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dead code unreachable code

 
veena bijur
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Why System.out.println("hello"); not considered as unreachable code rather considered as dead code?

Thanks.
 
Mansukhdeep Thind
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If my memory serves me correctly, you had asked the exact same question before.
 
Lalit Mehra
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Hello Veena,

It is Because this System.out.println() is compiled though never used.

cheers
 
fred rosenberger
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also, remember that the compiler is not as smart as humans. It doesn't/can't check everything that you can...so it doesn't realize that the break will always be executed
 
Mike Simmons
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As usual, the compiler is obeying the JLS, which specifically requires that an if statement be treated this way. Here's the explanation:
As an example, the following statement results in a compile-time error:

because the statement x=3; is not reachable; but the superficially similar case:

does not result in a compile-time error. An optimizing compiler may realize that the statement x=3; will never be executed and may choose to omit the code for that statement from the generated class file, but the statement x=3; is not regarded as "unreachable" in the technical sense specified here.

The rationale for this differing treatment is to allow programmers to define "flag variables" such as:

and then write code such as:

The idea is that it should be possible to change the value of DEBUG from false to true or from true to false and then compile the code correctly with no other changes to the program text.

In reality, no one uses DEBUG like this anymore; they use proper logging frameworks (which will work better once we get lamdas). But back in the mid-nineties, such code was not too uncommon among the C/C++ programmers that Java was targeting for its audience.
 
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