Question 450 of 481
1Z0-811 Arrays and Methods Practice Question
Which two statements about passing arrays to methods are correct? (Select two.)
⚠ Common exam trap
Many candidates confuse 'passing by reference' with 'passing the reference by value' — they think reassigning the parameter inside the method will update the caller's variable, but Java always passes references by value, so the original reference is unchanged.
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
✓
Modifying the array elements inside the method affects the original array.
In Java, arrays are objects, and when you pass an array to a method, you pass a copy of the reference to the array. This means the method can modify the contents of the array (e.g., change element values), and those changes are reflected in the original array because both the caller and the method refer to the same array object in heap memory.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
Passing an array is the same as passing each element individually.
Why it's wrong here
Passing an array passes a reference; passing each element individually would create copies of element values.
- ✗
The method can change the size of the original array.
Why it's wrong here
Array size is immutable once created; the method cannot change it.
- ✓
Modifying the array elements inside the method affects the original array.
Why this is correct
Since the method has a reference to the same array object, changes to elements are visible to the caller.
- ✗
If the method assigns a new array to the parameter, the original reference is updated outside the method.
Why it's wrong here
Assigning a new array to the parameter only changes the local copy of the reference; the caller's reference remains unchanged.
- ✓
The array reference is passed by value.
Why this is correct
The reference value is copied, so the method receives a copy of the reference to the same array object.
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Last reviewed: Jul 4, 2026
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