Looping questions in c and answers

Looping questions and answers with explanation for written test exam and interview in c programming language

(1)

What will be output of following c code?

#include<stdio.h>

extern int x;

int main(){

    do{

        do{

             printf("%o",x);

         }

         while(!-2);

    }

    while(0);

    return 0;

}

int x=8;

Explanation

Output: 10

Explanation:

Here variable x is extern type. So it will search the definition of variable x. which is present at the end of the code. So value of variable x =8

There are two do-while loops in the above code.  AS we know do-while executes at least one time even that condition is false.  So program control will reach  at printf statement at it will print octal number 10 which is equal to decimal number 8.

Note: %o is used to print the number in octal format.

In inner do- while loop while condition is ! -2 = 0

In C zero means false.  Hence program control will come out of the inner do-while loop.   In outer do-while loop while condition is 0. That is again false. So program control will also come out of the outer do-while loop.

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(2)

What will be output of following c code?

        

#include<stdio.h>

int main(){

    int i=2,j=2;

    while(i+1?--i:j++)

         printf("%d",i);

    return 0;

}

Explanation

Output: 1

Explanation:

Consider the while loop condition: i + 1 ? -- i : ++j

In first iteration:

i + 1 = 3 (True)

So ternary operator will return -–i i.e. 1

In c 1 means true so while condition is true. Hence printf statement will print 1

In second iteration:

i+ 1 = 2 (True)

So ternary operator will return -–i i.e. 0

In c zero means false so while condition is false. Hence program control will come out of the while loop.

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(3)

What will be output of following c code?

#include<stdio.h>

int main(){

    int x=011,i;

    for(i=0;i<x;i+=3){

         printf("Start ");

         continue;

         printf("End");

    }

    return 0;

}

Explanation

Output: Start Start Start

Explantion:

011 is octal number. Its equivalent decimal value is 9.

So, x = 9

First iteration:

i = 0

i < x i.e. 0 < 9  i.e. if loop condition is true.

Hence printf statement will print: Start

Due to continue keyword program control will come at the beginning of the for loop and value of variable i will be:

i += 3

i = i + 3 = 3

Second iteration:

i = 3

i < x i.e. 3 < 9 i.e. if loop condition is true.

Hence printf statement will print: Start

Due to continue keyword program control will come at the beginning of the for loop and value of variable i will be:

i += 3

i = i + 3 = 6

Third iteration:

i = 3

i < x i.e. 6 < 9 i.e. if loop condition is true.

Hence printf statement will print: Start

Due to continue keyword program control will come at the beginning of the for loop and value of variable i will be:

i += 3

i = i + 3 = 9

fourth iteration:

i = 6

i < x i.e. 9 < 9 i.e. if loop condition is false.

Hence program control will come out of the for loop.

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(4)What will be output of following c code?

#include<stdio.h>

int main(){

    int i,j;

    i=j=2,3;

    while(--i&&j++)

         printf("%d %d",i,j);

    return 0;

}

Explanation

Output: 13

Explanation:

Initial value of variable

i = 2

j = 2

Consider the while condition : --i && j++

In first iteration:

--i && j++

= 1 && 2 //In c any non-zero number represents true.

= 1 (True)

So while loop condition is true. Hence printf function will print value of i = 1 and j = 3 (Due to post increment operator)

In second iteration:

--i && j++

= 0 && 3  //In c zero represents false

= 0  //False

So while loop condition is false. Hence program control will come out of the for loop.

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(5)What will be output of following c code?

#include<stdio.h>

int main(){

    static int i;

    for(++i;++i;++i) {

         printf("%d ",i);

         if(i==4) break;

    }

    return 0;

}

Explanation

Output: 24

Explanation:

Default value of static int variable in c is zero. So, initial value of variable i = 0

First iteration:

For loop starts value: ++i i.e. i = 0 + 1 = 1

For loop condition: ++i i.e. i = 1 + 1 = 2 i.e. loop condition is true. Hence printf statement will print 2

Loop incrimination: ++I i.e. i = 2 + 1 =3

Second iteration:

For loop condition: ++i i.e. i = 3 + 1 = 4 i.e. loop condition is true. Hence printf statement will print 4.

Since is equal to for so if condition is also true. But due to break keyword program control will come out of the for loop.

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(6)What will be output of following c code?

#include<stdio.h>

int main(){

    int i=1;

    for(i=0;i=-1;i=1) {

         printf("%d ",i);

         if(i!=1) break;

    }

    return 0;

}

Explanation

Output: -1

Explanation:

Initial value of variable i is 1.

