C switch-case constants and duplicates: Which compilation error(s) will be reported for this switch that includes a computed case and a literal with the same value? #include <stdio.h> int main() { int a = 5; switch (a) { case 1: printf("First"); case 2: printf("Second"); case 3 + 2: printf("Third"); case 5: printf("Final"); break; } return 0; }
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AThere is no break statement in each case.
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BExpression as in case 3 + 2 is not allowed.
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CDuplicate case case 5:
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DNo error will be reported.
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ENone of the above
Answer
Correct Answer: Duplicate case case 5:
Explanation
Introduction / Context:This question focuses on constant expressions in case labels and the prohibition of duplicate case values. It also checks the misconception that every case must end with a break to compile.
Given Data / Assumptions:
- A case uses
3 + 2. - Another case uses the literal
5. - Several cases have no
break.
Concept / Approach:Case labels must be integer constant expressions, and duplicates are illegal. The expression 3 + 2 is a constant expression that evaluates to 5 at compile time, colliding with the later case 5:. Missing break may cause fall-through, but that is a runtime behavior choice, not a compile error.
Step-by-Step Solution:Evaluate 3 + 2 → 5.Detect duplicate labels: case 3 + 2 and case 5 both equal 5.Compiler emits duplicate case label error.
Verification / Alternative check:Replace one of the labels with a distinct constant (e.g., 6) and compilation succeeds, regardless of missing break statements.
Why Other Options Are Wrong:Expression not allowed — constant expressions are allowed. No break in each case — not a compile error. No error — incorrect due to duplication.
Common Pitfalls:Forgetting that constant expressions are folded at compile time; believing break is mandatory.
Final Answer:Duplicate case case 5: