is a pointer memset to zero necessarily eqal to NULL?

Hi all

Is this assertion guaranteed?
{
void *vptr;
memset( & vptr, 0, sizeof vptr);
assert( NULL == vptr );
}

What about these ones?
{
sometype *ptr;
memset( & ptr, 0, sizeof ptr);
assert( (sometype*)NULL == ptr );
assert( NULL == (void*)ptr );
}

Thanks,
viza

>Is this assertion guaranteed?
>{
void *vptr;
memset( & vptr, 0, sizeof vptr);
assert( NULL == vptr );
>}

No. It is not guaranteed that the internal representation of a
null pointer is 0xdeadbeef, even on a machine with 32-bit pointers.
But it might be.

>What about these ones?
>{
sometype *ptr;
memset( & ptr, 0, sizeof ptr);
assert( (sometype*)NULL == ptr );
No.
assert( NULL == (void*)ptr );
No.
>}

Sat, 05 Jul 2008 17:52:40 -0500, Gordon Burditt wrote:

>>Is this assertion guaranteed?
>>{
>void *vptr;
>memset( & vptr, 0, sizeof vptr);
>assert( NULL == vptr );
>>}
>
No. It is not guaranteed that the internal representation of a null
pointer is 0xdeadbeef, even on a machine with 32-bit pointers. But it
might be.

Is there then some library function to set an array of objects of
size 1 to the same value? Something like wmemset() but for things the
size of pointers, or even larger structs?

Thanks,
viza

viza wrote:
Hi all
>
Is this assertion guaranteed?
{
void *vptr;
memset( & vptr, 0, sizeof vptr);
assert( NULL == vptr );
}
>
What about these ones?
{
sometype *ptr;
memset( & ptr, 0, sizeof ptr);
assert( (sometype*)NULL == ptr );
assert( NULL == (void*)ptr );
}


None of those.

These assertions are guaranteed:
{
void *vptr = 0;
sometype *ptr = 0;

assert( NULL == vptr );
assert( NULL == 0 );
assert( 0 == vptr );
assert( ptr == vptr );
}

--
pete

viza wrote:
Is this assertion guaranteed?
{
void *vptr;
memset( & vptr, 0, sizeof vptr);
assert( NULL == vptr );
}

Have you tried reading the FAQ?

viza wrote:
Is there then some library function to set an array of
objects of size 1 to the same value?

Where does the size come into it?

Just use a loop

static const struct foo foo_zero; /* zero initialised */
for (i = 0; i < N; i++)
array[i] = foo_zero;

--
Peter

8 Jul 2008 at 8:40, Richard Heathfield wrote:
Perhaps when he said "heathfield clique" he actually means "the people
in clc who know C".

There really are no bounds to your arrogance, are there?