Question 425 of 481
1Z0-811 Arrays and Methods Practice Question
Exhibit
Refer to the exhibit.
public class ArrayTest {
public static void main(String[] args) {
int[] arr = new int[3];
arr[0] = 5;
arr[1] = 10;
arr[2] = 15;
modify(arr);
System.out.println(arr[0]);
}
public static void modify(int[] a) {
a = new int[3];
a[0] = 100;
}
}What is printed when the main method runs?
⚠ Common exam trap
Oracle often tests the distinction between modifying an object's state (which affects the caller) versus reassigning the reference (which does not), and the trap here is that candidates mistakenly think reassigning the parameter changes the original array, leading them to choose option D.
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
✓
5
The method `modifyArray` receives a reference to the array, and the assignment `arr = new int[]{100};` changes the local reference `arr` to point to a new array object. The original array in `main` remains unchanged, so `arr[0]` still prints 5. Java passes object references by value, meaning the reference itself is copied, and reassigning the local reference does not affect the caller's reference.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
5
Why this is correct
The original array is not modified; the method reassigned the reference.
- ✗
0
Why it's wrong here
Not the initial value.
- ✗
Compilation error: cannot assign new array to parameter
Why it's wrong here
Parameters can be reassigned inside the method.
- ✗
100
Why it's wrong here
The modification applies to the new local array, not the original.
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Last reviewed: Jun 30, 2026
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