*(l+b);



(f) Two numbers are input through the keyboard into two locations C and D. Write a program to interchange the contents of C and D.

#include<stdio.h>#include<conio.h>main() {int c,d,t;clrscr();printf("Enter value of C: ");scanf("%d",&c);printf("\n\nEnter value of D: ");scanf("%d",&d);t=c;c=d;d=t;printf("\n\nC = %d\nD = %d\n",c,d);getch();}

------------------------------------------------------------------------------------------------------------

(g) If a five-digit number is input through the keyboard, write a program to calculate the sum of its digits.(Hint: Use the modulus operator ‘%’)

#include<stdio.h>#include<conio.h>main() {long i,d1,d2,d3,d4,d5,a,b,c,d,sum;clrscr();printf("Please enter the five digit number: \n");scanf("%ld",&i);d1=i/10000;a=i%10000;d2=a/1000;b=a%1000;d3=b/100;c=b%100;d4=c/10;d=c%10;d5=d/1;sum=d1+d2+d3+d4+d5;printf("Sum of the digits is %ld\n",sum);getch();return 0;}

------------------------------------------------------------------------------------------------------------

(h) If a five-digit number is input through the keyboard, write a program to reverse the number.

printf("\n\nAREA = %f square cm.\t PERIMETER = %f cm.\n",area,peri);printf("\n\nPlease enter the radius of the circle:");scanf("%f",&r);ca=3.14*r*r;cr=2*3.14*r;printf("\n\nAREA = %f square cm.\t CIRCUMFERENCE = %f cm.\n",ca,cr);getch();return 0;}

------------------------------------------------------------------------------------------------------------

Solution:

Solution:



------------------------------------------------------------------------------------------------------------

(i) If a four-digit number is input through the keyboard, write a program to obtain the sum of the first and last digit of this number.

------------------------------------------------------------------------------------------------------------

(j) In a town, the percentage of men is 52. The percentage of total literacy is 48. If total percentage of literate men is 35 of the total population, write a program to find the total number of illiterate men and women if the population of the town is 80,000.

#include<stdio.h>#include<conio.h>main() {int i,d1,d4,a,b,c,sum;clrscr();printf("Please enter the four digit number: \n");scanf("%d",&i);d1=i/1000;

Solution:

#include<stdio.h>#include<conio.h>main() {long i,d1,d2,d3,d4,d5,a,b,c,d,n_num;clrscr();printf("Please enter the five digit number: \n");scanf("%ld",&i);d1=i/10000;a=i%10000;d2=a/1000;b=a%1000;d3=b/100;c=b%100;d4=c/10;d=c%10;d5=d/1;n_num=(d5*10000)+(d4*1000)+(d3*100)+(d2*10)+(d1*1);printf("reversed number = %ld \n",n_num);getch();return 0;}

Solution:

#include<stdio.h>#include<conio.h>main() {int i,d1,d4,a,b,c,sum;clrscr();printf("Please enter the four digit number: \n");scanf("%d",&i);d1=i/1000;a=i%1000;b=a%100;c=b%10;d4=c/1;sum=d1+d4;

printf("The sum of first and last digit is %d\n",sum);getch();return 0;}

Solution:



a=i%1000;b=a%100;c=b%10;d4=c/1;sum=d1+d4;printf("The sum of first and last digit is %d\n",sum);getch();return 0;}

------------------------------------------------------------------------------------------------------------

(k) A cashier has currency notes of denominations 10, 50 and 100. If the amount to be withdrawn is input through the keyboard in hundreds, find the total number of currency notes of each denomination the cashier will have to give to the withdrawer.

#include<stdio.h>#include<conio.h>main() {int a,t_10,f_50,h_100,b,c;clrscr();printf("Please enter the amount to be withdrawn: \n");scanf("%d",&a);h_100=a/100;b=a%100;f_50=b/50;c=b%50;t_10=c/10;

printf("\nCurrency notes of 100 should be = %d\n",h_100);printf("\nCurrency notes of 50 should be = %d\n",f_50);printf("\nCurrency notes of 10 should be = %d\n",t_10);getch();return 0;}

------------------------------------------------------------------------------------------------------------

(l) If the total selling price of 15 items and the total profit earned on them is input through the keyboard, write a program to find the cost price of one item.

#include<stdio.h>#include<conio.h>main() {int sp,p,cp;clrscr();

printf("Please enter the selling price of 15 items: \n");scanf("%d",&sp);printf("\nPlease enter the total profit of 15 items: \n");scanf("%d",&p);cp=(sp-p)/15;printf("Cost price of one item is %d\n",cp);getch();return 0;}

------------------------------------------------------------------------------------------------------------

(m) If a five-digit number is input through the keyboard, write a program

Solution:

Solution:






to print a new number by adding one to each of its digits. For example if the number that is input is 12391 then the output should be displayed as 23502.

#include<stdio.h>#include<conio.h>main() {long i,add;clrscr();printf("Please enter a five digit number: \n");scanf("%ld",&i);add=i+11111;printf("addition = %ld\n",add);getch();return 0;}

______________________________________________________________________

Solution:



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Let us C solutions ( By Chetan ): Let Us C / Chapter 3 (The Loop Control Structure)

http://letuscalllessons.blogspot.com/2014/01/chapter-3-loop-control-structure.html[07-Apr-14 8:51:29 PM]





(a) Write a program to calculate overtime pay of 10 employees. Overtime is paid at the rate of Rs. 12.00 per hour for every hour worked above 40 hours. Assume that employees do not work for fractional part of an hour.


main() {

int wrkh,hr,emp,pay,r=12;clrscr();

printf("Please enter working hours of 10 employees: \n\n");

emp=1;

while(emp<=10) {

printf("\n\nEnter working hours of employee: ",emp);

scanf("%d",&wrkh);

if(wrkh>40) {

hr=wrkh-40;pay=hr*r;

printf("\n%d\tOver-time pay %d rs\n",emp,pay);

}

else {

printf("\n%d\tNo over time pay\n",emp);

}

emp++;

}

Exercise [B]

Solution:

Chetan Raikwar


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getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(b) Write a program to find the factorial value of any number entered through the keyboard.


main() {

long i,j,fact=1;clrscr();

printf("Please enter any number: ");scanf("%ld",&i);

for(j=i;j>=1;j--) {

fact=fact*j;

}

printf("Factorial value = %ld ",fact);

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(c) Two numbers are entered through the keyboard. Write a program to find the value of one number raised to the power of another.


main() {

int num1,num2,i,rslt=1;clrscr();

printf("Enter two numbers: \n");scanf("%d%d",&num1,&num2);

for(i=1;i<=num2;i++) {rslt=rslt*num1;

if(i==num2) {break;}

}

Solution:

Solution:



printf("\n\nvalue of %d raised to the power of %d is = %d\n",num1,num2,rslt);

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(d) Write a program to print all the ASCII values and their equivalent characters using a while loop. The ASCII values vary from 0 to 255.


int i=0;clrscr();

while(i<=255) {

printf(" %d %c \n",i,i);i++;

}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(e) Write a program to print out all Armstrong numbers between 1 and 500. If sum of cubes of each digit of the number is equal to the number itself, then the number is called an Armstrong number. For example, 153 = ( 1 * 1 * 1 ) + ( 5 * 5 * 5 ) + ( 3 * 3 * 3 )


main() {

int i,a,b,c,d,e;clrscr();

printf("All Armstrong Numbers between 1-500 are: \n\n");

for(i=1;i<=500;i++) {

if(i<9) { /* for one digit numbers */a=i/1;

if((a*a*a)==(a*1)) {printf(" %d ",i);

} }

if(i>9 && i<100) { /* for two digit numbers */

a=i/10;b=i%10;

Solution:

Solution:



c=b/1;

if(((a*a*a)+(c*c*c))==((a*10)+(c*1))) {printf(" %d ",i);

}}

if(i>99) { /* for three digit numbers */

a=i/100;b=i%100;c=b/10;d=b%10;e=d/1;

if(((a*a*a)+(c*c*c)+(e*e*e))==((a*100)+(c*10)+(e*1))){

printf(" %d ",i);

} }

}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(f) Write a program for a matchstick game being played between the computer and a user. Your program should ensure that the computer always wins. Rules for the game are as follows:− There are 21 matchsticks.− The computer asks the player to pick 1, 2, 3, or 4 matchsticks.− After the person picks, the computer does its picking.− Whoever is forced to pick up the last matchstick loses the game.


main() {

int tms=21,lms=0,user,cpu,pms=0;char another;

clrscr();

do {

printf("\nPick up 1,2,3 or 4 matchsticks: ");scanf("%d",&user); /* Matchsticks picked by you */

cpu=user; /* Matchsticks picked by computer */

if(pms<20) {printf("\nyou picked = %d\tcomputer picked = %d\n",user,cpu);}else {printf("\nyou picked =%d\n",user);}

pms=pms+(cpu+user); /* Total picked Matchsticks */

Solution:



lms=tms-pms; /* Total left matchsticks */

if(pms<=21) {printf("\nleft matchsticks = %d\n",lms);}else {break;}

printf("\nPick again Y/N: ");scanf(" %c",&another);

