Quick Answer
The answer is 0 because the `labeled break in nested loops Java` construct immediately terminates the outer loop when encountered. In this code, the `break outer` statement executes during the first iteration of the inner `for` loop when `j` equals 1, which exits the `while` loop entirely before `i` is ever incremented. This tests your understanding of how labeled breaks differ from unlabeled breaks—an unlabeled break would only exit the inner loop, but a labeled break jumps directly to the end of the specified outer loop. On the Oracle Certified Professional Java SE 17 Developer 1Z0-829 exam, this concept frequently appears in loop control questions, often as a trap where candidates mistakenly assume `i` increments to 1 or 2. The key memory tip: a labeled break bypasses all intermediate loop iterations, so any variable incremented after the inner loop remains unchanged.
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. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. 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.
Given: int i=0; outer: while(i<3) { for(int j=0; j<3; j++) { if(j==1) break outer; } i++; } What is the value of i after the outer loop?
1
Why wrong: Incorrect; i++ is not executed because the outer loop is broken.
2
Why wrong: Incorrect; i never increments.
3
Why wrong: Incorrect; the outer loop does not complete.
0
Correct: i is still 0 when break outer executes.
Answer choices
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
0
The correct answer is D (0) because the `break outer` statement immediately terminates the outer `while` loop when `j` equals 1 during the first iteration of the inner `for` loop. Since `i` is incremented only after the inner loop completes, and the inner loop never finishes its first iteration, `i` remains 0.
Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Answer analysis
For each option: why learners choose it and why it is or isn't the right answer here.
1
Why it's wrong here
Incorrect; i++ is not executed because the outer loop is broken.
2
Why it's wrong here
Incorrect; i never increments.
3
Why it's wrong here
Incorrect; the outer loop does not complete.
0
Why this is correct
Correct: i is still 0 when break outer executes.
Related concept
Read the scenario before looking for a memorised answer.
Common exam traps
The trap here is that candidates often overlook that `i++` is never reached because the labeled break exits the outer loop before the increment executes, leading them to incorrectly assume `i` is 1 or higher.
Detailed technical explanation
The `break outer` statement in Java uses a labeled break to exit an outer loop from within a nested loop. The label `outer` must be placed before the outer loop statement. When `break outer` executes, control jumps to just after the labeled block, skipping any remaining code in both loops, including the increment `i++`. This is distinct from an unlabeled `break`, which would only exit the inner `for` loop.
Key takeaway
Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.
Real-world example
A small business has 20 workstations on the 192.168.1.0/24 network and one public IP from its ISP. The router uses PAT (NAT overload) so all 20 devices share one public address using different source ports. NAT questions test whether you understand the four address terms and which direction each translation applies.
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FAQ
Controlling Program Flow — This question tests Controlling Program Flow — Read the scenario before looking for a memorised answer..
The correct answer is: 0 — The correct answer is D (0) because the `break outer` statement immediately terminates the outer `while` loop when `j` equals 1 during the first iteration of the inner `for` loop. Since `i` is incremented only after the inner loop completes, and the inner loop never finishes its first iteration, `i` remains 0.
Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.
Read the scenario before looking for a memorised answer.
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Variation 1. Given: for(int i=0; i<3; i++) { for(int j=0; j<3; j++) { if(i==1 && j==1) break; } } What is the value of i after the outer loop completes?
easyWhy D: The outer loop variable `i` increments from 0 to 2. When `i` is 1 and `j` is 1, the `break` statement exits only the inner `for` loop, not the outer loop. The outer loop continues to its next iteration, incrementing `i` to 2, then to 3. When `i` becomes 3, the outer loop condition `i < 3` fails, so the loop terminates. Thus, after the outer loop completes, `i` is 3.
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Last reviewed: Jun 25, 2026
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