While loop in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a simple while loop

What is the output of this C program?

C

#include <stdio.h>

int main() {
    int i = 1;
    while (i <= 3) {
        printf("%d ", i);
        i++;
    }
    return 0;
}

A 0 1 2 3

B 1 2 3 4

C 1 2 3

D 1 2

💡 Hint

Look at the condition and how the variable i changes inside the loop.

The loop starts with i=1 and runs while i is less than or equal to 3. It prints i and then increases it by 1. So it prints 1, 2, and 3.

❓ Predict Output

intermediate

While loop with break statement

What will this program print?

C

#include <stdio.h>

int main() {
    int i = 0;
    while (1) {
        if (i == 3) {
            break;
        }
        printf("%d ", i);
        i++;
    }
    return 0;
}

A 0 1 2

B 1 2 3

C 0 1 2 3

D 0 1 2 3 4

💡 Hint

The loop runs forever until the break condition is met.

The loop prints i starting from 0. When i reaches 3, the break stops the loop before printing 3.

🔧 Debug

advanced

Identify the error in this while loop

What error does this code produce when compiled or run?

C

#include <stdio.h>

int main() {
    int i = 0;
    while (i < 5)
        printf("%d ", i);
        i++;
    return 0;
}

A Infinite loop printing 0 forever

B Runtime error: segmentation fault

C Compilation error: missing semicolon

D Prints 0 1 2 3 4 and exits normally

💡 Hint

Check which statements are inside the loop and which are not.

Only the printf is inside the loop because there are no braces. The i++ is outside, so i never changes inside the loop, causing an infinite loop printing 0.

🧠 Conceptual

advanced

While loop behavior with post-increment

What is the value of variable i after this loop finishes?

C

#include <stdio.h>

int main() {
    int i = 0;
    while (i++ < 5) {
        // empty body
    }
    printf("%d", i);
    return 0;
}

A 0

B 5

C 4

D 6

💡 Hint

Remember that i++ returns the value before incrementing.

The condition i++ < 5 uses post-increment: it checks the current value of i against 5, then increments i regardless. The loop runs 5 times (when i=0 to 4), setting i to 1-5. When i=5, 5 < 5 is false, but i increments to 6 before exiting. So i is 6.

🚀 Application

expert

Count digits in an integer using while loop

What is the output of this program?

C

#include <stdio.h>

int main() {
    int n = 12345, count = 0;
    while (n > 0) {
        n = n / 10;
        count++;
    }
    printf("%d", count);
    return 0;
}

A 0

B 5

C 6

D 4

💡 Hint

Each division by 10 removes one digit from the number.

The loop divides n by 10 until n becomes 0. Each division removes one digit. The count increments each time, so it counts digits.