Question 129 of 519
Controlling Program FlowhardMultiple ChoiceObjective-mapped

Unlabeled Break in Inner Loop — Hard Example | Oracle Certified Professional Java SE 17 Explained

This 1Z0-829 practice question tests your understanding of controlling program flow. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. A key principle to apply: unlabeled break. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.

Exhibit

Refer to the exhibit.
public class LoopTest {
    public static void main(String[] args) {
        outer: for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (i == 1 && j == 1) continue outer;
                System.out.print(i + "," + j + " ");
            }
        }
    }
}

What is the output?

Exhibit

Refer to the exhibit.
public class LoopTest {
    public static void main(String[] args) {
        outer: for (int i = 0; i < 3; i++) {
            for (int j = 0; j < 3; j++) {
                if (i == 1 && j == 1) continue outer;
                System.out.print(i + "," + j + " ");
            }
        }
    }
}

Answer choices

Why each option matters

Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

0,0 0,1 0,2 1,0 2,0 2,1 2,2

The code uses an unlabeled 'break;' inside the inner loop when i==1 and j==0. This break exits only the inner loop, not the outer loop. Therefore, for i=1, after printing '1,0', the inner loop terminates and the outer loop continues to i=2, where j iterates 0,1,2 printing '2,0 2,1 2,2'. For i=0, j prints all three values. Thus the output is '0,0 0,1 0,2 1,0 2,0 2,1 2,2'.

Key principle: unlabeled break

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

  • 0,0 0,1 0,2 1,0 1,1 1,2 2,0 2,1 2,2

    Why it's wrong here

    Incorrect. This option includes all combinations from i=0 to 2 and j=0 to 2, but the labeled break when i=1 and j=0 exits the outer loop, so combinations with i=1 and j>0 are not printed.

  • 0,0 0,1 0,2 1,0

    Why it's wrong here

    Incorrect. Although the text matches the correct output, this option is marked as wrong due to a duplication error. The actual correct output is given in option C.

  • 0,0 0,1 0,2 1,0 2,0 2,1 2,2

    Why this is correct

    Correct. The output correctly reflects the labeled break: i=0 prints all j, i=1 prints only j=0 then breaks, i=2 prints all j, resulting in '0,0 0,1 0,2 1,0 2,0 2,1 2,2'.

    Related concept

    unlabeled break

  • 0,0 1,0 2,0

    Why it's wrong here

    Incorrect. This option only prints combinations with j=0, missing the iterations where i=0 and j=1,2 and i=2 and j=1,2.

Common exam traps

Common exam trap: answer the scenario, not the keyword

Candidates often confuse unlabeled break with labeled break. An unlabeled break exits only the innermost enclosing loop, while a labeled break exits the loop marked by the label. Here, the break is unlabeled, so after '1,0' the outer loop continues.

Trap categories for this question

  • Command / output trap

    Incorrect. Although the text matches the correct output, this option is marked as wrong due to a duplication error. The actual correct output is given in option C.

Detailed technical explanation

How to think about this question

The labeled break statement in Java allows breaking out of an outer loop from within a nested loop. Under the hood, the JVM uses a jump instruction to the target label, skipping all remaining iterations of the outer loop. This is useful in scenarios like searching a 2D array for a specific value and stopping all loops once found, avoiding unnecessary iterations.

KKey Concepts to Remember

  • unlabeled break
  • nested loops

TExam Day Tips

  • Watch for words such as best, first, most likely and least administrative effort.
  • Review why wrong options are wrong, not only why the correct option is correct.

Key takeaway

unlabeled break

Real-world example

How this comes up in practice

A practitioner preparing for the 1Z0-829 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. unlabeled break Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.

What to study next

Got this wrong? Here's your next step.

Review unlabeled break, then practise related 1Z0-829 questions on the same topic to reinforce the concept.

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FAQ

Questions learners often ask

What does this 1Z0-829 question test?

Controlling Program Flow — This question tests Controlling Program Flow — unlabeled break.

What is the correct answer to this question?

The correct answer is: 0,0 0,1 0,2 1,0 2,0 2,1 2,2 — The code uses an unlabeled 'break;' inside the inner loop when i==1 and j==0. This break exits only the inner loop, not the outer loop. Therefore, for i=1, after printing '1,0', the inner loop terminates and the outer loop continues to i=2, where j iterates 0,1,2 printing '2,0 2,1 2,2'. For i=0, j prints all three values. Thus the output is '0,0 0,1 0,2 1,0 2,0 2,1 2,2'.

What should I do if I get this 1Z0-829 question wrong?

Review unlabeled break, then practise related 1Z0-829 questions on the same topic to reinforce the concept.

What is the key concept behind this question?

unlabeled break

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Last reviewed: Jun 25, 2026

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This 1Z0-829 practice question is part of Courseiva's free Oracle certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the 1Z0-829 exam.