Common macro to identify a UNIX derived system? (Linux, OSX, BSD, ...)

Question

I wonder why on MacOSX the macro __unix__ is not defined.

Isn't MacOSX a BSD UNIX derivative?

If I define the __unix__ macro in my code, could I have some issues?

In general, when checking the current platform, I prefer to do something like:

#ifdef __unix__
...
#endif

instead of:

#if defined(__unix__) || defined(__APPLE__) || defined(__linux__) || defined(BSD) ...
...
#endif

Could the best option be to define my own macro in a single place? E.g.:

#if defined(__unix__) || defined(__APPLE__) || defined(__linux__) || defined(BSD) ...
#define UNIX_
#endif

If you choose to do that, do not use double underscore in the name. — William Pursell, Oct 27 '11 at 14:36
you'll get more eyes on your problem if you change one of your tags to 'c'. Good luck. — shellter, Oct 27 '11 at 15:29
What sort of things would you surround with your `#ifdef __unix__` or `#ifdef UNIX_`? Perhaps you should be looking for better alternatives to those, like `__has_include()` or `__has_feature()` (depending on compiler support). Or autoconf. — Ken Thomases, Feb 20 '16 at 22:28
See also: [Macro `__unix__` not defined in MacOS X](https://stackoverflow.com/questions/7063303) — hippietrail, Apr 06 '21 at 04:31

score 4 · Answer 1 · answered Oct 27 '11 at 17:53

4

POSIX requires _POSIX_VERSION to be defined in <unistd.h> (also accessible via sysconf(_SC_VERSION)), so try that.

answered Oct 27 '11 at 17:53

ninjalj

Ok, but before including I have to check I am on a POSIX system, otherwise that file will not be available. And should not be a small file to include... – Pietro Oct 28 '11 at 23:59

1 Answers1