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.