Java keywords and for-loop tracing: identify the compile-time outcome of using “signed”. public class Test { public static void main(String[] args) { signed int x = 10; for (int y = 0; y < 5; y++, x--) System.out.print(x + ", "); } }
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A10, 9, 8, 7, 6,
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B9, 8, 7, 6, 5,
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CCompilation fails.
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DAn exception is thrown at runtime.
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EEmpty output
Answer
Correct Answer: Compilation fails.
Explanation
Introduction / Context: The snippet checks knowledge of Java’s primitive type system and reserved keywords. Unlike C/C++, Java does not use the “signed” or “unsigned” modifiers. All integer primitives (byte, short, int, long) are signed by definition except char, which is unsigned.
Given Data / Assumptions:
- Declaration attempts: signed int x = 10;
- Loop would otherwise print a countdown of x in tandem with y.
Concept / Approach: The presence of “signed” in a variable declaration is not valid Java syntax. The compiler flags this with an error such as “not a statement” or “illegal start of type,” preventing any bytecode generation or execution.
Step-by-Step Solution:
Compilation step parses “signed int x = 10;”.Java grammar has no modifier “signed” → syntax error.Build fails; the program never runs.Verification / Alternative check: Replace with int x = 10; to compile and run. Then the output would be “10, 9, 8, 7, 6,” because x is decremented after each print due to the for-loop’s update expression.
Why Other Options Are Wrong:
- Numeric outputs assume successful compilation and execution.
- Runtime exception presupposes the program ran.
- Empty output is not specific; the correct outcome is a compile-time failure.
Common Pitfalls: Porting C/C++ idioms to Java without adjusting for language differences; assuming signedness modifiers exist in Java.
Final Answer: Compilation fails.