Really newbie question about #ifdef _DEBUG

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3.00/5 (2 votes)

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Ok,

C++

#ifdef _DEBUG

SuperSecretVariable = 0;

#endif

Will SuperSecretVariable = 0; exist in any way in the release build assuming _DEBUG is not defined in the release configuration? Or in other words is it the equivalent of commented-out code?

Thanks.

:Mor-Ron

Posted 1-Nov-13 12:24pm

Ron Anders

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OriginalGriff · Answer 1 · 2013-11-01T12:44:00

Solution 2

#ifdef is a preprocessor directive - which means it is processed before your source code is passed to the actual compiler, and the #if (and all other #directives) are handled then and there and the substitutions made in the source at that point.
So if the test fails, the code is never passed anywhere and thus can't appear in the final output in any form.

Posted 1-Nov-13 12:44pm

OriginalGriff

Comments

Ron Anders 1-Nov-13 18:53pm

That's what I thought but in this instance I had to be certain.

Excellent.

Thank you all and have a super weekend.

Garth J Lancaster · Answer 2 · 2013-11-01T12:38:00

Solution 1

the answers you seek are no, and yes

(assuming there's no other place you've defined 'SuperSecretVariable' )

'g'

Posted 1-Nov-13 12:38pm

Garth J Lancaster

Comments

Ron Anders 1-Nov-13 18:41pm

Thanks. I realize that.

Carry on all. :-)