linux - Why does C99 complain about storage sizes?

Question

This is some code I'm compiling on Linux:

#include <net/if.h>

int main() {
  struct ifreq ifr;
}

gcc test.c is fine.

gcc -std=gnu99 test.c is fine.

gcc -std=c99 test.c fails with the following error:

test.c: In function ‘main’:
test.c:4:16: error: storage size of ‘ifr’ isn’t known

What's different about C99 that it doesn't like the definition of struct ifreq in Linux?

Answer 1

It's a chain of consequences of preprocessing and GNU C vs C99.

First up, net/if.h:

1. net/if.h includes features.h
2. Later on, it defines struct ifreq inside a #ifdef __USE_MISC block.

So:

1. What is __USE_MISC? -- it is stuff common to BSD and System V
2. Is it defined at this point? -- We need to check that out in features.h

So now, features.h:

1. When you use --std=c99 GCC by default defines __STRICT_ANSI__ (since thats what C99 is)
2. While preprocessing features.h, when __STRICT_ANSI__ is on, the BSD and System V features don't kick in. i.e. __USE_MISC is left undefined.

Back up to net/if.h: struct ifreq does not even exist after preprocessing! Therefore, the complaint about storage size.

You can catch the whole story by doing:

vimdiff <(cpp test.c --std=c99 -dD) <(cpp test.c --std=gnu99 -dD)

or diff'ing them in any other way (like diff --side-by-side) instead of vimdiff.

If you want this to cleanly compile with -std=c99, you must consider the inclusion of the _DEFAULT_SOURCE feature test macro (for glibc versions >= 2.19; for older glibc versions, use either _BSD_SOURCE or _SVID_SOURCE) so that the required functionality is enabled on top of what is offered by C99.