}while(another=='y');

printf("\n\n\n");printf("\n==============\n");printf("\nComputer wins!\n");printf("\n==============\n");

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(g) Write a program to enter the numbers till the user wants and at the end it should display the count of positive, negative and zeros entered.


main() {

int num,tnum=0,tp=0,tn=0,tz=0;char another;

clrscr();

do {printf("\nEnter any number: ");scanf("%d",&num);

tnum=tnum+1; /* count of total numbers entered by you */

if(num==0) /* count of total zeros */tz=tz+1;

if(num>0) /* count of total positive numbers */tp=tp+1;

if(num<0) /* count of total negative number */tn=tn+1;

printf("\nEnter another number Y/N: ");scanf(" %c",&another);


clrscr();

printf("\n\n\nTotal numbers entered = %d\n",tnum);

Solution:



printf("\nTotal zeros = %d\n",tz);printf("\nTotal positive numbers = %d\n",tp);printf("\nTotal negative numbers = %d\n",tn);

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(h) Write a program to find the octal equivalent of the entered number.


long num,n;clrscr();

printf("Please enter the number: ");scanf("%ld",&num);printf("\n\nOctal equivalent: ");

while(num!=0) {

n=num%8;

printf("%ld",n);

num=num/8; /* devision by 8 */

}

gotoxy(20,6);printf("<%c%c%c%c%c read from right to left",196,196,196,196,196);

getch();return 0;}

------------------------------------------------------------------------------------------------------------

(i) Write a program to find the range of a set of numbers. Range is the difference between the smallest and biggest number in the list


int min,max,num,range,flag=0;char ch;

clrscr();

do {

printf("\nEnter number: ");scanf("%d",&num);

if(flag==0) {

min=num;max=num;}

Solution:

Solution:



flag++; /* counting total numbers */

if(min>num) {min=num;}

if(max<num)max=num;

printf("\n\nEnter another number(Y/N): ");scanf(" %c",&ch);

clrscr();}while(ch=='Y' || ch=='y');

range=max-min;

printf("\nTotal number = %d\n",flag);printf("\nLargest number = %d\n",max);printf("\nSmallest number = %d\n",min);printf("\nRange = %d\n",range);

getch();return 0;}______________________________________________________________________

(a) Write a program to print all prime numbers from 1 to 300. (Hint: Use nested loops, break and continue)


int i=1,j;

clrscr();

printf("\nPrime numbers between 1 to 300 are:\n\n");

while(i!=300) {

j=2;

while(j<i) {

if(i%j==0) {break;}

j++;

}

if(j==i) {

printf(" %d",i);

}

i++;}

Exercise [E]

Solution:



getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(b) Write a program to fill the entire screen with a smiling face. The smiling face has an ASCII value 1.


int i;clrscr();

for(i=1;i<=10000;i++) {printf("%c",1);}

getch();return 0;

------------------------------------------------------------------------------------------------------------

(c) Write a program to add first seven terms of the following series using a for loop:

1/1! + 2/2! + 3/3! + .............


main() {

float i,j,k,fact=1,factsum,numsum;clrscr();

printf("\tTerm\t\tFactorial\n\n");

for(i=1;i<=7;i++) {

for(j=i;j>=2;j--) {

fact=fact*j;factsum=factsum+fact;break;}numsum=numsum+i;printf("\t%.0f\t\t%.0f\n",i,fact);}

k=numsum/factsum;

printf("\n\nsum of 7 terms = %.0f ",numsum);printf("\nsum of all factorials = %.0f",factsum);printf("\n\naddition of all terms / sum of factorials: %f",k);

getch();return 0;

Solution:

Solution:



}

------------------------------------------------------------------------------------------------------------

(d) Write a program to generate all combinations of 1, 2 and 3 using for loop.


int i,j,k;clrscr();

printf("All combinations of 1,2,3 are:\n\n");for(i=1;i<=3;i++) {for(j=1;j<=3;j++) {for(k=1;k<=3;k++) {printf(" %d%d%d ",i,j,k);} } }

getch();return 0;}

------------------------------------------------------------------------------------------------------------

(e) According to a study, the approximate level of intelligence of a person can be calculated using the following formula:i = 2 + ( y + 0.5 x )Write a program, which will produce a table of values of i, y and x, where y varies from 1 to 6, and, for each value of y, x varies from 5.5 to 12.5 in steps of 0.5.


float i=1,x,y;clrscr();

for(y=1.0;y<=6.0;y++) {

for(x=5.5;x<=12.5;x+=0.5) {

i=2+(y+(0.5*0.5));

printf("%.2f\t%.2f\t%.2f\n",i,y,x);

}

}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

Solution:

Solution:



(f) Write a program to produce the following output:

A B C D E F G F E D C B AA B C D E F F E D C B AA B C D E E D C B AA B C D D C B AA B C C B AA B B AA A


int i,j,k,l;clrscr();

for(i=71;i>=65;i--) {

/* loop for printing ascending letters */

for(j=65;j<=i;j++) {

printf("%c ",j);

}

/* loop for making a space between patterns */

for(k=i+1;k<=71;k++) {

if(k==71)printf(" ");

if(k<71)printf(" ");}

/* loop to print descending letters */

for(l=i;l>=65;l--) {

if(l==71) { /* to skip printing 'G' twice */continue;}

printf("%c ",l);

}printf("\n");}

getch();

return 0;}

------------------------------------------------------------------------------------------------------------

(g) Write a program to fill the entire screen with diamond and heart alternatively. The ASCII value for heart is 3 and that of diamond is 4.

Solution:

Solution:




int i,j;clrscr();

for(i=j=1;i<=10000,j<=10000;i++,j++) {printf("%c%c",4,3);}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(h) Write a program to print the multiplication table of the number entered by the user. The table should get displayed in the following form.29 * 1 = 2929 * 2 = 58…


int i,j,k;clrscr();

printf("Please enter the number:\n");scanf("%d",&i);

for(j=1;j<=10;j++) {k=i*j;printf("\n%d*%d = %d\n",i,j,k);}

getch();return 0;

}

------------------------------------------------------------------------------------------------------------

(i) Write a program to produce the following output: 1 2 3 4 5 67 8 9 10


int i,num=0,k,j;clrscr();

Solution:

Solution:



for(i=1;i<=4;i++) {

for(j=4;j>=i;j--) { /* loop for making a space from left corner */

printf(" ");

}

for(k=1;k<=i;k++) {

num=num+1; /* incrementing external number for printing */

printf(" %d ",num);

}

printf("\n\n");}

getch();

}

-----------------------------------------------------------------------------------------------------------

(j) Write a program to produce the following output:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1


int i,j,k;clrscr();

printf(" 1\n\n\n");

for(i=1;i<=4;i++) {

for(k=5;k>=i;k--) {printf(" ");}printf("1 ");

for(j=1;j<=i;j++) {

if(i==4 && j==3) {printf("%d ",i+j-1);continue;}

if(j>1)printf("%d ",i);

}

printf("1\n\n\n");

}

Solution:



getch();return 0;

}

-----------------------------------------------------------------------------------------------------------

(k) A machine is purchased which will produce earning of Rs. 1000 per year while it lasts. The machine costs Rs. 6000 and will have a salvage of Rs. 2000 when it is condemned. If 12 percent per annum can be earned on alternate investments what would be the minimum life of the machine to make it a more attractive investment compared to alternative investment?

