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.

Practice

1. What does a while loop do in C programming?

easy

A. Repeats a block of code while a condition is true

B. Executes code only once regardless of condition

C. Runs code a fixed number of times without checking condition

D. Stops the program immediately

Solution

1. Step 1: Understand the purpose of a while loop

   A while loop runs repeatedly as long as its condition remains true.

2. Step 2: Compare options with the definition

   Only Repeats a block of code while a condition is true correctly describes this behavior.

3. Final Answer:

   Repeats a block of code while a condition is true -> Option A

4. Quick Check:

   While loop = repeat while condition true [OK]

Hint: While loops run only if condition is true before each repeat [OK]

Common Mistakes:

• Thinking while loop runs at least once even if condition is false
• Confusing while loop with for loop fixed iterations
• Believing while loop stops program immediately

2. Which of the following is the correct syntax for a while loop in C?

easy

A. while (condition) { /* code */ }

B. while condition { /* code */ }

C. while [condition] { /* code */ }

D. while: (condition) { /* code */ }

Solution

1. Step 1: Recall C while loop syntax

   The condition must be inside parentheses and the code block inside braces.

2. Step 2: Check each option

   Only while (condition) { /* code */ } uses correct parentheses and braces syntax.

3. Final Answer:

   while (condition) { /* code */ } -> Option A

4. Quick Check:

   Correct syntax uses parentheses and braces [OK]

Hint: Remember: while(condition) with parentheses and braces [OK]

Common Mistakes:

• Omitting parentheses around condition
• Using square brackets instead of parentheses
• Adding colon after while keyword

3. What is the output of this C code?

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

medium

A. 0 1 2 3

B. 1 2 3 4

C. 3 2 1

D. 1 2 3

Solution

1. Step 1: Trace the loop variable values

   i starts at 1, prints 1, then increments to 2, prints 2, then 3, and stops after printing 3 because i becomes 4 and condition fails.

2. Step 2: Confirm printed output

   The printed numbers are "1 2 3 " with spaces after each number.

3. Final Answer:

   1 2 3 -> Option D

4. Quick Check:

   Prints 1 to 3 with spaces [OK]

Hint: Check loop start, end condition, and increments carefully [OK]

Common Mistakes:

• Including 4 in output by mistake
• Starting from 0 instead of 1
• Printing numbers in reverse order

4. Identify the error in this code snippet:

int i = 0;
while (i < 5)
    printf("%d ", i);

medium

A. Syntax error due to missing braces

B. Missing semicolon after while condition

C. Infinite loop because i is never updated

D. Variable i is not declared

Solution

1. Step 1: Analyze loop behavior

   Variable i starts at 0 and condition is i < 5, but i is never incremented inside the loop.

2. Step 2: Understand consequence

   Since i never changes, the condition stays true forever, causing an infinite loop.

3. Final Answer:

   Infinite loop because i is never updated -> Option C

4. Quick Check:

   Loop variable must change to avoid infinite loop [OK]

Hint: Always update loop variable inside while loop [OK]

Common Mistakes:

• Assuming loop ends without variable update
• Thinking missing braces cause syntax error here
• Ignoring infinite loop possibility

5. How can you modify this code to print all even numbers from 2 to 10 using a while loop?

int num = 1;
while (num <= 10) {
    printf("%d ", num);
    num++;
}

hard

A. Use a for loop instead of while loop

B. Initialize num to 2 and increment by 2 inside the loop

C. Initialize num to 0 and increment by 1 inside the loop

D. Keep num at 1 and print only if num % 2 == 0 without incrementing

Solution

1. Step 1: Set starting point and increment for even numbers

   Start num at 2 (first even number) and increase by 2 each time to get only even numbers.

2. Step 2: Modify loop accordingly

   Change initialization to num = 2 and inside loop do num += 2 instead of num++.

3. Final Answer:

   Initialize num to 2 and increment by 2 inside the loop -> Option B

4. Quick Check:

   Start at 2, add 2 each time for evens [OK]

Hint: Start at first even, add 2 each loop to get evens [OK]

Common Mistakes:

• Not incrementing num causing infinite loop
• Printing odd numbers by starting at 1
• Using for loop when question asks for while loop