Hi All,
As per the standard what is the result of passing NULL to both malloc and free?
Regards,
Mohan.
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Nov 14 '05
32  2677 
Stephen Clamage <St************ *@Sun.COM> wrote in message news:<r3******* *************** **********@4ax. com>... On Sun, 8 Feb 2004 15:32:59 +0000 (UTC), la************@ ugsplm.com
wrote:
In comp.std.c Steve Clamage <St************ *@sun.com> wrote:
I don't find anything in either standard that disallows always returning the
same non-null address for malloc(0), as long as that address is not othewise used.
C99 7.20.3:
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.
C90 contained slightly different language that was more easily misread,
but the intent was the same.
Yes, and as I said, I don't see anything that prohibits always
returning the same non-null address.
Suppose the malloc/free implementation reserves one byte, malloc(0)
always returns the address of that byte, and free() of that address is
a no-op. I don't see any rule violation.
I read the above section as requiring that if malloc(0) doesn't return
a null pointer, then there be some value of 'n', possibly different
for each call to malloc(0), for which the behavior of malloc(n) would
be identical to the actual behavior of malloc(0). Fow what size n is
the behavior of malloc(n) the same as the behavior you suggest for
malloc(0)?
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Stephen Clamage wrote: Yes, and as I said, I don't see anything that prohibits always
returning the same non-null address.
Since Standard C doesn't include 0-sized objects, if there
is a successful malloc(0) in a conforming implementation,
the associated object will occupy at least one byte,
although a s.c. program isn't allowed to access it.
It would be folly for an implementation to return a pointer
to *the same* 0-sized region for multiple concurrent live
allocations, because those allocations can be free()d and
how then would it know when the last free() occurs (unless
it adds the overhead of a reference count, which would have
no other purpose).
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[ FAQ: http://www.jamesd.demon.co.uk/csc/faq.html ] St************* @Sun.COM (Stephen Clamage) wrote (abridged): 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.
[...]
Suppose the malloc/free implementation reserves one byte, malloc(0)
always returns the address of that byte, and free() of that address is
a no-op. I don't see any rule violation.
That would not be allowed if the size were some nonzero value.
char *a = malloc( 1 );
char *b = malloc( 1 );
assert( a != b || a == NULL );
The assert must succeed. So it must also succeed if we replace the 1's
with 0's.
-- Dave Harris, Nottingham, UK
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In comp.std.c Stephen Clamage <St************ *@sun.com> wrote:
Yes, and as I said, I don't see anything that prohibits always
returning the same non-null address.
It says that malloc(0) behaves like, for example, malloc(1). Do you
think that malloc(1) is allowed to return the same value all the time?
-Larry Jones
They say winning isn't everything, and I've decided
to take their word for it. -- Calvin
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Stephen Clamage <St************ *@Sun.COM> wrote in message
news:<r9******* *************** **********@4ax. com>...
[...] I read section 5.3.4 to say that the expression
new T[0]
is ill-formed, for any type T. The standard requires a strictly
positive value for the constant-expression giving the array size.
I'm not sure. I presume you are referring to §5.3.4/6: "Every
_constant-expression_ in a _direct-new-declarator_ shal be an integral
constant expression evaluating to a strictly positive value." The
following sentence reads "The _expression_ in a _direct-new-declarator_
shall have integral type with a non-negative value." The grammar at the
top of §5.3.4 reads:
[...]
direct-new-declarator:
[ expression ]
direct-new-declarator [ constant-expression ]
Because of the hypen, and because the expression is written in italics,
I would interpret "constant-expression" in the first quote to refer to
the gramatical construct in the second production for
direct-new-declarator, and not to be the equivalent of a simple
"constant expression". In "new char[0]", the [0] matches the first
production above, and not the second, so the constraints of "expressin"
in §5.3.4/6 apply: it must have a non-negative value (but 0 is OK).
In this case, §5.3.4/7 applies: "When the value of the expression in a
direct-new-declarator is zero, the allocation function is called to
allocate an array with no elements. The pointer returned by the
new-expression is non-null."
But I tried a couple of compilers at full warning levels, and neither
complained about new char[0].
If the code is invalid, the standard imposes no requirements on the
behavior of the program. But it seems to me that an implemention
allowing the expression should also allow deleteing the pointer that
was returned.
The closest I could find is §5.3.5/2: "[...] In the second alternative
(delete array), the value of the operand of delete shall be the pointer
value which resulted from a previous array new-expression." I can find
nothing which requires that the number of elements be greater than 0, so
I would assume that the delete is legal.
