#ifdef DEBUG

Hi,
I was wondering in what case the following code defining ASSERT will be processed by gcc :

Code:

---

#ifdef DEBUG
#define ASSERT(x) if(!(x)) { \
  std::cerr << "ASSERT FAILED IN " << __FILE__ << ":" << __LINE__ << endl; \
  abort(); \
}
#else
#define ASSERT(x)
#endif

---

Is it when specifying -g to gcc?

Is ASSERT defined as such same as assert() in assert.h?

Thanks!

Quote:

---

Originally Posted by lehe

Hi,
I was wondering in what case the following code defining ASSERT will be processed by gcc :

---

Whenever DEBUG is defined. gcc will never define it itself (even with -g). You must define it manually.

The built-in assert will trigger by default, unless NDEBUG is defined. Again, the compiler DOES NOT DEFINE THIS FOR YOU. You must define it yourself when building for release.

I'm very much doubting that...

Code:

---

#include <stdio.h>

int main()
{
  #if DEBUG
  printf("Debug\n");
  #else
  printf("No Debug\n");
  #endif
  return 0;
}

---

compiled with gcc -g dbg.c, and then executing will show "No Debug" - at least using gcc-mingw 3.4.

There is a symbol "NDEBUG" that is normally set by the compiler in "NO Debug" mode.

--
Mats

Quote:

---

Originally Posted by matsp

There is a symbol "NDEBUG" that is normally set by the compiler in "NO Debug" mode.

---

I didn't think the compiler ever did that itself. I want to be able to enable asserts in release mode, or disable them for a debug build, if I want. Why should the compiler prevent me?

And anyway, "release" and "debug" are just conveniences -- all that's actually real are the set of compiler switches. The -g flag means symbolic debugging information, not "build for debug." I can ask for optimization and debug at the same time if I want..

Sorry, you are right, I'm wrong. I'm probably confusing this with the "regular build setup in Visual studio", which normally DOES set NDEBUG in the release mode.

--
Mats