Question

Given the following code:

int *a = NULL;
a = calloc(1, sizeof(*a));
printf("%d\n", a);
a = realloc(a, 0);

printf("%d\n", a);
return (0);

It returns:

4078904
0

Is this realloc equivalent to a free ?

NOTE:
I am using MinGW under WindowsXP.

Answer 1

Not necessarily.

It often does as with the link that munissor posted, but the Mac OS 10.5 man page says:

If size is zero and ptr is not NULL, a new, minimum sized object is allocated and the original object is freed.

What is a “minimum sized object”? Well, any allocator stores some information about the allocations, and that takes space which is often allotted in addition to the space reserved for the user. Presumably a “minimum sized object” is just one of these headers plus zero bytes of space reserved for the user.

I would guess that this provision is present to support implementations that existed at the time of standardization, and that those implementations are useful for debugging allocation behavior.

---

To address Jonathan’s comments

Consider the difference between

for (int i=0; i<VERY_BIG_NUMBER; ++i){
  char *p = malloc(sizeof(char[10]));
  free(p);
}

and

for (int i=0; i<VERY_BIG_NUMBER; ++i){
  char *p = malloc(sizeof(char[10]));
  realloc(p,0);
}

With a sane implementation of malloc and free the first clip does not consume memory without bound. But if the realloc implementation returns those “minimum sized objects” it might.

Certainly this example is contrived and it relies on understanding what is meant by “minimum sized object”, but I think that text allows it.

In short, if you mean free you should say free.