Which one of the following lists contains only valid Java language keywords (no identifiers or non-Java tokens)?
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Aclass, if, void, long, Int, continue
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Bgoto, instanceof, native, finally, default, throws
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Ctry, virtual, throw, final, volatile, transient
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Dstrictfp, constant, super, implements, do
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Ebyte, break, assert, switch, include
Answer
Correct Answer: goto, instanceof, native, finally, default, throws
Explanation
Introduction / Context:Knowing Java’s keyword set helps avoid using reserved words as identifiers and recognizes special meaning in code. Trick options often include almost-correct items (e.g., Int vs. int) or non-Java words.
Given Data / Assumptions:
- We must select a list where every token is a Java keyword.
Concept / Approach:Valid keywords include goto (reserved but unusable), instanceof, native, finally, default, throws. Non-keywords: Int (wrong case), virtual (not in Java), constant (not a keyword), include (not in Java). assert is a keyword since Java 1.4, but option E includes include, invalidating the entire set.
Step-by-Step Solution:
Option B → all six are keywords → correct.Option A → Int is not a keyword; should be lowercase int.Option C → virtual is not a Java keyword.Option D → constant is not a keyword (const exists but is reserved, not constant).Option E → include is not a Java keyword (C/C++ preprocessor concept).Verification / Alternative check:Consult any official list or attempt to compile variables named after these tokens.
Why Other Options Are Wrong:Each contains at least one non-keyword.
Common Pitfalls:Case sensitivity (Int vs. int) and language crossover terms like include or virtual.
Final Answer:goto, instanceof, native, finally, default, throws