#include <stdio.h>#include <conio.h>int main(){int year=0,inv,altn;

while(altn>inv) {

year+=1;altn=120*year;inv=(1000*year)-(6000-2000);

}

printf("Minimum life of the machine would be = %d",year); getch(); return 0;}------------------------------------------------------------------------------------------------------------

(l) When interest compounds q times per year at an annual rate of r % for n years, the principle p compounds to an amount a as per the following formulaa = p ( 1 + r / q ) nqWrite a program to read 10 sets of p, r, n & q and calculate the corresponding as.


int i,p,r,n,q,j,pw;float a;

clrscr();

for(i=0;i<10;i++) {

printf("Set number %d:\n\n\n\n",i+1);

printf("Enter p: ");scanf("%d",&p);

printf("\n\nEnter r: ");scanf("%d",&r);

printf("\n\nEnter n: ");scanf("%d",&n);

Solution:

Solution:




printf("\n\nEnter q: ");scanf("%d",&q);

pw=n*q;

for(j=0;j<pw;j++) {

a = a + p *(1+r/1);

}printf("\n\n\nA = %f\n",a);

printf("\n\n\n\npress any key to proceed...");getch();

clrscr();}

getch();return 0;}

------------------------------------------------------------------------------------------------------------

(m) The natural logarithm can be approximated by the following series.x-1/x + 1/2(x-1/x)2 + 1/2(x-1/x)3 + 1/2(x-1/x)4 + --------If x is input through the keyboard, write a program to calculate the sum of first seven terms of this series.


float x,nl,x_1,t1;int i;

clrscr();

printf("\n\nEnter the value of X: ");scanf("%f",&x);

t1 = (x-1)/x; /* fist term */

for(i=2;i<=7;i++) {

nl = (1.0/2.0) * pow((x_1),i); /* terms from 2 to 7 */

}

nl = nl + t1; /* adding all the terms */

printf("\n\nApproximated natural log = %f\n",nl);

getch();return 0;}___________________________________________________________________

Solution:














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Let us C solutions ( By Chetan ): Let Us C / Chapter 4 (The Case Control Structure)

http://letuscalllessons.blogspot.com/2014/01/chapter-4-case-control-structure.html[07-Apr-14 8:52:22 PM]





Write a menu driven program which has following options:

1. Factorial of a number.2. Prime or not3. Odd or even4. Exit


int num,i,j=0,k=0,choice,fact=1;clrscr();

printf("Please enter any number: ");scanf("%d",&num);

while(1){printf("\n\n1. Factorial");printf("\n2. Prime");printf("\n3. Odd/Even");printf("\n4. Exit");printf("\n\nPlease enter your choice: ");scanf("%d",&choice);

switch(choice){

case 1:

for(i=num;i>=1;i--) { fact=fact*i; } printf("\nFactorial = %d ",fact); break;

case 2:

for(i=2;i<num;i++) { j=num%i; if(j==0){ k=1;

Exercise [C]

Solution:

Chetan Raikwar


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break; } } if(k==0) { printf("\nPrime Number"); } else { printf("\nNot a Prime Number"); } break;

case 3:

if((num%2)==0) printf("\nEven Number"); else printf("\nOdd Number"); break;

case 4:

exit();

} }

}

------------------------------------------------------------------------------------------------------------

Write a program which to find the grace marks for a student using switch. The user should enter the class obtained by the student and the number of subjects he has failed in.

− If the student gets first class and the number of subjects he failed in is greater than 3, then he does not get any grace. If the number of subjects he failed in is less than or equal to 3 then the grace is of 5 marks per subject.

− If the student gets second class and the number of subjects he failed in is greater than 2, then he does not get any grace. If the number of subjects he failed in is less than or equal to 2 then the grace is of 4 marks per subject.

− If the student gets third class and the number of subjects he failed in is greater than 1, then he does not get any grace. If the number of subjects he failed in is equal to 1 then the grace is of 5 marks per subject


int _class,f_sub;clrscr();

printf("\nPlease enter the class obtained\n1=first 2=second 3=third: ");scanf("%d",&_class);printf("\n\nPlease enter the number of failed subjects: ");

Exercise [D]

Solution:




scanf("%d",&f_sub);

switch(_class) {

case 1: if(f_sub<=3) { printf("\nGrace marks = 5 marks per subject.\n"); } else { printf("\nNo Grace marks.\n"); } break;


case 3: if(f_sub==1) { printf("\nGrace marks = 5 marks per subject.\n"); } else { printf("\nNo Grace marks.\n"); } break;

default: printf("Error! wrong input.\n"); break; }

getch(); return 0;

}_____________________________________________________________________



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Let us C solutions ( By Chetan ): Let Us C / Chapter 5 (Functions & Pointers)

http://letuscalllessons.blogspot.com/2014/01/chapter-5-functions-pointers.html[07-Apr-14 8:53:09 PM]





(a) Write a function to calculate the factorial value of any integer entered through the keyboard.


void main() {

int num;void func();

clrscr();


func(num);

getch();

}

void func(int n) {

int fact=1;

for(;n>=1;n--) {

fact=fact*n;

}

printf("\n\nFactorial value = %d \n",fact);

}

------------------------------------------------------------------------------------------------------------

(b) Write a function power ( a, b ), to calculate the value of a raised to

Exercise [D]

Solution:

Chetan Raikwar


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b.


void main() {

int num1,num2 ;clrscr();

printf("Please enter the value of a: ");scanf("%d",&num1);

printf("\n\nPlease enter the value of b: ");scanf("%d",&num2);

power(num1,num2);

getch();

}

power(int a , int b) {

int c=1,i;

for(i=1;i<=b;i++) {c=c*a;

if(i==b) {break;}

}

printf("\n\n\nValue of %d raised to the power of %d is = %d\n",a,b,c);

return 0; }

------------------------------------------------------------------------------------------------------------

(c) Write a general-purpose function to convert any given year into its roman equivalent. The following table shows the roman equivalents of decimal numbers:

Decimal Roman

1 i5 v10 x50 l100 c500 d1000 m

Example:Roman equivalent of 1988 is mdcccclxxxviiiRoman equivalent of 1525 is mdxxv

Solution:




void main() {

int yr;void func();

clrscr();


printf("\n\nRoman Equivalent = ");func(yr);

getch();

}void func(int y) {

int d1,d2,d3,d4,d5,d6,d7,a,b,c,d,e,f;

char thsnd='m',hndr_5='d',hndr='c',ffty='l',tn='x',fv='v',one='i';

/******* Roman Convertion ********/

/* To find all thousands */

d1=y/1000;a=y%1000;

for(;d1>=1;d1--) {printf("%c",thsnd);

if(a==0)break;}

/* To find all five-hundreds */

d2=a/500;b=a%500;

for(;d2>=1;d2--) {printf("%c",hndr_5);

if(b==0)break;}

/* To find all hundreds */

d3=b/100;c=b%100;

for(;d3>=1;d3--) {printf("%c",hndr);

if(c==0)break;}

/* To find all fifties *//***********************/

Solution:



d4=c/50;d=c%50;

for(;d4>=1;d4--) {printf("%c",ffty);

if(d==0)break;}

/* To find all tens *//********************/

d5=d/10;e=d%10;

for(;d5>=1;d5--) {printf("%c",tn);

if(e==0)break;

}

/* To find all fives */

d6=e/5;f=e%5;

for(;d6>=1;d6--) {printf("%c",fv);

if(f==0)break;}

/* To find all ones */

for(d7=f;d7>=1;d7--) {printf("%c",one);

}

}

------------------------------------------------------------------------------------------------------------

(d) Any year is entered through the keyboard. Write a function to determine whether the year is a leap year or not.


int yr;void func();clrscr();


func(yr);

Solution:



getch();

}void func(int y) {

if((y%4)==0)

printf("\nThis is a LEAP YEAR.\n");

else

printf("\nThis is NOT A LEAP YEAR.\n");

}

------------------------------------------------------------------------------------------------------------

(e) A positive integer is entered through the keyboard. Write a function to obtain the prime factors of this number.

For example, prime factors of 24 are 2, 2, 2 and 3, whereas prime factors of 35 are 5 and 7.


void main() {

int i,j,k;clrscr();

printf("enter the number: ");

scanf("%d",&j);

printf("\n\nprime factors:");

for(i=2;i<=j;) {

if(j%i==0) { /* divide only if remainder is supposed to be 0 */

j=j/i;

printf(" %d ",i); /* print the divisor */

}

else /* if divisor cannot divide completely */

i=i+1; /* increase it's value and try again */

}

getch();

}

________________________________________________________________________

(a) Write a function which receives a float and an int from main( ), finds

Solution:

Exercise [F]



the product of these two and returns the product which is printed through main( ).


void main() {

int i;float j,k,product();clrscr();

printf("Please enter an integer number: ");scanf("%d",&i);

printf("\nPlease enter a decimal number: ");scanf("%f",&j);

k=product(&i,&j);

printf("\nProduct = %.1f\n",k);

getch();

}

float product(int *a,float *b) {

float *c;

*c=(*a)*(*b);

return *c;

}

------------------------------------------------------------------------------------------------------------

(b) Write a function that receives 5 integers and returns the sum, average and standard deviation of these numbers. Call this function from main( ) and print the results in main( ).

