chapter 05 control

8/12/2019 Chapter 05 Control

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Control Statements5.1. INTRODUCTION

The term flow of control refers to the order in which a programs statementsare executed. Every programming language supports three control flow structuresnamely :

1. Sequential control structure2. Selective control structure. !terative control structure

5.1.1. Sequential Control Structure

The normal flow of control of all programs is sequential. !n sequentialstructure" a sequence of programs statements are executed one after another inthe order in which they are placed. #oth selection and repetition statementsallows allow the programmer to alter the normal sequential flow of control.

Sequential programming can also $e called linear programming. Thesequential programs are non%modular in nature. That is" reusa$ility of code is notpossi$le. Thus" they are difficult to maintain and understand. Examples ofsequence control structure statements are" the program will have statements thatare placed sequentially and there is no decision involved in the process. &lso" theprogram does not require a specific tas' to $e repeated over and over again.

5.1.2. Selective Control Structure (or) Decision Control Structure

The selective structure allows the usual sequential order of execution to $emodified. !t consists of a test for a condition followed $y alternative paths that theprogram can follow. The program selects one of the alternative paths dependingupon the result of the test for condition. Examples of selective control structuresstatements are :

1. Simple if statement2. if . . . else statement

. (ested if . . . else statement). else if ladder*. switch . . . case . . .default statement


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5.2 Control Statements

5.1.3. Iterative Control Structure (or) oo! Control Structure

The iterative structure provides the a$ility to go $ac' and repeat a set ofstatements. !terative structure is otherwise referred to as repetitive structure.

The + language has the a$ility to repeat the same calculation or sequence ofinstructions over and over again" each time using different data. The iterativestructure consists of an entry point that includes initiali,ation of varia$le" a loopcontinuation condition" a loop $ody and an exit point. Examples of iterativecontrol structure statements are :

1. while statement2. do . . . while statement. for statement

5.2. i" ST#T$%$NTS

+ allows decisions to $e made $y evaluating a given expression as true orfalse. Such an expression involves the relational and logical operators. -ependingon the outcome of the decision" program execution proceeds in one direction oranother. The + statement that ena$les these tests to $e made is called the ifstatements.

The if statements may $e implemented in different forms depending on thecomplexity of conditions to $e tested. They are :

1. Simple if statement2. if . . . else statement

. (ested if . . . else statement). else if ladder

5.2.1. Sim!le i" Statement

The simple if statement is used to specify conditional execution of programstatement or a group of statements enclosed in $races. The syntax is :

if (test condition){

statement-block ; } statement-x ;


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or equal to $ * **

Truealse

Two or more conditions may $e com$ined in an if statement using a logical&(- operator 6778 or a logical 9 operator 6;;8. !t can compare any num$er ofvaria$les in a single if statement. Ta$le $elow shows the various expressions thatare used as conditions inside an if statement :

ConditionalExpression

MeaningFor

e.g.,al!eof a

Fore.g.,

al!e of

b

Fore.g.,

al!e of c

"es!lt

66a5$8 77 6$5c88a is greater than

$ &(- $ isgreater than c

< 2< 1< True< 1< 2< alse

1< < 2< alse1< 2< < alse

66a5$8 ;; 6$5c8a is greater than$ 9 $ is greater

than c

< 2< 1< True< 1< 2< True

1< < 2< True1< 2< < alse

+onsider the pro$lem for finding the tic'et fare for adult and children. Thepro$lem involves decision" that is" the tas's include :

&ssuming that the tic'et fare as 1


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Control Statements 5 .5

etc+() 09

RUN 1 4

$nter t+e a e 4 25

T+e tic8et "are is Rs. 1

RUN 2 4

$nter t+e a e 4 1

T+e tic8et "are is Rs. 5

Note 4 There is only one statement in the if $loc'" the $races are optional.#ut if there is more than one statement you must use the $races.

5.2.2. i" . . . else Statement

Sometimes" we can execute one group of statements if the condition is trueand another group of statements if the condition is false" in such a situation" theif . . . else statement can $e used.

The if . . . else statement is an extension of simple if statement. The syntaxis :

if (test condition){

tr!e-block-statement(s) ; }else{

false-block-statement(s) ; } statement-x ;

!f the test condition is true" then the true%$loc'%statement6s8 are executed. !fthe test condition is false" then the false%$loc'%statement6s8 are executed. !neither case" either true%$loc'%statement6s8 or false%$loc'%statement6s8 will $eexecuted" not $oth.


