Analyze this Java switch with fall-through and erroneous case label (letter l instead of 1). What is the compile-time result? public class SwitchTest { public static void main(String[] args) { System.out.println("value =" + switchIt(4)); } public static int switchIt(int x) { int j = 1; switch (x) { case l: j++; case 2: j++; case 3: j++; case 4: j++; case 5: j++; default: j++; } return j + x; } }
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Avalue = 2
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Bvalue = 4
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Cvalue = 6
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Dvalue = 8
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ECompilation fails.
Answer
Correct Answer: Compilation fails.
Explanation
Introduction / Context:This question mixes switch fall-through with an intentional typographical error in a case label to ensure you can spot invalid identifiers in constant labels.
Given Data / Assumptions:
- The first case label is
case l:using the letterl(ell), not the digit1. - Case labels in Java must be constant expressions of the correct type (e.g., integer literals) and resolvable at compile time.
Concept / Approach:l is an undefined identifier and not a constant integer literal. Therefore, the switch statement contains an invalid case label, causing compilation to fail before any runtime behavior is possible.
Step-by-Step Solution:Type-check case labels: l is not a valid constant expression.The compiler reports an error for the first case and stops.No output is produced.
Verification / Alternative check:Change case l: to case 1: and re-run; then evaluate fall-through to compute the final value.
Why Other Options Are Wrong:They all assume successful compilation and a numeric result.
Common Pitfalls:Misreading the letter l as the digit 1.
Final Answer:Compilation fails.