[Bug tree-optimization/67705] incorrect restrict interpretation

[rguenth at gcc dot gnu.org]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67705

Richard Biener <rguenth at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 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.