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test

cond


The figure $elow shows the flowchart of simple if statement :

Entry

True alse

To understand this programming construct" let us consider an example ofchec'ing whether the person is eligi$le to vote or not $y getting the age of theperson. The tas' include :

=etting the input 6age8 from the user. ?a'ing the decision if the age is greater than or equal to 1@" then print

the person is eligi$le to vote. Else print the person is not eligi$le to vote.

The following program demonstrates the a$ove pro$lem :

&' Demonstration o" i" . . . else statement '& inclu e *st io.+, inclu e *conio.+,

voi main()-

int a e 0clrscr() 0!rint"( $nter t+e a e 4 ) 0scan"( / 6a e) 0i"(a e , 1:)

!rint"( 7nT+e !erson is eli i;le to vote. ) 0else

false%$loc'%statement6s8

statement%x

true%$loc'%statement6s8


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!rint"( 7nT+e !erson is not eli i;le to vote. ) 0etc+() 0

9

RUN 1 4

$nter t+e a e 4 25

T+e !erson is eli i;le to vote.

RUN 2 4

$nter t+e a e 4 1asse in secon class ) 0else i"(mar8 , = )

!rint"( 7n>asse in t+ir class ) 0else

!rint"( 7n?aile in t+e e@am. ) 0etc+() 0

9

RUN 1 4

$nter t+e mar8 score 4 asse in "irst class

RUN 2 4

$nter t+e mar8 score 4 35

?aile in t+e e@am.

5.3. sAitc+ ST#T$%$NT

The control statement which allows us to ma'e a decision from the num$er ofchoices is called a switch" or more correctly a switch . . . case . . . default" sincethese three 'eywords go together to ma'e up the control statement.


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The switch statement tests the value of a given varia$le 6or expression8against a list of case values and when a match is found" a $loc' of statementsassociated with the case is executed. The syntax is :

s&itc' (expression){

case al!e-# statement-# ;break ;

case al!e-$

statement-$ ;break ;. . .. . .. . .defa!lt

defa!lt-statement ; } statement-x ;

/here the expression is an integer expression or characters. value%1" value%2are constants or constant expression 6valua$le to an integral constant8 and are'nown as case la$els. Each of these values should $e should $e unique with aswitch statement. statement%1" statement%2 are statement lists and may containone or more statements. There is no need to put $races around these $loc's. The'eyword case is followed $y an integer or a character constant. Each constant ineach case must $e different from all others and the case la$els end with asemicolon 6:8.

/hen the switch is executed" the value is computed for expression" thelist of possi$le constant expression values determined from all case statements issearched for a match. !f a match is found" execution continues after the matchingcase statement and continues until a $rea' statement is encountered or the endof statement is reached.

The $rea' statement at the end of each $loc' signals the end of a particularcase and causes an exit from the switch statement" transferring the control to thestatement%x following the switch. The default is an optional case. !f a match is notfound and the default statement prefix is found within switch" execution continuesat this point. 9therwise" switch is s'ipped entirely and the control goes to thestatement%x.


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switch

expr


Note 4 &t the end of every case" there should $e a $rea' statement.9therwise" it will result in causing the program execution to continue into the nextcase whenever case gets executed.

The figure $elow shows the flowchart of switch statement :

Entry

Expression value 1

Expression value 2

. . .

. . .-efault 6no match8 . . .

statement%1

statement%2

default statement

statement%x


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+onsider the following program to read a num$er from 1 to 1< and print thecorresponding num$er in words :

&' Demonstration o" sAitc+ ... case ... e"ault statement '& inclu e *st io.+, inclu e *conio.+,

voi main()-

int num 0

clrscr() 0!rint"( $nter a num;er *1 to 1 , 4 ) 0scan"( / 6num) 0sAitc+(num)-

case 1 4!rint"( 7nON$ ) 0;rea8 0

case 2 4!rint"( 7nTBO ) 0;rea8 0

case 3 4!rint"( 7nT R$$ ) 0;rea8 0

case = 4!rint"( 7n?OUR ) 0;rea8 0

case 5 4!rint"( 7n?I $ ) 0;rea8 0

case < 4!rint"( 7nSIE ) 0;rea8 0

case F 4!rint"( 7nS$ $N ) 0;rea8 0

case : 4!rint"( 7n$IG T ) 0;rea8 0

case H 4!rint"( 7nNIN$ ) 0;rea8 0

case 1 4!rint"( 7nT$N ) 0


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;rea8 0e"ault 4

!rint"( 7nInvali In!ut. ) 09

etc+() 09RUN 1 4

$nter a num;er *1 to 1 , 4 5

?I $

RUN 2 4

$nter a num;er *1 to 1 , 4 12

Invali In!ut.

