1Z0-811 Control Flow and Loops Practice Question
Exhibit
Refer to the exhibit.
public class NestedLoop {
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) {
break outer;
}
System.out.println(i + "," + j);
}
}
}
}What is printed by the program?
```java
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 3; j++) {System.out.print(i + "," + j + " ");
if (i == 1) break;
}
}```
⚠ Common exam trap
A common mix-up: candidates assume both loops run to completion without considering that a `break` statement (or a conditional early exit) may terminate the inner loop prematurely, leading them to select Option D or C instead of recognizing the truncated output.
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
The program uses nested for loops. The outer loop runs with i from 0 to 1 (inclusive). For each i, the inner loop runs with j from 0 to 2 (inclusive). However, the inner loop contains a break statement that executes when i equals 1, causing the inner loop to terminate after printing the first pair for i=1 (1,0). Thus, the output is: 0,0 0,1 0,2 1,0. The outer loop then increments i to 2, but the condition i < 2 fails, so the program ends.
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
Why this is correct
The program's output results from the precise control flow achieved by a labelled `break` statement. The outer loop iterates, and for `i=0`, the inner loop executes completely, printing "0,0 0,1 0,2". When `i` becomes `1` and `j` is `0`, the program prints "1,0" and then encounters a `break` statement targeting the *outer* loop label. This immediately terminates both loops, satisfying the constraint for early exit from nested iterations.
- ✗
0,0 0,1 0,2
Why it's wrong here
This would occur if break outer were placed differently.
- ✗
0,0 0,1 0,2 1,0 1,1 1,2 2,0 2,1 2,2
Why it's wrong here
That would happen without the break.
- ✗
0,0 0,1 0,2 1,0 1,1
Why it's wrong here
Break outer stops after i=1,j=1, so only 1,0 is printed before break.
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