#include<stdio.h>#include<conio.h>#include<math.h>

void main() {

int d1,d2,d3,d4,d5,i,sum,avg;float sd;

clrscr();

printf("enter five digits: \n\n");scanf("%d%d%d%d%d",&d1,&d2,&d3,&d4,&d5);

func(d1,d2,d3,d4,d5,&sum,&avg,&sd);

printf("\tsum = %d\n\taverage = %d\n\tstandard deviation = %f ",sum,avg,sd);

getch();

Solution:

Solution:



}func(int a, int b, int c, int d, int e, int *s, int *av, float *ssd) {

float temp;

*s=a+b+c+d+e; /* sum of digits */*av=(a+b+c+d+e)/5; /* average */

a=a-(*av);b=b-(*av);c=c-(*av);d=d-(*av);e=e-(*av);

temp=((a*a)+(b*b)+(c*c)+(d*d)+(e*e))/4; /* standard deviation */*ssd=sqrt(temp);

return 0;

}

------------------------------------------------------------------------------------------------------------

(c) Write a function that receives marks received by a student in 3 subjects and returns the average and percentage of these marks. Call this function from main( ) and print the results in main( ).


int s1,s2,s3,*avg,*prcnt;void func();clrscr();

printf("Please enter the marks of 3 subjects: \n");scanf("%d%d%d",&s1,&s2,&s3);

func(s1,s2,s3,&avg,&prcnt);

printf(" Average = %d\nPercentage = %d%\n",avg,prcnt);

getch();

}void func(int a, int b, int c, int *d, int *f) {

*d=(a+b+c)/3;*f=(a+b+c)/3;

}_____________________________________________________________________

a) A 5-digit positive integer is entered through the keyboard, write a function to calculate sum of digits of the 5-digit number:(1) Without using recursion(2) Using recursion

Solution:

Exercise [J]

Solution:




long num,s=0,ch;clrscr();

printf("Enter any number: ");scanf("%ld",&num);

printf("\n\nChoose: 1: obtain sum of digits non-recursively\n\n");printf(" 2: obtain sum of digits recursively\n\n\n");

printf("Your choice: ");ch=getche();

switch(ch) {

case '1':

while(num!=0) {

s=s+(num%10);

num=num/10;

}

printf("\n\n\nsum of digits = %d\n",s);

break;

case '2':

s=sum(num);

printf("\n\n\nSum of digits = %d\n",s);break;}

getch();

}

sum(long n) {

static s=0;

if(n==0)return s;

else {

s=s+n%10;n=sum(n/10);

return n;

}

}

-----------------------------------------------------------------------------------------------------------

(b) A positive integer is entered through the keyboard, write a program to obtain the prime factors of the number. Modify the function suitably to obtain the prime factors recursively.




int d,ch,b=2;clrscr();

printf("Enter the number: ");

scanf("%d",&d);

printf("\n\nObtain prime factors:\n\n");printf("1: non-recursively\n\n");printf("2: recursively\n\n");

printf("your choice: ");scanf("%d",&ch);

switch(ch) {

case 1:

printf("\n\n\nPrime factors: ");while(d!=1) {

if(d%b==0) { /* non recursive method */d=d/b;printf("%d ",b);}elseb++;}break;

case 2:

printf("\n\n\nPrime factors: ");

factors(d);

break;

default:

printf("\n\nwrong input!");

break;}

getch();

}

int factors (int n) {

int b=2;

if(n==1)return 1;

else {

while(n!=1) {

if((n%b)==0) {

Solution:



n=factors(n/b); /* recursive function */

printf("%d ",b);}

elseb++;

}return n;}

}

------------------------------------------------------------------------------------------------------------

(c) Write a recursive function to obtain the first 25 numbers of a Fibonacci sequence. In a Fibonacci sequence the sum of two successive terms gives the third term. Following are the first few terms of the Fibonacci sequence:1 1 2 3 5 8 13 21 34 55 89...


unsigned i,num=25,c=1;clrscr();

for(i=0;i<num;i++) {

printf("%u ",fib(c));

c++;}

getch();}

fib(unsigned n) {

if(n==0)return 0;

if(n==1)return 1;

else

return fib(n-1)+fib(n-2);

}

------------------------------------------------------------------------------------------------------------

(d) A positive integer is entered through the keyboard, write a function to find the binary equivalent of this number using recursion.


int num;

Solution:

Solution:



clrscr();

printf("Enter the number: ");scanf("%d",&num);

printf("\n\n\nBinary equivalent: ");

binary(num);

gotoxy(20,7);printf("<%c%c%c%c%c read right to left",196,196,196,196,196);

getch();

}

binary(int n) {

if(n==0)return 0;

else {

printf("%d",n%2);

n= binary( n/2 ); /* recursive function */

return n;

}

}------------------------------------------------------------------------------------------------------------

(e) Write a recursive function to obtain the running sum of first 25 natural numbers.


int i=25,j;clrscr();

j=recsum(i);

printf("Addition of 25 natural numbers = %d",j);

getch();

}

int recsum(int n) {

if(n==1)return 1;

else

n = n + recsum(n-1); /* recursive addition */

return n;

}

------------------------------------------------------------------------------------------------------------

(f) Write a C function to evaluate the series

Solution:



sin(x) = x - (x3/3!) + ( x5/5!) - (x7/7!) + ........to five significant digits.

------------------------------------------------------------------------------------------------------------

(g) Given three variables x, y, z write a function to circularly shift their values to right. In other words if x = 5, y = 8, z = 10 after circular shift y = 5, z = 8, x =10 after circular shift y = 5, z = 8 and x = 10. Call the function with variables a, b, c to circularly shift values.


int x,y,z;char choice;void func();clrscr();

printf("Please enter values of X,Y,Z\n");scanf("%d",&x);scanf("%d",&y);scanf("%d",&z);

printf("\n\nBefore shift: X=%d Y=%d Z=%d\n",x,y,z);

func(&x,&y,&z); /* Call by reference */

printf("\n\nAfter shift: X=%d Y=%d Z=%d\n\n",x,y,z);

/* Circular Shifting continuously */

do {

printf("\nShift again(Y/N): ");scanf(" %c",&choice);

clrscr();

func(&x,&y,&z);printf("\nAfter another shift: X=%d Y=%d Z=%d\n\n",x,y,z);

}while(choice=='y');

}

void func(int *a,int *b,int *c) {

int d,e,f;

d=*a;e=*b;f=*c;*a=f;*b=d;*c=e;

}

Solution:

Solution:



------------------------------------------------------------------------------------------------------------

(h) Write a function to find the binary equivalent of a given decimal integer and display it.


int num;void binary();

clrscr();

printf("\t\tDecimal Integer to Binary conversion\n\n\n");printf("Enter the number: ");scanf("%d",&num);


binary(num);

gotoxy(20,10);printf("<%c%c%c%c%c",196,196,196,196,196);printf(" read right to left");

getch();}

void binary(int n) {

while(n!=0) {

printf("%d",n%2);

n=n/2;

}

}

------------------------------------------------------------------------------------------------------------

(i) If the lengths of the sides of a triangle are denoted by a, b, and c, then area of triangle is given byrootover ( S * (S-a) * (S-b) * (S-c))where, S = ( a + b + c ) / 2

#include<stdio.h>#include<conio.h>#include<math.h>void main() {

int s1,s2,s3,s;int area;clrscr();

printf("enter 3 sides of triangle: \n\n");scanf("%d%d%d",&s1,&s2,&s3);

s=s1+s2+s3/2;

area=func(s1,s2,s3,s);

Solution:

Solution:



printf("\narea = %d",area);

getch();

}func(int i, int j, int k, int h) {

int ar,area;

ar=sqrt(h*((h-i)*(h-j)*(h-k)));

return (ar);

}

------------------------------------------------------------------------------------------------------------

(j) Write a function to compute the distance between two points and use it to develop another function that will compute the area of the triangle whose vertices are A(x1, y1), B(x2, y2), and C(x3, y3). Use these functions to develop a function which returns a value 1 if the point (x, y) lines inside the triangle ABC, otherwise a value 0.

------------------------------------------------------------------------------------------------------------

(k) Write a function to compute the greatest common divisor given by Euclid’s algorithm, exemplified for J = 1980, K = 1617 as follows:1980 / 1617 = 11980 – 1 * 1617 = 3631617 / 363 = 41617 – 4 * 363 = 165363 / 165 = 2363 – 2 * 165 = 335 / 33 = 5165 – 5 * 33 = 0Thus, the greatest common divisor is 33.


int a,b,r,d1,d2,temp;clrscr();

printf("Enter first number: ");scanf("%d",&a);

printf("\n\nEnter second number: ");scanf("%d",&b);

while(b!=0) {

r=a%b;

a=b;b=r;

Solution:

Solution:






}

d1=a; /* devisor of first number */

temp=a;a=b;b=temp;

while(b!=0) {

r=a%b;a=b;b=r;

}

d2=a; /* devisor of second number */

printf("\n\n\nGreatest common devisor: ");

if(d1==d2) {

printf("%d",d1);

}

getch();

}______________________________________________________________________



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Let us C solutions ( By Chetan ): Let Us C / Chapter 6 (Data Types Revisited)

http://letuscalllessons.blogspot.com/2014/01/chapter-6-data-types-revisited.html[07-Apr-14 8:53:56 PM]





Following program calculates the sum of digits of the number 12345. Go through it and find out why is it necessary to declare the storage class of the variable sum as static.main( ){int a ;a = sumdig ( 12345 ) ;printf ( "\n%d", a ) ;}sumdig ( int num ){static int sum ;int a, b ;a = num % 10 ;b = ( num - a ) / 10 ;sum = sum + a ;if ( b != 0 )sumdig ( b ) ;elsereturn ( sum ) ;}


void main() {

int a;

clrscr();

a=sumdig ( 12345 ) ;

printf("\n %d ",a) ;

getch();

Exercise [D]

Solution:

Chetan Raikwar


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Let us C solutions ( By Chetan ): Let Us C / Chapter 6 (Data Types Revisited)

http://letuscalllessons.blogspot.com/2014/01/chapter-6-data-types-revisited.html[07-Apr-14 8:53:56 PM]




}sumdig(int num){

static int sum;

/*

It is necessary to declare the storage class of sum as static because it is withina recursive funtion. It is necessary to keep the value of 'num' same for all releventrecursions of it. And to obtain running sum of all of them.

