Memory question

[Mohanasundaram]

Hi All,

As per the standard what is the result of passing NULL to both malloc and free?

Regards,
Mohan.

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[Wojtek Lerch]

Steve Clamage <St************ *@Sun.COM> wrote in message news:<c0******* ***@news1nwk.SF bay.Sun.COM>...

The expression malloc(1) returns a pointer to an object, whereas malloc(0) does
not. That is, there are no zero-sized objects, and you are not allowed to
dereference the result of malloc(0).
No, malloc(0) either returns NULL or a pointer to an n-byte object,
for some unspecified nonzero n. This object must be disjoint from any
other object. You're not allowed to access the object, but you can
compare its address to addresses of other objects, and you can expect
them to turn out different.
So I think the standard has a loophole that allows an implementation like I
suggested in another posting: Reserve one byte. Have malloc(0) return the
address of that byte, and have free() of that address do nothing. The returns
from malloc never point to any object that was not already free'd.

I'm not seriously recommending this approach. I just think that it doesn't
violate any requirement in the standard.

Except the one that it must be an object of a non-zero size, disjoint
from any other object.
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[Dan Pop]

In <c0**********@n ews1nwk.SFbay.S un.COM> Steve Clamage <St************ *@Sun.COM> writes:

So I think the standard has a loophole that allows an implementation like I
suggested in another posting: Reserve one byte. Have malloc(0) return the
address of that byte, and have free() of that address do nothing. The returns
from malloc never point to any object that was not already free'd.

I'm not seriously recommending this approach. I just think that it doesn't
violate any requirement in the standard.

It does:

If the size
of the space requested is zero, the behavior is implementation-
defined: either a null pointer is returned, or the behavior is
as if the size were some nonzero value, except that the returned
^^^^^
pointer shall not be used to access an object.

This text guarantees that after:

char *p = malloc(0);
char *q = malloc(0);

p == q *only* if both are null pointers (see the underlined "as if"
in the above quoted text), because this guarantee stands for nonzero
values of the argument. A strictly conforming program can trivially
check if this is the case and expose your hypothetical implementation
as non-conforming.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
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[Douglas A. Gwyn]

Fergus Henderson wrote:

Why would the implementation _care_ when the last free() occurs?

Almost all malloc implementations have storage overhead
associated with each live allocation, and they want to
liberate the overhead storage when they can. One could
concoct an implementation that had some weird special
handling for 0-sized allocations, but there is no reason
for doing so, and all implementations I know of that
support 0-sized allocations treat it the same way as any
positive-sized request (after adding 1 or not, depending
on the implementation) .
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