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The strange ways of compile time constants

 
Steli Niculescu
Greenhorn
Posts: 14
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This, obviously, does not compile.


This compiles:


This, not so obviously, does not compile :


This DOES COMPILE :


Why ??!
 
Seetharaman Venkatasamy
Ranch Hand
Posts: 5575
Eclipse IDE Java Windows XP
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there is a possibility to b can be false, so local variable x could not be initialize before it get use.


compiler make sure that x will get initialize here. so no error.


here after the exception, code cannot be reached.hence error.


[I am not sure]I think this is because of java compiler(javac) dont have pre-processor
 
Steli Niculescu
Greenhorn
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Yes, I'm not sure either, actually my question was only for the last code sample.
I thought the compiler is sure that the exception will be thrown and the last statement will be unreachable, but it seems that's not true.
 
Lester Burnham
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It may be easy for a human being to spot that the last two code examples are identical, but it's not so easy for a compiler. You could make its analysis of which code will definitely be executed (or which won't) as complex as you want, and it will still miss opportunities for proving that some code can't be reached. It has to draw the line somewhere, and that line seems to be trying to prove the constant-ness of "if" condition values.
 
Wouter Oet
Saloon Keeper
Posts: 2700
IntelliJ IDE Opera
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The compiler allows this to build code like this:

It would be very annoying to get unreachable statement errors if Debug.enabled is set to false.
 
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