*/

int a,b;

a=num%10;b=(num-a)/10;sum=sum+a;

if(b!=0)sumdig(b);

elsereturn(sum);

}______________________________________________________________________



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Let us C solutions ( By Chetan ): Let Us C / Chapter 7 (The C Pre-processor)

http://letuscalllessons.blogspot.com/2014/01/chapter-7-c-pre-processor.html[07-Apr-14 8:54:48 PM]





(a) Write down macro definitions for the following:1. To test whether a character entered is a small case letter or not.2. To test whether a character entered is a upper case letter or not.3. To test whether a character is an alphabet or not. Make use of the macros you defined in (1) and (2) above.4. To obtain the bigger of two numbers.

#include<stdio.h>#include<conio.h>#define UPPER(x) (x>=65 && x<=90)#define SMALL(y) (y>=97 && y<=123)#define ALPHABET(z) (z>=65 && z<=90 || z>=97 && z<=123)#define BIGGER(a,b) (a>b)

void main() {

int i,d1,d2;char ch,ch1;clrscr();

printf("\t enter your choice: \n");printf("\t===================\n\n");

printf("1: to test if character is small case letter or not\n\n");printf("2: to test if character is upper case letter or not\n\n");printf("3: to test if character is an alphabet or not\n\n");printf("4: to find bigger of two number\n\n\n");

printf("choice: ");scanf("%d",&i);

switch (i) {

case 1:

clrscr();

printf("enter any character\n\n");

scanf(" %c",&ch);

if SMALL(ch)printf("\n\n it is a small case letter.\n");

Exercise [C]

Solution:

Chetan Raikwar


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elseprintf("\n\nit is not a small case letter.\n");

break;

case 2:

clrscr();


scanf(" %c",&ch);

if UPPER(ch)printf("\n\nit is an upper case letter.\n");elseprintf("\n\nit is not an upper case letter.\n");

break;

case 3:

clrscr();


scanf(" %c",&ch);

if ALPHABET(ch)printf("\n\nit is an alphabet.\n");

elseprintf("\n\nit is not an alphabet.\n");

break;

case 4:

clrscr();

printf("enter two numbers\n\n");

scanf("%d%d",&d1,&d2);

if BIGGER(d1,d2)

printf("\n\n%d is bigger\n",d1);

elseprintf("\n\n%d is bigger \n",d2);

break;

default:

printf("\n\n\nwrong choice entered\n");

}

getch();}

--------------------------------------------------------------------

(b) Write macro definitions with arguments for calculation of area and perimeter of a triangle, a square and a circle. Store these macro definitions in a file called “areaperi.h”. Include this file in your program,



and call the macro definitions for calculating area and perimeter for different squares, triangles and circles.

#define TP(a,b,c) (a+b+c) /* perimeter of triangle */#define TA(a,b,c,d) (d*((d-a)*(d-b)*(d-c))) /* area of triangle *#define PS(x) (4*x) /* perimeter of square */#define SA(x) (x*x) /* area of square */#define CP(x) (2*3.14*r) /* perimeter of circle */#define CA(x) (3.14*r*r) /* area of circle */

#include"areaperi.h" /* inclusion of custom header file */#include<stdio.h>#include<conio.h>#include<math.h>void main() {

float t1,t2,t3,hpt; /* sides of triangle,half of perimeter */float ss,r; /* side of square, radius of circle */int ch; float tp,ta,ta1,sp,sa,cp,ca; /* perimeter & area of triangle,square,circle */

clrscr();

printf("\tFind area and perimeter of: \n");printf(" *****************************\n\n");

printf("1: triangle\n\n");printf("2: square\n\n");printf("3: circle\n\n\n\n\n");

printf("enter your choice: ");scanf("%d",&ch);

switch(ch)

{

case 1:

/* triangle */

clrscr(); printf("enter side 1 of triangle: "); scanf("%f",&t1);

printf("\n\nenter side 2 of triangle: "); scanf("%f",&t2);


hpt=t1+t2+t3/2;

tp=TP(t1,t2,t3); ta=TA(t1,t2,t3,hpt);

ta1=sqrt(ta);

printf("\n\n\tperimeter of triangle = %f",tp); printf("\n\tarea of triangle = %f",ta1);

Note: write these macro definitions in a new file and save it as "areaperi.h", compile it and then you can include it during inclusion of libraries as - include"areaperi.h".

Solution:



break;

case 2:

/* square */

clrscr();

printf("\n\nenter side of a square: "); scanf("%f",&ss);

sp=PS(ss); sa=SA(ss);

printf("\n\n\tperimeter of square = %f",sp); printf("\n\tarea of square = %f",sa);

break;

case 3:

/* circle */

clrscr();

printf("\n\nenter radius of circle: "); scanf("%f",&r);

cp=CP(r); ca=CA(r);

printf("\n\n\tperimeter of circle = %f",cp); printf("\n\tarea of circle = %f",ca);

break;

default:

exit();}

getch();

}

-----------------------------------------------------------------------

(c) Write down macro definitions for the following:1. To find arithmetic mean of two numbers.2. To find absolute value of a number.3. To convert a uppercase alphabet to lowercase.4. To obtain the bigger of two numbers.

#include<stdio.h>#include<conio.h> #define AM(x,y) ((x+y)/2) #define ABS(x,y) (x<y) #define LOWER(x)(122-(90-x)) #define BIG(x,y) (x>y)

Solution:



void main() {

char ch,ch1; int choice,a,b,abs,c=0; float am;

clrscr();

printf("\tMacro Definitions to:\n"); printf(" ***************************\n\n");

printf("\n1: Find airthmetic mean of two numbers: \n\n"); printf("2: find absolute value of a number: \n\n"); printf("3: convert an uppercase character to lowercase: \n\n"); printf("4: obtain the bigger of two numbers: \n\n\n\n\n\n\n\n\n\n");

printf("enter your choice: "); scanf("%d",&choice);

switch(choice) {

case 1:

/* finding arithmetic mean of two numbers */

clrscr();

printf("\nenter two numbers:\n "); scanf("%d%d",&a,&b); am=AM(a,b); /* airthmetic mean */ printf("\nairthmetic mean = %f ",am); break;

case 2:

/* finding the absolute value of a number */

clrscr();

printf("\nenter a number: "); scanf("%d",&a);

if ABS(a,c)

abs=a*(-1);

else abs=a;

printf("\nabsolute value = %d",abs); break;

case 3:

/* converting an uppercase character to equivalent lowercase */

clrscr();

printf("\nenter an uppercase character: "); scanf(" %c",&ch);

ch1=LOWER(ch);

printf("\nequivalent lowercase character = %c",ch1); break;

case 4:



/* finding the bigger of two numbers */

clrscr();

printf("\nenter two numbers:\n "); scanf("%d%d",&a,&b);

if BIG(a,b) printf("\n\n%d is bigger.\n",a);

else printf("\n\n%d is bigger.\n",b);

break;

default:

exit(); }

getch();

}

-----------------------------------------------------------------------

(d) Write macro definitions with arguments for calculation of Simple Interest and Amount. Store these macro definitions in a file called “interest.h”. Include this file in your program, and use the macro definitions for calculating simple interest and amount.

#define INTEREST(x,y,z) (x*y*z/100)#define AMOUNT(x,y) (x+y)

#include<stdio.h>#include<conio.h> #include"interest.h" /* inclusion of custom header file */

void main() {

int p,r,t,a; float si;

clrscr();

printf("enter the principle: "); scanf("%d",&p);

printf("\n\nenter the rate: "); scanf("%d",&r);

printf("\n\nenter the time: "); scanf("%d",&t);

si=INTEREST(p,r,t);

Note: write these macro definitions in a new file and save it as

"interest.h", compile it and then you can include it during

inclusion of libraries as - include"interest.h".

