After importing unistd.h, compiler states that sbrk() is an implicit declaration. Why is this?

Question

I'm trying to implement malloc on CentOS, but I keep getting the error:

malloc.c: In function ‘malloc’:
malloc.c:11:5: error: implicit declaration of function ‘sbrk’ [-Werror=implicit-function-declaration]
     mem_ptr = sbrk(SIXTY_FOUR_K); /* Allocate 64 kB of memory */

Here is the code that the compiler warning is referencing:

#include "malloc.h"
#include <unistd.h>

void * malloc(size_t bytes) {
    uintptr_t mem_ptr;

    if (bytes <= 0) { /* If user passes in bad value, return NULL */
        return NULL;
    }

    mem_ptr = sbrk(SIXTY_FOUR_K); /* Allocate 64 kB of memory */

    if (mem_ptr == -1) { /* sbrk() failed */
        return NULL;
    }

    return (void *)mem_ptr;
}

According to the documentation on sbrk, you should just have to import unistd.h, which I do. Is there something I'm doing wrong?

Answer 1

Did you take a look at the Feature Test Macro requirements?

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

   brk(), sbrk():
       Since glibc 2.12:
           _BSD_SOURCE || _SVID_SOURCE ||
               (_XOPEN_SOURCE >= 500 ||
                   _XOPEN_SOURCE && _XOPEN_SOURCE_EXTENDED) &&
               !(_POSIX_C_SOURCE >= 200112L || _XOPEN_SOURCE >= 600)
       Before glibc 2.12:
           _BSD_SOURCE || _SVID_SOURCE || _XOPEN_SOURCE >= 500 || _XOPEN_SOURCE && _XOPEN_SOURCE_EXTENDED

See if compiling with something like -D_SVID_SOURCE will work (though it looks like there are a number of options based on that macro list)

As of glibc 2.19, a new feature test macro was added, _DEFAULT_SOURCE which is meant to replace _BSD_SOURCE and _SVID_SOURCE. For more information on _DEFAULT_SOURCE, see this question: What does -D_DEFAULT_SOURCE do?