First iteration:

For loop initial value: i = 0

For loop condition: i = -1 . Since -1 is non- zero number. So loop condition true. Hence printf function will print value of variable i i.e. -1

Since variable i is not equal to 1. So, if condition is true. Due to break keyword program control will come out of the for loop.

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(7)What will be output of following c code?

#include<stdio.h>

int main(){

    for(;;) {

         printf("%d ",10);

    }

    return 0;

}

Explanation

Output: Infinite loop

Explanation:

In for loop each part is optional.

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(8)What will be output of following c code?

        

#include<stdio.h>

int r();

int main(){

    for(r();r();r()) {

         printf("%d ",r());

    }

    return 0;

}

int r(){

    int static num=7;

    return num--;

}

Explanation

Output: 5 2

Explanation:

First iteration:

Loop initial value: r() = 7

Loop condition: r() = 6

Since condition is true so printf function will print r() i.e. 5

Loop incrimination: r() = 4

Second iteration:

Loop condition: r() = 3

Since condition is true so printf function will print r() i.e. 2

Loop incrimination: r() = 1

Third iteration:

Loop condition: r() = 0

Since condition is false so program control will come out of the for loop.

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(9)What will be output of following c code?

        

#include<stdio.h>

#define p(a,b) a##b

#define call(x) #x

int main(){

    do{

         int i=15,j=3;

         printf("%d",p(i-+,+j));

    }

    while(*(call(625)+3));

    return 0;

}

Explanation

Output: 11

Explanation:

First iteration:

p(i-+,+j)

=i-++j   // a##b

=i - ++j

=15 – 4

= 11

While condition is : *(call(625)+ 3)

= *(“625” + 3)

Note: # preprocessor operator convert the operand into the string.

=*(It will return the memory address of character ‘\0’)

= ‘\0’

= 0  //ASCII value of character null character

Since loop condition is false so program control will come out of the for loop.

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(10)

#include<stdio.h>

int main(){

    int i;

    for(i=0;i<=5;i++);

    printf("%d",i)

    return 0;

}

Explanation

Output: 6

Explanation:

It possible for loop without any body.

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(11)What will be output of following c code?

#include<stdio.h>

int i=40;

extern int i;

int main(){

    do{

         printf("%d",i++);

    }

    while(5,4,3,2,1,0);

    return 0;

}

Explanation

Output: 40

Explanation:

Initial value of variable i is 40

First iteration:

printf function will print i++ i.e. 40

do - while condition is : (5,4,3,2,1,0)

Here comma is behaving as operator and it will return 0. So while condition is false hence program control will come out of the for loop.

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(12)What will be output of following c code?

#include<stdio.h>

char _x_(int,...);

int main(){

    char (*p)(int,...)=&_x_;

    for(;(*p)(0,1,2,3,4); )

         printf("%d",!+2);

    return 0;

}

char _x_(int a,...){

    static i=-1;

    return i+++a;

}

Explanation

Output: 0

Explanation:

In c three continuous dot represents variable number of arguments.

p is the pointer to the function _x_

First iteration of for loop:

Initial value: Nothing // In c it is optional

Loop condition: (*p)(0,1,2,3,4)

= *(&_x_)(0,1,2,3,4)  // p = &_x_

= _x_(0,1,2,3,4)    //* and & always cancel to each other

= return i+++a

= return i+ ++a

= return -1 + 1

= 0

Since condition is false. But printf function will print 0. It is bug of c language.

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(13)What will be output of following c code?

#include<stdio.h>

int main(){

    int i;

    for(i=10;i<=15;i++){

         while(i){

             do{

                 printf("%d ",1);

                 if(i>>1)

                      continue;

             }while(0);

             break;

         }

    }

    return 0;

}

Explanation

Output: 1 1 1 1 1 1

For loop will execute six times.

Note: continue keyword in do-while loop bring the program its while condition (while(0)) which is always false.

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(14)How many times this loop will execute?

#include<stdio.h>

int main(){

    char c=125;

    do

         printf("%d ",c);

    while(c++);

    return 0;

}

Explanation

Output: Finite times

Explanation:

If we will increment the char variable c it will increment as:

126,127,-128,-127,126 . . . .   , 3, 2, 1, 0

When variable c = 0 then loop will terminate.    

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(15)What will be output of following c code?

        

#include<stdio.h>

int main(){

    int x=123;

    int i={

         printf("c" "++")

    };

    for(x=0;x<=i;x++){

         printf("%x ",x);

    }

    return 0;

}

Explanation

Output: c++0 1 2 3

Explanation:

First printf function will print: c++ and return 3 to variable i.

For loop will execute three time and printf function will print 0, 1, 2 respectively.

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