Solution:






a=AMOUNT(p,si);

printf("\n\n\n\n\t\tsimple interest = %f\n\n\t\tamount = %d",si,a);

getch(); }

_______________________________________________________________________



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Let us C solutions ( By Chetan ): Let Us C / Chapter 8 (Arrays)

http://letuscalllessons.blogspot.com/2014/01/chapter-8-arrays.html[07-Apr-14 8:56:03 PM]





(a) Twenty-five numbers are entered from the keyboard into an array. The number to be searched is entered through the keyboard by the user. Write a program to find if the number to be searched is present in the array and if it is present, display the number of times it appears in the array.


void main() {

int i,arr[25],prsnt=0,num; clrscr();

printf("Please enter 25 numbers: \n");

for(i=0;i<25;i++) { scanf("%d",&arr[i]); }

printf("\n\nPlease enter the number to be searched: "); scanf("%d",&num);

for(i=0;i<25;i++) {

if(num==arr[i]) prsnt=prsnt+1;

}

if(prsnt==0) { printf("\n\nNumber does not present in the array.\n"); }

else { printf("\n\nNumber presents in the array.\n"); printf("\nNumber of times it appears = %d.\n",prsnt); }

Exercise [D]

Solution:

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getch();

}

---------------------------------------------------------------------------------------------------------------

b) Twenty-five numbers are entered from the keyboard into an array. Write a program to find out how many of them are positive, how many are negative, how many are even and how many odd.


void main() {

int i,arr[25],tz=0,tp=0,tn=0;clrscr();

printf("Enter numbers in the array: \n");

for(i=0;i<25;i++) {scanf("%d",&arr[i]);

if(arr[i]<0){tn=tn+1;}

if(arr[i]==0){tz=tz+1;}

if(arr[i]>0){tp=tp+1;}

}

printf("\n\n\nTotal zeros = %d\n",tz);printf("Total positive numbers = %d\n",tp);printf("Total negative numbers = %d\n",tn);

getch();

}

----------------------------------------------------------------------------------------------------------------

(c) Implement the Selection Sort, Bubble Sort and Insertion sort algorithms on a set of 25 numbers. (Refer Figure 8.11 for the logic of the algorithms)− Selection sort− Bubble Sort− Insertion Sort

----------------------------------------------------------------------------------------------------------------

(d) Implement the following procedure to generate prime numbers from

Solution:

Solution:



1 to 100 into a program. This procedure is called sieve of Eratosthenes.step 1Fill an array num[100] with numbers from 1 to 100step 2Starting with the second entry in the array, set all its multiples to zero.step 3Proceed to the next non-zero element and set all its multiples to zero.step 4Repeat step 3 till you have set up the multiples of all the non-zero elements to zerostep 5At the conclusion of step 4, all the non-zero entries left in the array would be prime numbers, so print out these numbers.


void main() {

int i,j,a[100];clrscr();

for(i=0;i<100;i++) {

a[i]=i+1;

}

printf("\n100 numbers in the array:\n\n");

for(i=0;i<100;i++) {

printf("%3d ",a[i]);

}

printf("\n\nafter implementing eratothene's sieve:\n\n");

for(i=2;i<100;i++) {

for(j=2;j<a[i];j++) {

if(a[i]%j==0)a[i]=0;}}

i=a[0];for(;i<100;i++) {


}

printf("\n\nprime numbers are: \n\n");

for(i=a[0];i<100;i++) {

if(a[i]!=0)


Solution:



}

getch();}

---------------------------------------------------------------------------------------------------------------

(a) Write a program to copy the contents of one array into another in the reverse order.


int i,j,k,a1[5],a2[5]; clrscr();

for(i=1;i<=5;i++) { scanf("%d",&a1[i]); }

printf("\n\nThe elements you enterd are:\n");

for(i=1;i<=5;i++) { printf(" %d",a1[i]); }

printf("\n\nElements in reversed order:\n");

for(i=5,j=1;i>=1,j<=5;i--,j++) {

k=a1[i]; a2[j]=k;

printf(" %d",a2[j]); }

getch(); }

---------------------------------------------------------------------------------------------------------------

(b) If an array arr contains n elements, then write a program to check if arr[0] = arr[n-1], arr[1] = arr[n-2] and so on.


int arr[100];int n,i,f=0;clrscr();

printf("enter total elements of array(n): ");scanf("%d",&n);

printf("\n\nenter \"n\" elements of array: \n\n");

Exercise [I]

Solution:

Solution:



for(i=0;i<n;i++) {

scanf("%d",&arr[i]);

}

clrscr();

for(i=0;i<n;i++) {

if(arr[i]==arr[n-(i+1)]) { /* if element is equal, according to the problem, it will be printed */f=f+1;

printf("element no: %d = %d ",i,arr[i]);}

}

if(f==0)printf("\n\nNo such element found.\n");

getch();

}

---------------------------------------------------------------------------------------------------------------

(c) Find the smallest number in an array using pointers.


int i,n,*p,*s,a[100]; clrscr();

printf("enter how many numbers you want to save in array: "); scanf("%d",&n);

printf("\n\nenter %d number in array:\n",n);

for(i=0;i<n;i++) {

scanf("%d",&a[i]);

}

clrscr();

printf("array you entered: \n\n");

for(i=0;i<n;i++) {


}

printf("\n\n");

p=&a[0]; /* first pointer points 0th element */

for(i=0;i<n;i++) {

s=&a[i]; /* second pointer points every element one by one */

Solution:



if(*p>*s) /* if first is bigger than second */ *p=*s; /* first becomes second */

s++; /* second is incremented to check with other elements */

}

printf("smallest digit in array is %d\n",*p);

getch(); }

----------------------------------------------------------------------------------------------------------------

(d) Write a program which performs the following tasks:− initialize an integer array of 10 elements in main( )− pass the entire array to a function modify( )− in modify( ) multiply each element of array by 3− return the control to main( ) and print the new array elements in main( )


int i,j,a[10]={1,2,3,4,5,6,7,8,9,10}; modify(); clrscr();

printf("array before modification: \n\n");

for(i=0;i<10;i++) {

printf(" %d ",a[i]);

}

modify(a); /* passing only the name of array */

printf("\n\n\narray after modification:\n\n");

for(i=0;i<10;i++) { /* printing the array in main(); */


}

getch();

}

modify(int b[10]) {

int c;

for(c=0;c<10;c++) {

b[c]=b[c]*3; /* multiplying each element with 3 */

} return b[c]; /* returning the whole array to main(); */

Solution:



}

----------------------------------------------------------------------------------------------------------

(e) The screen is divided into 25 rows and 80 columns. The characters that are displayed on the screen are stored in a special memory called VDU memory (not to be confused with ordinary memory). Each character displayed on the screen occupies two bytes in VDU memory. The first of these bytes contains the ASCII value of the character being displayed, whereas, the second byte contains the colour in which the character is displayed.For example, the ASCII value of the character present on zeroth row and zeroth column on the screen is stored at location number 0xB8000000. Therefore the colour of this character would be present at location number 0xB8000000 + 1. Similarly ASCII value of character in row 0, col 1 will be at location 0xB8000000 + 2, and its colour at 0xB8000000 + 3.With this knowledge write a program which when executed would keep converting every capital letter on the screen to small case letter and every small case letter to capital letter. The procedure should stop the moment the user hits a key from the keyboard.This is an activity of a rampant Virus called Dancing Dolls. (For monochrome adapter, use 0xB0000000 instead of 0xB8000000).

_______________________________________________________________________

(a) How will you initialize a three-dimensional array threed[3][2][3]? How will you refer the first and last element in this array?


/* initialization of a 3 dimensional array */

int threed[3][2][3]={ { {100,2,3}, {1,2,3} }, { {8,5,6}, {4,5,6} }, { {7,8,9}, {7,8,200} } };int *f,*l;clrscr();

f=&threed[0][0][0]; /* reference to first element */

Solution:

Exercise [L]

Solution:



l=&threed[2][1][2]; /* reference to second element */

printf("\n\nfirst element = %d",*f);printf("\n\nlast element = %d",*l);

getch();

}

---------------------------------------------------------------------------------------------------------------

(b) Write a program to pick up the largest number from any 5 row by 5 column matrix.


void main() {

int i,j,a[5][5];clrscr();

printf("\nType the numbers to in matrix:\n");

for(i=0;i<5;i++) {

for(j=0;j<5;j++) {

scanf("%d",&a[i][j]);

} }

clrscr();

printf("matrix you entered is:\n\n");

for(i=0;i<5;i++) {

for(j=0;j<5;j++) {

printf(" %2d ",a[i][j]);

}

printf("\n");

}

/* finding the largest number */

for(i=0;i<5;i++) {

for(j=0;j<5;j++) {

if(a[0][0]<a[i][j]) /* if number is larger than first element */

a[0][0]=a[i][j]; /* larger number is placed as the first element */

}

}

Solution:



printf("\n\nThe largest number in matrix is: %d",a[0][0]);

getch();

}

---------------------------------------------------------------------------------------------------------------

(c) Write a program to obtain transpose of a 4 x 4 matrix. The transpose of a matrix is obtained by exchanging the elements of each row with the elements of the corresponding column.

