Continue statement - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of loop with continue

What is the output of this C program?

C

#include <stdio.h>

int main() {
    for (int i = 1; i <= 5; i++) {
        if (i == 3) {
            continue;
        }
        printf("%d ", i);
    }
    return 0;
}

A 1 2 4 5

B 2 3 4 5

C 1 2 3 4 5

D 1 2 3 4

💡 Hint

Remember that continue skips the rest of the loop body for the current iteration.

When i equals 3, the continue statement skips the printf call, so 3 is not printed.

🧠 Conceptual

intermediate

Effect of continue in nested loops

In the following nested loops, how many times will the inner printf execute?

C

#include <stdio.h>

int main() {
    int count = 0;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            if (j == 1) {
                continue;
            }
            count++;
        }
    }
    printf("%d", count);
    return 0;
}

A 0

B 6

C 3

D 9

💡 Hint

Count how many times the inner loop skips printing when j == 1.

For each i, j runs 0 to 2. When j == 1, continue skips count++. So count++ runs for j=0 and j=2, two times per i. With i from 0 to 2 (3 times), total is 3*2=6.

🔧 Debug

advanced

Identify the error caused by continue

What error will this code produce when compiled or run?

C

#include <stdio.h>

int main() {
    int i = 0;
    while (i < 5) {
        i++;
        if (i == 3) {
            continue;
        }
        printf("%d ", i);
    }
    return 0;
}

A Prints: 1 2 4 5

B Compilation error

C Infinite loop

D Prints: 1 2 3 4 5

💡 Hint

Check how i is incremented and when continue is called.

i is incremented before continue, so loop progresses normally. When i == 3, continue skips printf, so 3 is not printed. Output is 1 2 4 5.

📝 Syntax

advanced

Which option causes a syntax error?

Which of the following code snippets will cause a syntax error due to incorrect use of continue?

A do { if (0) continue; } while(0);

B for (int i = 0; i < 5; i++) { if (i == 2) continue; printf("%d", i); }

C while (0) { continue; }

D if (1) { continue; }

💡 Hint

Remember continue can only be used inside loops.

Option D uses continue inside an if statement that is not inside any loop, causing a syntax error.

🚀 Application

expert

Count numbers skipping multiples of 4

What is the output of this program that uses continue to skip multiples of 4?

C

#include <stdio.h>

int main() {
    int sum = 0;
    for (int i = 1; i <= 10; i++) {
        if (i % 4 == 0) {
            continue;
        }
        sum += i;
    }
    printf("%d", sum);
    return 0;
}

A 40

B 50

C 43

D 55

💡 Hint

Add numbers from 1 to 10 but skip those divisible by 4.

Numbers 4 and 8 are skipped. Sum of 1 to 10 is 55. Subtract 4 and 8 gives 55 - 12 = 43.

Practice

1. What does the continue statement do inside a loop in C?

easy

A. Skips the rest of the current loop iteration and moves to the next iteration

B. Stops the entire loop immediately

C. Exits the current function

D. Restarts the program from the beginning

Solution

1. Step 1: Understand the role of continue

   The continue statement does not stop the loop but skips the remaining code in the current iteration.

2. Step 2: Effect on loop execution

   After continue, the loop immediately proceeds to the next iteration, ignoring any code below it in the current cycle.

3. Final Answer:

   Skips the rest of the current loop iteration and moves to the next iteration -> Option A

4. Quick Check:

   continue skips current iteration [OK]

Hint: Remember: continue skips current loop step, not whole loop [OK]

Common Mistakes:

• Confusing continue with break
• Thinking continue exits the function
• Assuming continue restarts the program

2. Which of the following is the correct syntax to use continue inside a for loop in C?

easy

A. continue;

B. continue()

C. continue{}

D. continue[]

Solution

1. Step 1: Recall correct statement syntax

   In C, continue is a statement and must end with a semicolon.

2. Step 2: Identify valid syntax

   Only continue; is valid; parentheses, braces, or brackets are incorrect.

3. Final Answer:

   continue; -> Option A

4. Quick Check:

   Statement ends with semicolon [OK]

Hint: Statements end with semicolon; continue is no exception [OK]

Common Mistakes:

• Adding parentheses after continue
• Using braces or brackets with continue
• Omitting the semicolon

3. What is the output of the following C code?

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    printf("%d ", i);
}

medium

A. 1 2 3 4 5

B. 1 2 4 5

C. 1 2 3 4

D. 2 3 4 5

Solution

1. Step 1: Analyze loop iterations

   The loop runs i from 1 to 5. When i equals 3, the continue statement skips printing.

2. Step 2: Determine printed values

   Values 1, 2, 4, and 5 are printed with spaces. The number 3 is skipped.

3. Final Answer:

   1 2 4 5 -> Option B

4. Quick Check:

   continue skips printing 3 [OK]

Hint: Skip printing when continue triggers, others print normally [OK]

Common Mistakes:

• Including the skipped number in output
• Confusing continue with break
• Ignoring the loop condition

4. Identify the error in this code snippet using continue:

int i = 0;
while (i < 5) {
    if (i == 2)
        continue;
    printf("%d ", i);
    i++;
}

medium

A. Syntax error: continue cannot be used in while loops

B. The printf statement is unreachable

C. The loop will run infinitely because 'i' is not incremented when i == 2

D. The variable 'i' is not declared properly

Solution

1. Step 1: Understand loop flow with continue

   When i equals 2, continue skips the rest of the loop body, so i++ is not executed.

2. Step 2: Consequence of skipping i++

   Since i is not incremented at i == 2, the loop condition remains true forever, causing an infinite loop.

3. Final Answer:

   The loop will run infinitely because 'i' is not incremented when i == 2 -> Option C

4. Quick Check:

   continue skips i++ causing infinite loop [OK]

Hint: Always increment loop variable before continue to avoid infinite loops [OK]

Common Mistakes:

• Thinking continue causes syntax error
• Assuming printf is unreachable
• Ignoring loop variable increment

5. How can you modify this code to print only odd numbers from 1 to 10 using continue?

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

hard

A. Add if (i == 2) continue; before printf

B. Add if (i % 2 != 0) continue; before printf

C. Replace printf with if (i % 2 != 0) printf("%d ", i);

D. Add if (i % 2 == 0) continue; before printf

Solution

1. Step 1: Identify condition to skip even numbers

   Even numbers satisfy i % 2 == 0. Using continue here skips printing even numbers.

2. Step 2: Place continue before printf

   Adding if (i % 2 == 0) continue; before printf skips printing even numbers, so only odd numbers print.

3. Final Answer:

   Add if (i % 2 == 0) continue; before printf -> Option D

4. Quick Check:

   continue skips even numbers, prints odd [OK]

Hint: Skip even numbers with continue to print only odds [OK]

Common Mistakes:

• Skipping odd numbers instead of even
• Using continue after printf
• Incorrect condition for even numbers