Question 149 of 481
1Z0-811 Arrays and Methods Practice Question
Which TWO methods correctly modify the passed array in place?
⚠ Common exam trap
The 1Z0-811 exam often tests the misconception that reassigning the method parameter (e.g., a = new int[10]) modifies the original array, when in fact it only changes the local reference, leaving the original array untouched.
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
✓
public void zeroOut(int[] a) { for (int i = 0; i < a.length; i++) a[i] = 0; }
It directly modifies each element of the passed array by assigning 0 to a[i] within the loop. In Java, when an array is passed to a method, the method receives a reference to the same array object, so changes to the array's elements are reflected in the caller's array.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
public void zeroOut(int[] a) { for (int i = 0; i < a.length; i++) a[i] = 0; }
Why this is correct
Sets each element to 0, modifying the original array.
- ✓
public void incrementEach(int[] a) { for (int i = 0; i < a.length; i++) a[i]++; }
Why this is correct
Modifies each element in place.
- ✗
public void reset(int[] a) { a = new int[10]; }
Why it's wrong here
Creates a new array local to the method; original unchanged.
- ✗
public void swap(int[] a) { int[] temp = a; a = new int[1]; a[0] = temp[0]; }
Why it's wrong here
Reassigns the reference; original array unchanged.
- ✗
public void setFirst(int[][] a) { a = new int[1][1]; a[0][0] = 5; }
Why it's wrong here
Reassigns the reference, not modifying original.
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Last reviewed: Jun 25, 2026
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