C#.NET — Validity of case labels in switch with variables. int i, j, id = 0; switch (id) { case i: Console.WriteLine("I am in Case i"); break; case j: Console.WriteLine("I am in Case j"); break; }
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AThe compiler will report case i and case j as errors since variables cannot be used in cases.
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BCompiler will report an error since there is no default case in the switch case statement.
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CThe code snippet prints the result as "I am in Case i".
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DThe code snippet prints the result as "I am in Case j".
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EThere is no error in the code snippet.
Answer
Correct Answer: The compiler will report case i and case j as errors since variables cannot be used in cases.
Explanation
Introduction / Context:This question checks the requirement that C# switch case labels be compile-time constants, not variables whose values are only known at runtime.
Given Data / Assumptions:
- Switch discriminant: id (an int).
- Case labels: i and j, both variables.
Concept / Approach:In C#, case labels must be constant expressions (e.g., literals, const fields, or enum members). Using non-const variables as case labels is a compile-time error.
Step-by-Step Solution:
Identify the nature of case labels: i and j are plain int variables, not const. C# requires "case <constant>" for each label. Therefore, the compiler flags both case i and case j as errors.Verification / Alternative check:Replace i or j with const int K = 5; case K: ... and the code would compile. Similarly, enum members are allowed because they are constants.
Why Other Options Are Wrong:
- No default case is not itself an error.
- Printing "I am in Case i" or "Case j" cannot occur because compilation fails.
- "There is no error" contradicts the language rule.
Common Pitfalls:Confusing switch-case rules from other languages or assuming any expression is permitted in labels. C# enforces constants for determinism at compile time.
Final Answer:The compiler will report case i and case j as errors since variables cannot be used in cases.