__STDC__
Why does gcc not predefine the __STDC__ macro now? Almost anything I try to compile is broken by this. kronstadt:/usr/include$ gcc -E -dM -xc /dev/null #define __linux__ 1 #define linux 1 #define __i386__ 1 #define __i386 1 #define __GNUC_MINOR__ 95 #define i386 1 #define __unix 1 #define __unix__ 1 #define __GNUC__ 2 #define __linux 1 #define __ELF__ 1 #define unix 1 kronstadt:/usr/include$ -- Ian Zimmerman, Oakland, California, U.S.A. In his own soul a man bears the source from which he draws all his sorrows and his joys. Sophocles.
Re: __STDC__
On Sun, Jul 09, 2000 at 09:00:55PM -0700, Ian Zimmerman wrote: kronstadt:/usr/include$ gcc -E -dM -xc /dev/null #define __linux__ 1 #define linux 1 #define __i386__ 1 #define __i386 1 #define __GNUC_MINOR__ 95 #define i386 1 #define __unix 1 #define __unix__ 1 #define __GNUC__ 2 #define __linux 1 #define __ELF__ 1 #define unix 1 That doesn't define __STDC__ on Slink either. -- Mark Brown mailto:[EMAIL PROTECTED] (Trying to avoid grumpiness) http://www.tardis.ed.ac.uk/~broonie/ EUFShttp://www.eusa.ed.ac.uk/societies/filmsoc/ pgprsta4gZ7gj.pgp Description: PGP signature
Re: __STDC__
kronstadt:/usr/include$ gcc -E -dM -xc /dev/null #define __linux__ 1 #define linux 1 #define __i386__ 1 #define __i386 1 #define __GNUC_MINOR__ 95 #define i386 1 #define __unix 1 #define __unix__ 1 #define __GNUC__ 2 #define __linux 1 #define __ELF__ 1 #define unix 1 Mark That doesn't define __STDC__ on Slink either. Hmmm. I have actually solved the compilation problem, and it was unrelated; in other words, __STDC__ _is_ defined during normal compiles. Any gcc guru here to explain why it doesn't show in the output of the above command? -- Ian Zimmerman, Oakland, California, U.S.A. In his own soul a man bears the source from which he draws all his sorrows and his joys. Sophocles.
Re: __STDC__
Note I've sent a carbon to the list as I think this is still relevant. itz Hmmm. I have actually solved the compilation problem, and it was itz unrelated; in other words, __STDC__ _is_ defined during normal itz compiles. Any gcc guru here to explain why it doesn't show in itz the output of the above command? John I'm not a guru, but... I can only guess that this is defined John internally by the compiler. Only things that may change between John machines and compiler versions would be defined externally John like this. These defines are normally in a spec file John somewhere. gcc -v will tell you where the spec file is, under John /usr/lib/gcc-lib somewhere IIRC. The useful ones are the John architecture/CPU/binary formats, although we don't normally need John to worry about these unless you're embedding assembly code or John doing some cunning tweak. I know where my specs it, I checked it and this symbol is not mentioned. John I initially thought that maybe the -ansi flag was needed, but John this defines __STRICT_ANSI__. I travelled the same road .. John so gcc's pre-processor defines a whole lot of macros. Yes, but according to the info file spitting _all of them_ out is exactly what the -dM flag should do. -- Ian Zimmerman, Oakland, California, U.S.A. In his own soul a man bears the source from which he draws all his sorrows and his joys. Sophocles.