#include<stdio.h>#include<conio.h>#define MAX 4

void main() {

int i,j,a[4][4],b[4][4];clrscr();

printf("\nenter numbers in 5x5 matrix: \n\n");

for(i=0;i<MAX;i++) {

for(j=0;j<MAX;j++) {


}

}

clrscr();

printf("\nmatrix you entered is: \n\n");



printf("%2d ",a[i][j]);

}printf("\n");}

/* transpose of matrix */



b[j][i]=a[i][j];

}}

printf("\n\n");

/* printing the transpose */

printf("Transpose of matrix is: \n\n");


Solution:




printf("%2d ",b[i][j]);

}printf("\n");}

getch();}

----------------------------------------------------------------------------------------------------------------

(d) Very often in fairs we come across a puzzle that contains 15 numbered square pieces mounted on a frame. These pieces can be moved horizontally or vertically. A possible arrangement of these pieces is shown below:

1 4 15 7 8 10 2 1114 3 6 1312 9 5

As you can see there is a blank at bottom right corner. Implement the following procedure through a program:

Draw the boxes as shown above. Display the numbers in the above order. Allow the user to hit any of the arrow keys (up, down, left, or right). If user hits say, right arrow key then the piece with a number 5 should move to the right and blank should replace the original position of 5. Similarly, if down arrow key is hit, then 13 should move down and blank should replace the original position of 13. If left arrow key or up arrow key is hit then no action should be taken. The user would continue hitting the arrow keys till the numbers aren’t arranged in ascending order. Keep track of the number of moves in which the user manages to arrange the numbers in ascending order. The user who manages it in minimum number of moves is the one who wins. How do we tackle the arrow keys? We cannot receive them using scanf( ) function. Arrow keys are special keys which are identified by their ‘scan codes’. Use the following function in your program. It would return the scan code of the arrow key being hit. Don’t worry about how this function is written. We are going to deal with it later. The scan codes for the arrow keys are: up arrow key – 72 down arrow key – 80 left arrow key – 75 right arrow key – 77 /* Returns scan code of the key that has been hit */ #include "dos.h" getkey( ) { union REGS i, o ; while ( !kbhit( ) ) ;



i.h.ah = 0 ; int86 ( 22, &i, &o ) ; return ( o.h.ah ) ; }

Solution:


/* returns scan code of the key that has been hit */getkey(){union REGS i,o;while(!kbhit() );i.h.ah=0;int86(22,&i,&o);return(o.h.ah);}

void main() {

int i,j,a[16]={1,4,15,7,8,10,2,11,14,3,6,13,12,9,5,0};int temp,h,moves=0,won=0;

clrscr();

/****************************************************/

do { /**************/ /* to move up */ /**************/

if(h==72) {

for(i=0;i<16;i++) {if(a[i]==0){if(a[0]==0 || a[1]==0 || a[2]==0 || a[3]==0) {break;}temp=a[i];a[i]=a[i-4];a[i-4]=temp;moves=moves+1;break;}} } /****************/ /* to move left */ /****************/

if(h==75) {

for(i=0;i<16;i++) {

if(a[i]==0){if(a[0]==0 || a[4]==0 || a[8]==0 || a[12]==0) {break;}



temp=a[i];a[i]=a[i-1];a[i-1]=temp;moves=moves+1;break;}} }

/****************/ /* to move down */ /****************/

if(h==80) {for(i=0;i<16;i++) {if(a[i]==0){if(a[12]==0 || a[13]==0 || a[14]==0 || a[15]==0) {break;}temp=a[i];a[i]=a[i+4];a[i+4]=temp;moves=moves+1;break;}} } /*****************/ /* to move right */ /*****************/

if(h==77) {

for(i=0;i<16;i++) {

if(a[i]==0) {

if(a[3]==0 || a[7]==0 || a[11]==0 || a[15]==0 ) {break;}

temp=a[i];a[i]=a[i+1];a[i+1]=temp;moves=moves+1;break;} } }

/***********************************************************/

/**********************************/ /* printing the puzzle with boxes */ /**********************************/

printf("\n%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c\n",218,196,196,196,194,196,196,196,194,196,196,196,194,196,196,196,191);for(i=0;i<=15;i++) {

printf("%c",179);

if(a[i]==0) {printf("%c ",32); /* printing a blank space in the puzzle */}if(a[i]!=0)

printf(" %2d",a[i]);

if(a[i]==a[3] || a[i]==a[7] || a[i]==a[11] || a[i]==a[15])printf("%c",179);



if(i==3||i==7||i==11)printf("\n%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c\n",195,196,196,196,197,196,196,196,197,196,196,196,197,196,196,196,180);

if(a[0]==1 && a[1]==2 && a[2]==3 && a[3]==4 && a[4]==5 && a[5]==6&&a[6]==7 && a[7]==8 && a[8]==9 && a[9]==10 && a[11]==12 && a[12]==13&& a[13]==14 && a[14]==15 && a[15]==0 ) {

won=1;}

}printf("\n%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c\n",192,196,196,196,193,196,196,196,193,196,196,196,193,196,196,196,217);

/***************************************************/if(won==1) {printf("\n\n\tCongratulations! you have won.");break;} /**********************************/ /* to print instructions for user */ /**********************************/

printf("\n\n\n\n\n\n Total Moves: %d\t\t\t\t Use arrow keys to move blank:\n\n",moves);printf("\t\t\t\t\t\t %c to move up\n",30);printf("\t\t\t\t\t\t %c to move down\n",31);printf("\t\t\t\t\t\t %c to move left\n",17);printf("\t\t\t\t\t\t %c to move right\n",16);

/****************************************************/

/**********************/ /* to take user input */ /**********************/

h=getkey();clrscr();

/****************************************************/

}while(h==72 || h==75 || h==77 ||h==80);

getch();

}

-----------------------------------------------------------------------------------------------------------------

(e) Those readers who are from an Engineering/Science background may try writing programs for following problems.(1) Write a program to add two 6 x 6 matrices.(2) Write a program to multiply any two 3 x 3 matrices.(3) Write a program to sort all the elements of a 4 x 4 matrix.(4) Write a program to obtain the determinant value of a 5 x 5 matrix.

--------------------------------------------------------------------------------------------------------------

(j) Write a program that interchanges the odd and even components of an array.

Solution:




void main() {

int i,j,a[6],even,temp; clrscr();

printf("enter the numbers: \n\n");

for(i=0;i<6;i++) {

scanf("%d",&a[i]);

}

clrscr();

printf("\narray without exchanging even and odd numbers:\n\n");

for(i=0;i<6;i++) {


}

printf("\n\narray after exchanging even and odd numbers: \n\n");

for(i=0;i<6;i++) {

for(j=i+1;j<6;j++) {

/* if one element is even and another after that is odd \ ,they will be exchanged */

if((a[i]%2)!=0 && (a[j]%2)==0) {

temp=a[j]; a[j]=a[i]; a[i]=temp; } } }

for(i=0;i<6;i++) {


}

getch(); }

--------------------------------------------------------------------------------------------------------------

(k) Write a program to find if a square matrix is symmetric.


void main() {

int i,j,r,c,sym;

Solution:

Solution:



int a[100][100],b[100][100];

clrscr();

printf(" enter number of rows of matrix: ");scanf("%d",&r);

printf("\n\nenter number of columns of matrix: ");scanf("%d",&c);

clrscr();

printf("enter the elements in matrix: \n");

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {

scanf("%d",&a[i][j]);}

}

clrscr();

printf("\nmatrix you entered is\n\n");

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {

printf("%d ",a[i][j]);

}printf("\n");}

printf("\n\ntranspose of matrix is \n\n");

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {

b[j][i]=a[i][j];

}}

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {

printf("%d ",b[i][j]);

}printf("\n");}

/* finding if square matrix is equal to it's transpose to be symmetric */

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {

if(a[i][j]!=b[i][j])

sym=1;

}



}

if(sym==1)printf("\nSquare matrix is not symmetric.\n");

elseprintf("\nSquare matrix is symmetric.\n");

getch();

}

----------------------------------------------------------------------------------------------------------------

(l) Write a function to find the norm of a matrix. The norm is defined as the square root of the sum of squares of all elements in the matrix.


void main() {

int i,j,r,c,a[100][100];int norm=0,sum=0;

clrscr();

printf("\nEnter the number of rows: ");scanf("%d",&r);

printf("\n\nEnter the number of coloumns: ");scanf("%d",&c);

clrscr();

printf("Enter elements of %d x %d array: \n\n",r,c);

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {


}}

clrscr();


for(i=0;i<r;i++) {

for(j=0;j<c;j++) {


}printf("\n");}

/* norm of the matrix */

for(i=0;i<r;i++) {

for(j=0;j<c;j++) {

Solution:



sum=sum+(a[i][j]*a[i][j]);

