While loop in C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print numbers from 1 to 5 using a while loop.

C

#include <stdio.h>

int main() {
    int i = 1;
    while ([1]) {
        printf("%d\n", i);
        i++;
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

Ai >= 6

Bi > 6

Ci == 6

Di < 6

Attempts:

3 left

💡 Hint

Common Mistakes

Using a condition that never becomes false, causing an infinite loop.

Using the wrong comparison operator.

The while loop continues as long as the condition is true. To print numbers 1 to 5, the condition should be i < 6.

2fill in blank

medium

Complete the code to sum numbers from 1 to 10 using a while loop.

C

#include <stdio.h>

int main() {
    int sum = 0;
    int num = 1;
    while ([1]) {
        sum += num;
        num++;
    }
    printf("Sum is %d\n", sum);
    return 0;
}

Drag options to blanks, or click blank then click option'

Anum < 10

Bnum <= 10

Cnum == 10

Dnum > 10

Attempts:

3 left

💡 Hint

Common Mistakes

Stopping the loop before reaching 10, missing the last number.

Using a condition that never ends.

The loop should run while num is less than or equal to 10 to include 10 in the sum.

3fill in blank

hard

Fix the error in the while loop condition to avoid an infinite loop.

C

#include <stdio.h>

int main() {
    int count = 5;
    while ([1]) {
        printf("%d\n", count);
        count--;
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

Acount > 0

Bcount >= 5

Ccount < 5

Dcount == 5

Attempts:

3 left

💡 Hint

Common Mistakes

Using a condition that never becomes false.

Using equality which causes the loop to run only once.

The loop should continue while count is greater than 0 to avoid infinite looping.

4fill in blank

hard

Fill both blanks to create a while loop that prints even numbers from 2 to 10.

C

#include <stdio.h>

int main() {
    int num = [1];
    while (num [2] 10) {
        printf("%d\n", num);
        num += 2;
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

A2

B<=

C<

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 1 instead of 2.

Using a condition that excludes 10.

Start from 2 and continue while num is less than or equal to 10 to print even numbers up to 10.

5fill in blank

hard

Fill all three blanks to create a while loop that counts down from 10 to 1 and prints each number.

C

#include <stdio.h>

int main() {
    int count = [1];
    while (count [2] 0) {
        printf("%d\n", [3]);
        count--;
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

A0

Bcount

C>

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect start value.

Using wrong comparison operator.

Printing a wrong variable.

Start counting from 10, continue while count is greater than 0, and print the current count.

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