_DEBUG definition

[caykahve]

Hi, I want to disable some functions in the release build. This is the fucntion:

void func1(...)
{
#ifdef _DEBUG
func2();
#endif //_DEBUG
}

I still see the outputs of func2() in the release build. My preprocessor definitions for the release build: WIN32;_CONSOLE;NDEBUG debug build: WIN32;_CONSOLE;_DEBUG I don't understand what is wrong. How can i solve the problem? Any help appreciated. Thanks in advance

[toxcct]

caykahve wrote: I still see the outputs of func2() in the release build. i don't understand well. does func2() execute ? what happens when you explicitely write a #undef _DEBUG (at the begining of stdafx.h for example) ?

---

TOXCCT >>> GEII power
[toxcct][VisualCalc]

[caykahve]

Yes func2() executes. I wrote #undef _DEBUG in stdafx.h and it still executes.

[toxcct]

did you try rebuilding the whole project ?

---

TOXCCT >>> GEII power
[toxcct][VisualCalc]

[caykahve]

I always choose "Rebuild Solution" from build menu. i am using Visual Studio .net 2003 BTW.

[toxcct]

what about using DEBUG instead ?

---

TOXCCT >>> GEII power
[toxcct][VisualCalc]

[caykahve]

doesn't work

[David Crow]

Use the /P compiler switch on this file. It will preprocess the file into a file with a .i extension. Open that file and look for func1(). That will let you know if _DEBUG is being seen by the preprocessor or not.

---

"Ideas are a dime a dozen. People who put them into action are priceless." - Unknown

[Bob Stanneveld]

It's hard to believe that your function gets executed. Maybe you call func2 somewhere else? Try using the debugger and set a breakpoint in func2. Then see the callstack for the caller. Behind every great black man...             ... is the police. - Conspiracy brother Blog[^]