I might add that, unlike the poster you were responding to, when I
tried:
char* x = new char[0] ;
delete [] x ;
with VC++ 6.0, it worked as expected.
--
James Kanze GABI Software mailto:ka***@ga bi-soft.fr
Conseils en informatique orientée objet/ http://www.gabi-soft.fr
Beratung in objektorientier ter Datenverarbeitu ng
11 rue de Rambouillet, 78460 Chevreuse, France, +33 (0)1 30 23 45 16
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James Kuyper <ku****@saicmod is.com> wrote in message
news:<40******* ********@saicmo dis.com>...
2. NULL can be an integer constant expression with a value of 0, cast
to void* (this is not permitted in C++). In that case, malloc(NULL) is
equivalent to malloc((void*)0 ), which in turn is equivalent to
malloc((size_t) (void*)0). The result of (size_t)(void*) 0 is
implementation-defined; it can be any valid size_t value. On a great
many implementations , it will be 0, but there are real implementations
where it might not be, and portable code should not count on that.
However, the conversion of ((void*)0) to an integral type requires an
explicit cast. If this is the definition of NULL, then malloc(NULL)
should not compile.
More precisely, a diagnostic message is required, a fact that I forgot
to mention, in my haste to cover the NULL issue properly. However,
having issued that message, an implementation is free to continue with
translation and even execution of the program.
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[ FAQ: http://www.jamesd.demon.co.uk/csc/faq.html ] la************@ ugsplm.com wrote: In comp.std.c Stephen Clamage <St************ *@sun.com> wrote:
Yes, and as I said, I don't see anything that prohibits always
returning the same non-null address.
It says that malloc(0) behaves like, for example, malloc(1). Do you
think that malloc(1) is allowed to return the same value all the time?
I take your point, but I think there is a difference between malloc(0) and
malloc(1).
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).
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.
--
Steve Clamage, st************* @sun.com
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"Douglas A. Gwyn" <DA****@null.ne t> writes: Stephen Clamage wrote: Yes, and as I said, I don't see anything that prohibits always
returning the same non-null address.
Since Standard C doesn't include 0-sized objects, if there
is a successful malloc(0) in a conforming implementation,
the associated object will occupy at least one byte,
although a s.c. program isn't allowed to access it.
It would be folly for an implementation to return a pointer
to *the same* 0-sized region for multiple concurrent live
allocations, because those allocations can be free()d and
how then would it know when the last free() occurs (unless
it adds the overhead of a reference count, which would have
no other purpose).
Why would the implementation _care_ when the last free() occurs?
--
Fergus Henderson <fj*@cs.mu.oz.a u> | "I have always known that the pursuit
The University of Melbourne | of excellence is a lethal habit"
WWW: <http://www.cs.mu.oz.au/~fjh> | -- the last words of T. S. Garp.
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Steve Clamage wrote:
la************@ ugsplm.com wrote: > In comp.std.c Stephen Clamage <St************ *@sun.com> wrote:
> >>Yes, and as I said, I don't see anything that prohibits always
>>returning the same non-null address.
>
>
> It says that malloc(0) behaves like, for example, malloc(1). Do you
> think that malloc(1) is allowed to return the same value all the time?
>
I take your point, but I think there is a difference between malloc(0) and
malloc(1).
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).
The standard specifies that the behavior of malloc(0), when it does not
return a null pointer, must be the same as the behavior of malloc(n) for
some non-zero value of n. That means it does return a pointer to an
object which is NOT zero-sized. It also says that it's illegal to
dereference the pointer to access that object. However, it's still
perfectly legal to use it for equality comparisons, and it must behave
with respect to those comparisons exactly like the result of malloc(n) -
which means it must compare unequal to any other pointer returned by a
malloc(), unless one of the two pointer values being compared has been
passed to free(), in which case all bets are off.
Switching from issues of what the standard does say, to what it should
say: I think there's a fair amount of code out there written to call
malloc(size) for size values where you don't want to bother having
special handling for size==0, but where you do want to assume that each
non-null return value is unique. That doesn't seem to me to be an
unreasonable style of coding, given the current standard, and such code
would be broken if the standard were changed to allow malloc(0) to
always return the same non-null pointer value.
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In comp.std.c Steve Clamage <St************ *@sun.com> wrote:
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).
I don't agree with your conclusion -- malloc(0) most certainly *does*
return a pointer to an object. Said object is not zero sized ("the
behavior is as if the size were some nonzero value"), but you are not
permitted to access it.
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.
That does not meet the requirement that each allocation yield a pointer
to an object distinct from any other (live) object.
-Larry Jones
It's not denial. I'm just very selective about the reality I accept.
-- Calvin
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