Labeled break and continue in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of labeled break in nested loops

What is the output of the following Java code using a labeled break?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 1; i <= 3; i++) {
            for (int j = 1; j <= 3; j++) {
                if (i * j > 3) {
                    break outer;
                }
                System.out.print(i * j + " ");
            }
        }
    }
}

A 1 2 3 2 4 6

B 1 2 3

C 1 2 3 2 4 6 3 6 9

D 1 2 3 2

💡 Hint

The labeled break exits the outer loop immediately when the condition is met.

For i=1, j=1 prints 1, j=2 prints 2, j=3 prints 3. For i=2, j=1: 2*1=2 <=3 prints 2, j=2: 2*2=4 >3 break outer. Output: 1 2 3 2

❓ Predict Output

intermediate

Output of labeled continue in nested loops

What is the output of this Java code using a labeled continue?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 1; i <= 2; i++) {
            for (int j = 1; j <= 3; j++) {
                if (j == 2) {
                    continue outer;
                }
                System.out.print(i * j + " ");
            }
        }
    }
}

A 1 3 2 3

B 1 2

C 1 3 3 6

D 1 3 2 4

💡 Hint

The labeled continue skips to the next iteration of the outer loop when j == 2, skipping the print for j=2 and remaining inner iterations.

i=1, j=1: prints 1, j=2: continue outer (skips j=3, print), goes to i=2. i=2, j=1: prints 2, j=2: continue outer (end). Output: 1 2

🔧 Debug

advanced

Identify error with labeled continue usage

What error does this Java code produce?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 0; i < 3; i++) {
            if (i == 1) {
                continue outer;
            }
            System.out.print(i + " ");
        }
    }
}

A Output: 0 2

B Runtime exception: Illegal continue label

C Output: 0 1 2

D Compile-time error: continue label not allowed here

💡 Hint

Labeled continue can be used within the body of its own labeled loop and acts like an unlabeled continue.

No compile error. The labeled continue 'continue outer;' skips the print statement when i==1, equivalent to unlabeled continue. Prints 0 (i=0), skips i=1, prints 2 (i=2).

❓ Predict Output

advanced

Output with nested loops and labeled continue

What is the output of this Java program using labeled continue inside nested loops?

Java

public class Test {
    public static void main(String[] args) {
        outer:
        for (int i = 1; i <= 3; i++) {
            inner:
            for (int j = 1; j <= 3; j++) {
                if (i == j) {
                    continue outer;
                }
                System.out.print(i + "-" + j + " ");
            }
        }
    }
}

A 1-2 1-3 2-1 2-3 3-1 3-2

B 1-2 1-3 2-1 2-3 3-1

C 2-1 3-1 3-2

D 1-2 1-3 2-1 3-1

💡 Hint

The continue outer skips the rest of the inner loop and continues the next iteration of the outer loop when i == j.

i=1 j=1: i==j continue outer (no print, skips j=2,3) -> i=2. i=2 j=1: != print 2-1, j=2== continue outer (skips j=3) -> i=3. i=3 j=1 != print 3-1, j=2 != print 3-2, j=3== continue outer (end). Output: 2-1 3-1 3-2

🧠 Conceptual

expert

Effect of labeled break on loop variables

Consider the following Java code snippet. What is the final value of variable 'count' after execution?

Java

public class Test {
    public static void main(String[] args) {
        int count = 0;
        outer:
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < 5; j++) {
                if (i + j > 5) {
                    break outer;
                }
                count++;
            }
        }
        System.out.println(count);
    }
}

A 14

B 21

C 25

D 10

💡 Hint

The labeled break exits both loops immediately when i + j > 5.

Counts for i=0 all j=0-4 (5), i=1 all j=0-4 (5), i=2 j=0-3 (4, since j=4: 2+4=6>5 break outer). Total 14.