1Z0-829 Controlling Program Flow Practice Question
Exhibit
Refer to the exhibit.
```
int count = 0;
for (int i = 0; i < 5; i++) {
if (i == 3) {
break;
}
count++;
}
System.out.println(count);
```What is the output?
⚠ Common exam trap
The trap here is that candidates often miscount the number of iterations by including the value where the condition becomes false, leading to an off-by-one error, or they confuse the loop's upper bound (3) with the number of iterations.
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
✓
3
3 because the code uses a `for` loop that increments `i` from 0 to 4, but the loop body only executes when `i < 3` due to the condition `i < 3`. The loop runs for `i = 0, 1, 2` (three iterations), and each time `count` is incremented by 1, resulting in `count = 3`. The final value of `count` is printed.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
3
Why this is correct
Loop runs for i=0,1,2 then breaks.
- ✗
0
Why it's wrong here
count is incremented.
- ✗
4
Why it's wrong here
Break at i=3 prevents increment.
- ✗
5
Why it's wrong here
Loop exits early.
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