Compiler DesignAim: Program for 3-address code.

#include<conio.h>#include<stdio.h>#include<ctype.h>#include<string.h>void main(){char s[20],st[20];int t=0,i,num[20],top=0;clrscr();printf(“\n Enter postfix notation”);scanf(“%s”,s);for(i=0;i<strlen(s);i++)if isalnum(s[i]);st[top++]=s[i];else{printf(”\n t %d=”,t);if((str[top-2]!=’t’)&&(top>=2))printf(“%c”,st[top-2]);else if(st[top-2]==’t’)printf(“%c %d”,st[top-2],num[top-2]);printf(“%c”,s[i]);if(st[top-1]==’t’)printf(“%c %d”,st[top-1],num[top-1]);elseprintf(“%c”,st[top-1]);if(top>=2)top=top-2;

elsetop=top-1;st[top]=’t’;num[top++]=t++;}}

Output: Input : Enter postfix notation; ab+cd+* Output: T0=a+b T1=c+d T2=T0*T1

Write a program for Brute Force Technique

Aim: Program to implement Brute Force Tchnique

SA|AB A ab|adb B cd

Source Code:

#include<stdio.h>

#include<string.h>void S();void A();void AB();char str[10];int i=0,count=0;int main(){printf(“Enter the string:”);scanf(“%s”,str);S();if(count<0){Printf(“\nString can’t be delivered from given grammar”);}}

void S(){A();if(count==0)AB();}void A(){if(str[i]==’a’&&str[i+1]==’b’&&str[i+2]==’$’){printf(“ \n Parsing is successful for ab”);

count++;return ;}if(str[i]==’a’&&str[i+1]==’b’&&str[i+3]==’d’&&str[i+4]==’$’){printf(“\n Parsing is successful for abd”);count++;return ;}}Void AB(){

if(str[i]==’a’ && str[i+1]==’b’&& str[i+2]==’c’&& str[i+4]==’d’ && str[i+5]==’$’){print(“\n Parsing is successful for abcd”);count++;return ;}if(str[i]==’a’ && str[i+1]==’b’&& str[i+2]==’d&& str[i+3]=’c’&& str[i+4]=’d && str[i+5]==’$’){print(“\n Parsing is successful for abdcd”);count++;return ;}}

Output:

Enter the String: abcd$ Parsing is successful for abcd.

Write a Program for Predictive Parser

Aim: To implement Predictive Parser.

Source Code:

#include<stdio.h>#include<string.h>int top=-1;char s[20], stk[20],item, ch;char push(char);void error();char pop();int main(){int ip=0;printf(“Enter the input string”);scanf(“%s”, s);push($);push(E);while(stk[top]!=’$’){

switch(stk[top]){

case ‘E’: if(s[ip]==’a’|| s[ip]==’(‘){pop();push(‘A’);push(‘T’);}

else error();break;case ‘T’: if(s[ip]==’a’|| s[ip]==’(‘)

{pop();push(‘B’);push(‘F’);}

elseerror();

break;

case ‘A’: if(s[ip]==’+’){push(‘A’);push(‘T’);ip++;}

elseif(s[ip]==’$’|| s[ip]==’)’)pop();

break;case ‘B’: if(s[ip]==’*’)

{push(‘B’);push(‘F’);ip++;}

elseif(s[ip]==’+’|| s[ip]==’)’|| s[ip]==’$’)pop();

break;else

error();break;case ‘F’: if(s[ip]==’a’)

{pop();ip++;}

elseif(s[ip]==’(‘){push(‘)’);push(‘E’);ip++;}

else

error();break;case ‘)’: if(s[ip]==’)’)

{pop();ip++;}

elseerror();

break;}}if((s[ip]==’$’)&&(stk[top]==’$’))printf(“Input String is Processed”);elseerror();}void error(){printf(“error”);exit(0);}char pop(){top--;}char push(char item){top++;stk[top]=item;

}

Output:

Enter the input string: a+a*aInput String is Processed.

