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

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?