Re: [RFH] Folding of &a[-1]

[Theodore Papadopoulo <Theodore dot Papadopoulo at sophia dot inria dot fr>]

On Wed, 2005-02-16 at 14:25 +0100, Richard Guenther wrote:
> On Wed, 16 Feb 2005, Paul Schlie wrote:

> Yes, of course, but it is the C frontent that is producing
> &a + (int *)-4, not me.  I'm just trying to work around this...
> 
> In fact, it is c-common.c:2289 that does -4  ->  (int *)-4
> conversion, but pointer_int_sum is already called with PLUS_EXPR.
> build_unary_op unconditionally expands &x[y] to x+y, regardless
> of the sign of y.  Of course the standard says that they are equal.
> But is &x[-1] == x + (int *)4*(int *)-1 ?  From this follows that
> we have no way to convert this back to &x[-1], as we loose the
> sign information by the (int *) cast.

Reading this thread, I keep wondering why this cast to an (int*). It
should be a simple int no ??  If the intent is to have both operands of
the PLUS to have the same type then it might be a counter-example for
this rule (but I do understand that it would require specific patterns
for pointer addition which is fairly stupid in general)....

Now, I'm not sure I understand the premisses of the discussion: &a
[-4U/4] from &a[-1] are not written the same but still refer to the same
value since the overflow happens the same when multiplying
(unsigned)(-4) by 4... so I do not really see where is the problem.

Then, I must add that I do not know much about the compiler's
internals...