Write a c program to implement Recursive Descent parser.

char is,char str[20];int i=0;main(){clrscr();printf(“enter a string to be parse”);scanf(“%s”,str);is=str[0];E();getch();}E(){T();E_prime();if(is==’$’)printf(“successful parsed”);

elseprintf(“parsing fail”);}

E_prime(){if(is==’+’){Advance();T();E_prime(); }}

T(){F();T_prime();}

T_prime(){if (is==’*’){ Advance();F();T_prime();}}

F(){ if (is >=’a’&& is<=’z’) { Advance(); } else if(is==’(‘) { Advance(); E(); if(is==’)’) Advance(); else Error(); } else Error();}

Advance(){is=str[++i];}Error(){printf(“there are some syntactic errors”);}

Input: a+a*a$

Output: Successful parsed

1. Write a Lex program to find length of a string.%{int length;%}%%[a-z A-Z 0-9]+ {length=yyleng; }%%main(){yylex();printf(“length of given string is %d”,length);}int yywrap(){return 1;}

Compile: lex str_length.l cc lex.yy.c Run: ./a.out Input: sree chaitanya Output: lenth of given string is 14

2. Write a Lex program to find Predecessor & successor of a character

%{%}%%[a-z A-Z 0-9] {printf(“given character is %c”,yytext); printf (“printf(“ its predecessor is %c”,yytext[0]-1);printf(“its successor is %c”,yytext[0]+1);}%%main(){printf(“ enter a character”);yylex();}int yywrap(){return 1;}

Compile: lex ps.l cc lex.yy.c Run: ./a.out Input: s Output: given character is S Its predecessor is r Its successor is t

3. Write a Lex program to print no.of words ,numbers and Lines in a Text.

%{Int num_words=0,num_numbers=0,num_lines=0;%}Word [a-zA-Z]Number [0-9]+%%{word} {++num_words;}{number} {++num_numbers;}\n {++num_lines;}. { }%%main(){yylex();Printf(“no.of words=%d\n no.of numbers=%d\n no.of lines=%d\n”,num_wors,num_numbers,num_lines);}int yywrap(){return 1;}

Compile: lex count.l cc lex.yy.c

Run: ./a.out Input: sree chaitanya 123 34 College 56. Output: no.of words=3 No.of numbers=3 No.of lines=2

4. Write a Lex specification for identifying positive integers,negative integers,positive real numbers,negative real numbers,positive exponential,negative exponential.

%{%}

%%[0-9]+ { printf(“positive integer”); } [-][0-9]+ { printf(“negative integer”); }[0-9]+[.] [0-9]+ { printf(“positive real numbers”); }[-][0-9]+[.] [0-9]+ { printf(“negative real numbers”);

}[0-9]+”e”[.] [0-9]+ {

printf(“positive exponential”); }

[0-9]+”e”[.] [0-9]+ { printf(“negative exponential”); } %% Main(){yylex();}int yywrap(){return 1;}

Output : lex idennum.l cc lex.yy.c –ll ./a.out

Input: 3 Output : positive number Input: -0.1 Output : negative number Input: 1e.1 Output : positive exponential

5. Lex specification for palindrome or not.

%{int i,j,flag;%} %% [a-z A-z 0-9]* {for(i=0,j=yyleng-1;i<=j;i++,j--)

{if (yylext[i]==yytext[j]) {

flag=1; }

else {

flag=0; break; } } if (flag==1) printf(“given string is polyndrome”); else printf(“given string is not polyndrome”); }%% main(){printf(“enter a string:”);yylex();

}int yywrap(){return 1;} Output : lex poly.l cc lex.yy.c ./a.out madam palindrome tree

not palindrome6. Write a Lex program to identify whethere given

string is word or number or alphanumeric.

%{%}%%[a-zA-Z]* {printf(“given string is WORD”);}[0-9]* {printf(“given string is NUMBER”);}[a-zA-Z0-9]* {printf(“given string is alphanumeric”);}%%main(){printf(“enter a string”);yylex();}

int yywrap(){return 1;}

Compile: lex str.l cc lex.yy.c Run: ./a.out Input: sree Output: given string is WORDInput:123Output: given string is numberInput: sree123Output: given string is alphanumeric