1Z0-829 Working with Arrays and Collections Practice Question
A developer creates an unmodifiable list via Collections.unmodifiableList(originalList). Later, originalList is modified by adding an element. Which statement is true?
⚠ Common exam trap
Watch out — candidates often assume `Collections.unmodifiableList()` creates an immutable copy, similar to `List.copyOf()`, but it actually creates a dynamic view that changes with the original list.
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
✓
The unmodifiable list reflects the addition
D is correct because `Collections.unmodifiableList()` returns a view-backed unmodifiable list that reflects any changes made to the original list. The unmodifiable list does not create a snapshot; it wraps the original list, so modifications to the original are visible through the unmodifiable wrapper.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
The unmodifiable list remains unchanged
Why it's wrong here
It reflects changes to the original.
- ✗
The unmodifiable list becomes inconsistent
Why it's wrong here
It remains consistent as a view.
- ✗
An UnsupportedOperationException is thrown
Why it's wrong here
Modifying the original does not affect the view's state directly; exception only on direct modification attempts.
- ✓
The unmodifiable list reflects the addition
Why this is correct
Unmodifiable wrapper is a view; changes to backing list are visible.
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