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POWhy else case of if doesn't make any error when the if argument is already set as true in java
 

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  1. POWhy else case of if doesn't make any error when the if argument is already set as true in java
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    <p>Why java program doesn't make error in the if else case when the if argument is true. Why doesn't it make any exception. for example, here method1 and method2 do not make any (compilation) error even if it is having unreachable statements, but method3 makes compilation error. First read the code carefully and provide the answer.</p> <pre><code> public int method1() { if(true) { return 1; } else { return 2;//unreachable statement but doesn't make exception } } public int method2() { if(true) { return 1;//unreachable statement but doesn't make exception } else if (true) { return 2;//unreachable statement but doesn't make exception } else { return 3;//unreachable statement but doesn't make exception } } public int method3() { if(true) { return 1;//unreachable statement but doesn't make exception } else if (true) { return 2;//unreachable statement but doesn't make exception } else { return 3;//unreachable statement but doesn't make exception } return 3;//unreachable statement but makes exception } </code></pre> <p>doesn't java support a strict compilation? What is the priciple behind this problem? </p>
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    1. COIf you're interested in how compiler writers think and make decisions on what features to add, you might find Eric Lippert's blog, http://blogs.msdn.com/b/ericlippert/, worth reading. C#, not Java, but still interesting from a compiler development perspective.
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    2. COThis seems a reasonable question with a single factual answer. Closing it seems rather trigger-happy.
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      1. USAndy Thomas
 

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