Chris Barts wrote:
john blackburn wrote:
|
| free() is the companion function to malloc() and is definitely a
function.
Maybe, maybe not. It's up to the implementation.
| Look up the library documentation; malloc is used to request memory space
| and initialize a pointer to it and free() is used to release that memory
| space.
free() could be a wrapper around a function that programmers aren't
supposed to call directly, or it might not use a function at all, just
odd compiler magic.
free() is always a function, in any conforming
implementation. free() may *also* be provided as a
macro, at the implementation' s discretion, but it must
in any case exist as a function. The same is true of
all the other Standard library functions (other than
those "functions" that are specifically described as
macros, of course).
The following program must compile and run:
#include <stdio.h>
#include <stdlib.h>
static void do_it( void (*func)(void*), void *arg) {
func(arg);
}
static void not_free(void *arg) {
printf ("not_free: arg = %p\n", arg);
}
int main(void) {
void *ptr = malloc(42);
do_it (not_free, ptr);
do_it (free, ptr);
return 0;
}
This example is both contrived and pointless, but
it's not too hard to come up with a scenario in which
using a pointer to free() makes sense. At the risk of
straying from topicality, imagine a suite of programs
that share a single region of memory. The ordinary
malloc() and free() functions won't manipulate the
shared region, so you might well write shared_malloc()
and shared_free() with identical signatures and similar
semantics. Now, let's suppose you want to write a handy
function to release a linked list which might reside in
either shared or ordinary memory:
void liberate_list(N ode *list, void (*liberator)(vo id*)) {
Node *head;
while ((head = list) != NULL) {
list = head->next;
liberator (list);
}
}
Assuming that you know (somehow) whether the list is in
ordinary or in shared memory, you could write
if (in_ordinary_me mory)
liberate_list (list, free);
else
liberate_list (list, shared_free);
.... and the C Standard requires that the pointer to the
function free() work as expected.
--
Er*********@sun .com 