5.=. oto ST#T$%$NT

The goto statement is used to alter the normal sequence of programexecution $y unconditionally transferring control to some part of the program. Thesyntax is :

goto label

/here la$el is an identifier used to la$el the target statement to which thecontrol would $e transferred. +ontrol may $e transferred to any other statementwithin the current function. The target function must $e la$eled followed $y acolon. The syntax is :

label statement ;

+onsider the following program for finding winning team in a cric'et match :

&' Demonstration o" oto statement '& inclu e *st io.+, inclu e *conio.+,

voi main()-

int in / !a8 0clrscr() 0!rint"( $nter t+e In ia s score 4 ) 0scan"( / 6in ) 0


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testcond


control is transferred out of the loop. 9n exit" the program continues with thestatement%x which is immediately after the $ody of the loop.

Note 4 The test condition specified in the while loop should eventually$ecome false at one point of the program" otherwise the loop will $ecome aninfinite loop.

The figure $elow shows the flowchart of while statement :

Entry

alse

True

$ody of the loop

statement%x


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+onsider the following program to find the sum of first n num$ers :

&' Demonstration o" A+ile statement '& inclu e *st io.+, inclu e *conio.+,

voi main()-

int n/ count 1/ sum 0clrscr() 0!rint"( $nter t+e value "or n 4 ) 0scan"( / 6n) 0A+ile(count * n)-

sum sum J count 0count count J 1 0

9!rint"( 7nT+e sum o" "irst num;ers is 4 / n/ sum) 0

etc+() 09

RUN 1 4

$nter t+e value "or n 4 5

T+e sum o" "irst 5 num;ers is 4 15

5.


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testcond


do{

bod* of t'e loop ; } &'ile(test condition) ; statement-x ;

/here the $ody of the loop may have one or more statements. 9n reachingthe do statement" the program proceeds to evaluate the $ody of the loop first. &tthe end of the loop" the test condition in the while statement is evaluated. !f thecondition is true" then the program continues to evaluate the $ody of the looponce again. This process continues as long as the condition is true. /hen thecondition $ecomes false" the loop will $e terminated and the control goes to the

statement%x that appears immediately after the while statement.

The figure $elow shows the flowchart of do . . . while statement :

Entry

True

alse

statement%x

$ody of the loop


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+onsider the following program to find the sum of two num$ers for n num$erof times :

&' Demonstration o" o . . . A+ile statement '& inclu e *st io.+, inclu e *conio.+,

voi main()-

int num1/ num2 0

c+ar c+oice 0clrscr() 0o

-!rint"( 7n$nter t+e "irst num;er 4 ) 0scan"( / 6num1) 0!rint"( 7n$nter t+e secon num;er 4 ) 0scan"( / 6num2) 0!rint"( 7n J / num1/ num2/ num1 J num2) 0""lus+(st in) 0!rint"( 7n7nDo Kou Aant to continue (L&N) 4 ) 0scan"( c / 6c+oice) 0

9 A+ile((c+oice L ) MM (c+oice K )) 0etc+() 0

9

RUN 1 4

$nter t+e "irst num;er 4 2

$nter t+e secon num;er 4 5

2 J 5 F

Do Kou Aant to continue (L&N) 4 K

$nter t+e "irst num;er 4 1

$nter t+e secon num;er 4 2

1 J 2 3

Do Kou Aant to continue (L&N) 4 n


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5.F. "or ST#T$%$NT

The for loop is another entry controlled loop that provides a more conciseloop control structure. The syntax is :

for(exp# ; exp$ ; exp%){

bod* of t'e loop ; }

#efore the first iteration" 4expr15 is evaluated. This is usually used toinitiali,e varia$les for the loop that is used to set the loop control varia$le. The4expr25 is a relational expression that determines when the loop shouldterminate. &fter each iteration of the loop" 4expr 5 is evaluated. This is usuallyused to increment or decrement the loop counters. The $ody of the loop isexecuted repeatedly till the condition in 4expr25 is satisfied. The expressionsmust $e separated $y a semicolon. &ll the expressions are optional. !f 4expr25 isleft out" it is assumed to $e 1 6i.e. True8.

+onsider the pro$lem for printing the num$ers from 1 to n.

The following program demonstrates the a$ove pro$lem :

&' Demonstration o" "or statement '& inclu e *st io.+, inclu e *conio.+,

voi main()-

int i/ n 0clrscr() 0!rint"( $nter t+e value "or n 4 ) 0scan"( / 6n) 0!rint"( 7nT+e num;ers are 4 7n7n ) 0"or(i 1 0 i * n 0 iJJ)

!rint"( 7t / i) 0etc+() 0

9

RUN 1 4

$nter t+e value "or n 4 F

T+e num;ers are 4


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1 2 3 = 5 < F

chapter 05 control

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