}

}

norm=sqrt(sum);

printf("\nNorm of matrix = %d",norm);

getch();}

----------------------------------------------------------------------------------------------------------------

(m) Given an array p[5], write a function to shift it circularly left by two positions. Thus, if p[0] = 15, p[1]= 30, p[2] = 28, p[3]= 19 and p[4] = 61 then after the shift p[0] = 28, p[1] = 19, p[2] = 61, p[3] = 15 and p[4] = 30. Call this function for a (4 x 5 ) matrix and get its rows left shifted.


void main() {

int i,j,p[]={15,30,28,19,61}; int a[4][5];

clrscr();

printf("Array before shift:\n\n"); for(i=0;i<5;i++) {

printf("%2d ",p[i]);

}

func(p);

printf("\n\nArray after shift:\n\n");

for(i=0;i<5;i++) {


}

printf("\n\n\nenter the elements of 4x5 matrix: \n\n");

for(i=0;i<4;i++) {

for(j=0;j<5;j++) {


}

}

clrscr();

printf("matrix you enterd before shift: \n\n");

for(i=0;i<4;i++) {

Solution:



for(j=0;j<5;j++) {


}

printf("\n"); }

printf("\n\nafter shift:\n\n");

/* shift the rows of matrix */

for(i=0;i<4;i++) {

func(a[i]);

}

for(i=0;i<4;i++) {

for(j=0;j<5;j++) {


}

printf("\n");

}

getch();

}

func(int q[5]) {

int a,t1,t2,t3;

t1=q[0]; t2=q[1];

q[0]=q[2]; q[1]=q[3]; q[2]=q[4]; q[3]=t1; q[4]=t2;

return q[5];

}----------------------------------------------------------------------------------------------------------

(n) A 6 x 6 matrix is entered through the keyboard and stored in a 2-dimensional array mat[7][7]. Write a program to obtain the Determinant values of this matrix.

------------------------------------------------------------------------------

(o) For the following set of sample data, compute the standard deviation and the mean.-6, -12, 8, 13, 11, 6, 7, 2, -6, -9, -10, 11, 10, 9, 2

Solution:



#include<stdio.h> #include<conio.h> #include<math.h>

void main() {

int a[15]={-6,-12,8,13,11,6,7,2,-6,-9,-10,11,10,9,2}; int i,j; float temp,sd,sum=0,mean,x;

clrscr();

printf("\ndata set: \n\n");

for(i=0;i<15;i++) {

printf(" %3d ",a[i]);

}

printf("\n");

for(i=0;i<15;i++) {

sum=sum+a[i]; /* adding all the numbers */

}

mean=sum/15; /* calculating the mean */

/* computing standard deviation */

for(i=0;i<15;i++) {

a[i]=pow((a[i]-mean),2); x=x+a[i]; }

temp=x/15; sd=sqrt(temp);

printf("\n\n\t\tmean= %f\n\t\tstandard deviation = %f\n",mean,sd);

getch(); }

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(p) The area of a triangle can be computed by the sine law when 2 sides of the triangle and the angle between them are known.Area = (1 / 2 ) ab sin ( angle )Given the following 6 triangular pieces of land, write a program to find their area and determine which is largest,Plot No. a b angle

137.4 80.9 0.78

Solution:



155.2 92.62 0.89

149.3 97.93 1.35160.0 100.25 9.00

155.6 68.95 1.25

149.7 120.0 1.75

#include<stdio.h> #include<conio.h> #include<math.h> void main() {

float a[6][3]={ {137.4, 80.9, 0.78}, {155.2, 92.62, 0.89}, {149.3, 97.93, 1.35}, {160.0, 100.25, 9.00}, {155.6, 68.95, 1.25}, {149.7, 120.0, 1.75} };

float big=0,area; int sr=0,i; clrscr();

for(i=0;i<6;i++) {

area=(1.0/2.0)*a[i][0]*a[i][1]*sin(a[i][2]);

if(area>big) { big=area; sr=i; }

}

printf("\n\nPlot no. %d is the biggest.\n",sr); printf("\nArea of plot no. %d = %f\n",sr,big);

getch();

}

- ------------------------------------------------------------------------------------------------------------

(q) For the following set of n data points (x, y), compute the correlation coefficient r,

x y34.22 102.4339.87 100.93

41.85 97.43

Solution:



43.23 97.8140.06 98.3253.29 98.3253.29 100.0754.14 97.0849.12 91.5940.71 94.8555.15 94.65


void main() {

float a[][2]={ {34.22,102.43}, {39.87,100.93}, {41.85,97.43}, {43.23,97.81}, {40.06,98.32}, {53.29,98.32}, {53.29,100.07}, {54.14,97.08}, {49.12,91.59}, {40.71,94.85}, {55.15,94.65} };

int i,n=0; float x2,y2,x,y,x_y,n_x2,n_y2,r; clrscr();

for(i=0;i<11;i++) {

x2= x2 + ( a[i][0] * a[i][0] ); /* computing square of x */

y2= y2 + ( a[i][1] * a[i][1] ); /* computing square of y */

x= x + a[i][0]; /* computing total of x */

y= y + a[i][1]; /* computing total of y */

x_y= x_y + ( a[i][0] * a[i][1] ); /* computing total of x * y */

n++;

}

n_x2= n * x2;

n_y2= n * y2;

r= ( x_y - x*y )/sqrt((n_x2-x2) * (n_y2-y2));

Solution:



printf(" sum of square of x = %f\n\n",x2); printf(" sum of square of y = %f\n\n",y2); printf(" sum of x = %f\n\n",x); printf(" sum of y = %f\n\n",y); printf(" sum of x * y = %f\n\n",x_y); printf(" sum of n*x2 = %f\n\n",n_x2); printf(" sum of n*y2 = %f\n\n",n_y2);

printf("\n\n\nCorrelation cofficient = %f\n",r);

getch();

} -------------------------------------------------------------------------------------------------------------

(r) For the following set of point given by (x, y) fit a straight line given by y = a + bx

x y 3.0 1.5 4.5 2.0 5.5 3.5 6.5 5.0 7.5 6.0 8.5 7.5 8.0 9.0 9.0 10.5 9.5 12.0 10.0 14.0


Solution:



void main() {

float data[][2]= { {3.0,1.5}, {4.5,2.0}, {5.5,3.5}, {6.5,5.0}, {7.5,6.0}, {8.5,7.5}, {8.0,9.0}, {9.0,10.5}, {9.5,12.0}, {10.0,14.0} };

int i,n=0;float sx,sy,x2,y2,xy,a,b,Y;

clrscr();

for(i=0;i<10;i++) {

sx = sx + data[i][0];

sy = sy + data[i][1];

x2= x2 + ( data[i][0] * data[i][0] );

y2= y2 + ( data[i][1] * data[i][1] );

xy = xy + ( data[i][0] * data[i][1] );

n++;

}

printf(" sum of x = %f\n",sx);printf(" sum of y = %f\n",sy);printf(" sum of x2 = %f\n",x2);printf(" sum of y2 = %f\n",y2);printf(" sum of x*y = %f\n",xy);printf(" total number = %d\n",n);

b = ( (n*xy) - (sx*sy) ) / ( n*x2 - (sx*sx) );

a = (sy/n) - b*(sx/n);

Y= a + b*sx ;

printf("\n\nvalue of a = %f\n\n",a);

printf("value of b = %f\n\n",b);

printf(" Y = %f \n\n",Y);

getch();

}

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(s) The X and Y coordinates of 10 different points are entered through the keyboard. Write a program to find the distance of last point from the first point (sum of distance between consecutive points).





float a[10][2],sx,sy; int i;

clrscr();

for(i=0;i<10;i++) {

printf("Enter coordinates of point number %d:\n\n",i+1);

printf("Value of X coordinate: "); scanf("%f",&a[i][0]);

printf("\nValue of Y coordinate: "); scanf("%f",&a[i][1]);

clrscr(); }

for(i=0;i<10;i++) {

if(i>0 && i<10-1) {

sx = sx + a[i][0];

sy = sy = a[i][1];

} }

printf(" First coordinate: X = %f\tY = %f\n\n",a[0][0],a[0][1]);

printf(" Last coordinate: X = %f\tY = %f\n\n",a[9][0],a[9][1]);

printf("\nDistance between them: X = %f\tY = %f\n",sx,sy);

getch(); } ______________________________________________________________________

Solution:



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Let us C solutions ( By Chetan ): Let Us C / Chapter 9 (Puppeting on Strings)

http://letuscalllessons.blogspot.com/2014/01/chapter-9-puppeting-on-strings.html[07-Apr-14 8:57:15 PM]







char str[100];

gets(str);

printf("\n\n");


getch();

}

-----------------------------------------------------------------------------------------------------------



void main() {


Exercise [D]

Solution:

Solution:

Chetan Raikwar


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