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:

The variable x is declared as an extern type, prompting the compiler to search for its definition. In this code, the definition is found at the end, where x is assigned a value of 8. Consequently, the value of x is 8.

The code features two do-while loops, known for executing at least once even if the condition is initially false. The printf statement within the outer do-while loop prints the octal number 10, equivalent to the decimal number 8 (%o is used for octal format).

Inside the inner do-while loop, the condition is !-2, which evaluates to 0 (false) in C. Consequently, the program control exits the inner loop. The outer do-while loop's condition is now 0 (false) as well, leading to the program control exiting the outer loop as well.

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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:

n the first iteration of the while loop with the condition i + 1 ? --i : ++j:

• 1. i + 1 evaluates to 3 (True), causing the ternary operator to return --i, resulting in 1.
• 2. In C, 1 is considered true, so the while condition is true, and the printf statement prints 1.

In the second iteration:

• 1. i + 1 evaluates to 2 (True), causing the ternary operator to return --i, resulting in 0.
• 2. In C, 0 is considered false, so the while condition is false, and the program control exits 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:

The default value of a static int variable in C is zero, so the initial value of variable 'i' is 0.

In the first iteration:

• 1. The for loop initializes with ++i, setting 'i' to 1.
• 2. The loop condition, ++i, evaluates to 2 (true), resulting in the printf statement printing 2.
• 3. The loop increments with ++i, updating 'i' to 3.

In the second iteration:

• 1. The loop condition, ++i, evaluates to 4 (true), and the printf statement prints 4.
• 2. Although there is an 'if' condition that is also true, the program control exits the for loop due to the break keyword.

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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:

With an initial value of 1 for variable 'i':

In the first iteration:

• The for loop initializes 'i' to 0.
• The loop condition, 'i = -1', is true (non-zero), resulting in the printf function printing the value of 'i' (-1).
• The if condition is true since 'i' is not equal to 1, causing the program control to exit the for loop due to the break keyword.

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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 a for loop, each part (initialization, condition, and iteration) 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:

In the first iteration:

• 1. Loop initialization: r() = 7
• 2. Loop condition: r() = 6 (true), leading to the printf function printing r() (5).
• 3. Loop increment: r() = 4

In the second iteration:

• 1. Loop condition: r() = 3 (true), resulting in the printf function printing r() (2).
• 2. Loop increment: r() = 1

In the third iteration:

• 1. Loop condition: r() = 0 (false), prompting the program control to exit 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:

A for loop without any body is indeed possible.

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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:

With an initial value of 40 for variable 'i':

In the first iteration, the printf function prints i++ (40).

The do-while condition is (5,4,3,2,1,0), where the comma operator returns 0. As the while condition evaluates to false, the program control exits 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 dots (...) represent a variable number of arguments.

Considering a pointer to the function x, denoted as 'p':

In the first iteration of the for loop:

• 1. Initial value: Nothing (In C, it is optional)
• 2. Loop condition: (*p)(0,1,2,3,4) where p = &_x_
• 3. This evaluates to _x_(0,1,2,3,4) since * and & cancel each other.
• 4. The function returns i+++a, equivalent to i + ++a, resulting in -1 + 1 and finally, 0.
• 5. Although the condition is false, the printf function prints 0, highlighting a quirk in the 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

The output is: 1 1 1 1 1 1.

Note : The for loop executes six times. It's important to note that the continue keyword in the do-while loop brings the program control to 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 increment the char variable 'c', it will follow the sequence: 126, 127, -128, -127, 126, ..., 3, 2, 1, 0. The loop will terminate when 'c' reaches 0.

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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:

The first printf function prints 'c++' and returns 3 to variable 'i'. The for loop executes three times, and the printf function within it prints 0, 1, and 2, respectively.

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