P
pete
Robert said:
No. I was emphasizing the fact that they compare equal.
The non-NULL return value of malloc points to memory.
The NULL return value of malloc, doesn't.
No. I was emphasizing the fact that they compare equal.
The non-NULL return value of malloc points to memory.
The NULL return value of malloc, doesn't.
K
Keith Thompson
pete said:
You wrote:
void *pointer = &pointer;
if (pointer == &pointer) {
printf("&pointer is %p\n", pointer);
printf("&pointer is %p\n", (void *)&pointer);
}
I think what you meant was:
void *pointer = &pointer;
if (pointer == &pointer) {
printf("pointer is %p\n", pointer);
printf("&pointer is %p\n", (void *)&pointer);
}
P
pete
Keith said:
(pointer) and ((void *)&pointer) are two expressions
which have the exact same type and value,
therefore they are interchangable as arguments
in a function call.
W
Walter Roberson
Pointers can have different sizes. A pointer to a char
is not necessarily the same size as a pointer to an int,
for example -- and from what I've read in other postings,
there have been real implementations in which they were
quite different.
Look at all the verbage the standard spends on talking about
-converting- pointer types (which it distinguishes from
removing type qualifications.)
K
Keith Thompson
pete said:
Yes, but by printing "&pointer is ..." on both lines, you're
*asserting* that they're the same thing, not demonstrating it. The
output of your program doesn't demonstrate the point you're trying to
make.
F
Flash Gordon
Walter said:
In general you are correct, but in the message pete was replying to
pointer was declared with the line
| void *pointer = &pointer;
making peter's statement correct in *this* instance, just not in the
general case.
pointer is of type void* and contains the address of pointer. &pointer
is obviously the address of pointer, and after the cast the value has
been converted to type void* just as it was when pointer was initialised.
C
Chris Torek
Keith Thompson wrote:
[snippage]
Keith's, er, "point", is not that the second argument to printf()
is wrong, but that the first one -- the string -- prints something
other than what he think you meant to print. It might be
more obvious if the first line read:
printf("Mozambique Itemize Sasquatch! %p!!\n", pointer);
and the second then read:
printf("pointer is %p\n", pointer);
[snippage]
Keith's, er, "point", is not that the second argument to printf()
is wrong, but that the first one -- the string -- prints something
other than what he think you meant to print. It might be
more obvious if the first line read:
printf("Mozambique Itemize Sasquatch! %p!!\n", pointer);
and the second then read:
printf("pointer is %p\n", pointer);
P
pete
Keith said:
Yes, that's my assertion.
P
pete
Flash said:
That's it.
K
Keith Thompson
pete said:
Sigh. I give up.
P
pete
Chris said:
Keith's, er, "point", is not that the second argument to printf()
is wrong, but that the first one -- the string -- prints something
other than what he think you meant to print. It might be
more obvious if the first line read:
printf("Mozambique Itemize Sasquatch! %p!!\n", pointer);
and the second then read:
printf("pointer is %p\n", pointer);
My point is that
printf("&pointer is %p\n", pointer);
is a valid way to print the value of &pointer,
given that (pointer == &pointer) is true.
In some contexts, such as with the equality operator, and %p,
there is no difference between pointer and &pointer,
even though they are expressions of different types.
When (pointer == &pointer), then pointer points to itself
as much as the return value of malloc can point to anything.
T
tanvir
D
Dave Thompson
Actually, I did once use it as the boundary case for a type-unsafely
generic linked list. I'd call that _very little_ real use, however.
- David.Thompson1 at worldnet.att.net
generic linked list. I'd call that _very little_ real use, however.
- David.Thompson1 at worldnet.att.net
J
Jordan Abel
I'm not sure what's so subtle about void *p = &p; - or is this wrong and
you had something else in mind?