Re: sizeof(array) with variable-length array parametery

[Andrew Haley <aph at redhat dot com>]

peter.kourzanov@xs4all.nl wrote:
> On Wed, Apr 09, 2008 at 06:34:05PM +0100, Andrew Haley wrote:
>> But it's *not* always easily accessible.  In this case,
>>
>> int bar(int p[n])
> 
>   Well, n must be referring a symbol that is in scope, so
> it must be bound to a value at run-time. Otherwise, if its undefined,
> the sizeof should also be undefined...
> 
>> n is somewhere in global scope, maybe in another translation unit,
>> and as I point out below you'd have to copy it at the time the array
>> was created.
> 
>   What's wrong with taking a snapshot of "n" when the function is
> entered? After all, that's what p[n] specifies in the signature?

I was trying to find out if you were serious.

>> So, you're saying it *should* create a hidden copy of the size
>> parameter for use by sizeof?  While this would work, it's not very
>> C.
> 
>   Hhm, not strictly following C99 standard, sure, but following common
> sense, yes. That's what seems to be happening in this valid C99
> snippet:
> 
> int main() {
> 	int s=10,a[s];
>         s=11;
>        	assert(sizeof a == 40);
> }

Mmm, yes.  

>   I do understand that our idle rumblings will not change the C99
> standard, but, frankly speaking, its very disappointing having to
> write things such as:
> 
> void foo(int s,int a[s])
> {
> 	/* work-around the C99 always returning
> 	 * sizeof(int*) instead of s*sizeof(int)
> 	 */
> 	int _a[s];
> 	s=11;
> 	for (int i=0; i<sizeof _a/sizeof _a[0]; i++);
> }
> 
>   This one also generates slightly less optimal code...
> 
> Maybe it's a good canditate for a GCC extension?

I'm not sure how.  C99 says "6.7.5.3 Function
declarators (including prototypes) ...  A declaration of a
parameter as ââarray of typeââ shall be adjusted to ââqualified
pointer to typeââ, where the type qualifiers (if any) are those
specified within the [ and ] of the array type derivation."

Andrew.