Java switch with constant vs non-constant case labels and fall-through: what happens here? public class Switch2 { final static short x = 2; public static int y = 0; // not constant expression public static void main(String [] args) { for (int z…

Question

Java switch with constant vs non-constant case labels and fall-through: what happens here?  public class Switch2 {  final static short x = 2;  public static int y = 0; // not constant expression  public static void main(String [] args) {  for (int z…

Accepted Answer

Introduction / Context: This question tests the requirement that case labels in a switch must be constant expressions known at compile time, and also hints at fall-through behavior (though compilation never gets that far here). Given Data / Assumptions: x is final static short x = 2 → a compile-time constant. y is public static int y = 0 → not final, thus not a constant expression. Switch variable is the loop variable z. Concept / Approach: Java requires each case label to be a constant expression. case y: uses a non-final variable; the compiler cannot fold it into a constant, so it rejects the code at that line before execution begins. Step-by-Step Solution: Check label at Line 11: case y: ← invalid because y is not a constant expression. Compiler raises an error referencing Line 11 (or similar wording). Program does not compile; no runtime behavior occurs. Verification / Alternative check: Change y to final static int y = 0; and the code will compile. Then, with no breaks, output would fall through for each matching case. Why Other Options Are Wrong: Any output assumes successful compilation. Failing at Line 12 would require an issue with x-1, which is valid since x is final and known at compile time. Common Pitfalls: Forgetting the final requirement; confusing runtime variable values with compile-time constants; ignoring fall-through only after a successful compile. Final Answer: Compilation fails at line 11.

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