Re: Can I Trust Pointer Arithmetic In Re-Allocated Memory?

On Thu, 17 Aug 2006 23:54:41 UTC, "Bill Reid"
<hormelfree@xxxxxxxxxxxxxxxx> wrote:

When realloc fails you needs to either work with the memory (p)
you
have already or to cleanup (free(p). Overwriting p with NULL gives you
a memory leak as you lost the address of the memory you have laready
allocated.

Hmmm...if this is the case this should be in the NON-documentation.
Actually, I guess it kind of is, but after a quick read of this I always
thought realloc() freed the original memory block and returned NULL
if it couldn't reallocate the new block:

Yes, but it does NOT free() the old block then. That block leaves
unchanged.

...

Syntax

#include <stdlib.h>
void *realloc(void *block, size_t size);

...

If the block cannot be reallocated, realloc returns NULL.

If the value of size is 0, the memory block is freed and realloc returns
NULL.

---end of NON-documentation

I guess I got the last two conditions conflated in my mind; it just seemed
logical to me that if realloc() failed it would free the previous block.
It SEEMS like it should.

But it does not so because you may need to continue your work with the
old block.
When you have no need for the old block you have to free() that
yourself, else you should free() it. In any case you needs its
address.

As an orthogonal point, what horrible things happen if you try
to free() a NULL pointer?

Nothing. free(NULL); works like a noop. Nothing occures. That is
guaranteed.

--
Tschau/Bye
Herbert

Visit http://www.ecomstation.de the home of german eComStation
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