1Z0-811 Exception Handling and Development Tools Practice Question
Given the following try-catch block: try { // some code
} catch (IOException | NumberFormatException e) {e = new IOException("wrapper"); // throw e;
}
Which line causes a compilation error?
⚠ Common exam trap
It's easy for candidates to assume the multi-catch syntax itself is invalid or that the commented `throw` statement causes the error, overlooking the implicit `final` restriction on the exception parameter.
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 assignment to e
In a multi-catch block, the exception parameter `e` is implicitly `final` and cannot be reassigned. The assignment `e = new IOException("wrapper")` violates this rule, causing a compilation error at that line.
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 multi-catch syntax
Why it's wrong here
Multi-catch syntax is valid since Java 7.
- ✓
The assignment to e
Why this is correct
In a multi-catch clause, the exception parameter is implicitly final, so reassignment is not allowed.
- ✗
No error
Why it's wrong here
There is a compilation error at the assignment.
- ✗
The throw statement
Why it's wrong here
The throw statement is syntactically correct, but it would only compile if e were not final.
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Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
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