X3J11 Pleasanton meeting summary

[Mark Brader]

To Doug Gwyn's summary:
> 	int a[4][5];
> 	a[1][7] = 0;	/* undefined behavior */
> Dave Prosser (our Redactor) vigorously protested ...

I wrote:
@ My opinion is that the protest was right and the ruling wrong.

Doug then posts a longer form of the ruling:
: For an array of arrays, the permitted pointer arithmetic in
: Standard section 3.3.6 Semantics (p. 48, ll. 12-40) is to be
: understood by interpreting the use of the word "object" as
: denoting the specific object determined directly by the pointer's
: type and value, NOT other objects related to that one by contiguity.

I now see that this is right, I was wrong, and it is undefined behavior.
But this leads to a problem, which I'll get to in a moment.

First, I should explain my error.  In reply to Karl Heuer's comment:
! I presume that this ruling (if upheld) also means that strictly conforming
! programs may not use extensible structs via the usual overmalloc hack?

Alan Rosenthal writes:
| Alternatively, (int (*)[5])malloc(4 * 5 * sizeof(int)) may be deemed to
| create an object viewable as being 20 ints even though int a[4][5] does
| not, in which case the overmalloc hack would still be fine (though
| blecherous).

And I agree with this also.  This was in fact the source of my original
confusion: I had previously examined the wording carefully to ensure that
the "overmalloc hack" was legal, remembered the conclusion, and forgotten
the reasoning behind it.  I then assumed that as I knew the conclusion
I didn't need to go back to the Standard and see the actual wording.
Stupid.


Now the problem.  Doug also notes:
> Note that even such a subterfuge as the following would not be strictly
> conforming:
> 	void func( int *p ) { p[7] = 0; }
> 	int main( void ) { int a[4][5]; func( a[1] ); return 0; }

And I agree with this also.  p points to the element a[1][0] in the array
a[1], and there is no element [7] in that array, so it's undefined.
But what if func was written as follows?

	void func (void *p) {int *ip = p; ip[7] = 0; }

Does ip also point to the element a[1][0] in the array a[1], and is
this therefore still non-conforming?  If so... can memcpy() really
be implemented in ANSI C, and used to copy arbitrary objects?

-- 
Mark Brader				"It's easier to deal with 'opposite
SoftQuad Inc., Toronto			 numbers' when you know you cannot
utzoo!sq!msb, msb at sq.com		 trust them."		-- Chess

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