https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67705
Richard Biener <rguenth at gcc dot gnu.org> changed:
What |Removed |Added
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CC| |jsm28 at gcc dot gnu.org,
| |rguenth at gcc dot gnu.org
--- Comment #2 from Richard Biener <rguenth at gcc dot gnu.org> ---
Indeed the wording contains "and X is also modified (by any means)." thus
fp == fp2 are valid even though they are marked restrict. Btw, lets forgo
the use of references here and talk plain C:
void
f (int *__restrict__ *__restrict__ fp,
int *__restrict__ *__restrict__ fp2)
{
int *p = *fp; // 1
int *p2 = *fp2; // 2
*p = 1; // 3
*p2 = 2; // 4
}
void
g (int *__restrict__ gp)
{
f (gp, gp);
}
void
h (void)
{
int ha;
g (&ha);
}
which better explains the issue you are rising. The standard doesn't talk
about stmts 1 and 2 (they do not modify anything). We are interested on
whether the writes 3 and 4 may alias or not. *p and *p2 would have to
be based on the same pointer: "Every other lvalue used to access the value
of X shall also have its address based on P."
The "pointer expression" here are *fp or *fp2 and either p needs to be
dependent on p2 or p2 needs to be dependent on p. Both are not the
case IMHO - that both p and p2 are dependent on "*gp" does not count.
I don't think writing
**fp = 1;
**fp2 = 2;
changes any of this. If it does then this is worth a DR I think.
I think the standard refers to 'P' as rvalue, thus "modifying P
to point to a copy of the array object..." would not modify the lvalue *fp
but the rvalue *fp.
But yes, my original reasoning that *fp and *fp2 cannot refer to the same
storage is moot.
Adding Joseph.
