Question

This simple .c file:

#include <unistd.h>

void test() {
   char string[40];
   gethostname(string,40);
}

… when compiled normally, works fine:

$ cc  -Wall -c -o tmp.o tmp.c
$

… but when compiled in C99 mode, gives a warning:

$ cc -Wall -std=c99 -c -o tmp.o tmp.c 
tmp.c: In function `test':
tmp.c:5: warning: implicit declaration of function `gethostname'
$

The resultant .o file is fine, and linking works. I’d just like to get rid of the warning. I can achieve this in a hacky way, by putting declarations in my own .h file.

What is it about C99 that means the declarations in unistd.h don’t get included?
Can this be overcome, without giving up the niceness of C99?

I see the same problem for other standard libs.

Answer 1

You may need to define some macros in a particluar way to get the prototype for gethostname()

From man gethostname:

Feature Test Macro Requirements for
glibc (see feature_test_macros(7)):

   gethostname(): _BSD_SOURCE || _XOPEN_SOURCE >= 500
   sethostname(): _BSD_SOURCE || (_XOPEN_SOURCE && _XOPEN_SOURCE < 500)

So:

#define _BSD_SOURCE

#include <unistd.h>

void test() {
   char string[40];
   gethostname(string,40);
}

The gory details:

If you don’t specify the -std-c99 option, then features.h (which is implicitly included by unistd.h) will default to setting _BSD_SOURCE in such a way that the prototype for gethostname() gets included. However, specifying -std=c99 causes the compiler to automatically define __STRICT_ANSI__, which in turn causes features.h to not define _BSD_SOURCE, unless you force it with your own feature macro definition (as above).