1Z0-811 Primitives, Strings and Operators Practice Question
Given: short s = 10; s = s + 5; What is the result?
⚠ Common exam trap
Oracle often tests the misconception that arithmetic on smaller numeric types (like `short` or `byte`) stays within that type, when in fact Java promotes them to `int` before the operation, causing a compilation error on assignment back without a cast.
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
✓
Compilation fails: possible lossy conversion from int to short
The expression `s + 5` performs arithmetic on a `short` and an `int` literal, so the result is promoted to `int`. Assigning that `int` back to a `short` variable without an explicit cast causes a compilation error because an `int` may be larger than a `short` (16-bit range), leading to possible lossy conversion. Therefore, option D is correct.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✗
s = 10
Why it's wrong here
No change, but code fails to compile.
- ✗
Runtime exception
Why it's wrong here
Compilation error occurs before runtime.
- ✗
s = 15
Why it's wrong here
Assignment requires cast; code does not compile.
- ✓
Compilation fails: possible lossy conversion from int to short
Why this is correct
s + 5 is int, cannot assign to short.
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