https://gcc.gnu.org/bugzilla/show_bug.cgi?id=67705
--- Comment #3 from vries at gcc dot gnu.org ---
(In reply to Richard Biener from comment #2)
> 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:
>
Agreed, that makes it easier to name the individual bits.
> 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);
After forgoing the references, that should be
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.
Agreed.
> *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."
>
I interpret this as follows.
Restrict declaration D: "int *__restrict__ gp"
Type T: int
Object P: gp
Block B: g function block
Lvalues: *p and *p2
Objects designated by lvalues: ha in both cases.
Object ha modified during execution of B: yes
&Lvalues: pointer expressions p and p2 (or equivalently,
pointer expressions *fp and *fp2)
Now, the question is: are pointer expressions p and p2 based on gp? I think so.
If we modify gp to point to a copy of ha before calling f in g, then both
pointer expressions p and p2 (and *fp and *fp2) will change value.
> The "pointer expression" here are *fp or *fp2
Agreed.
> and either p needs to be
> dependent on p2 or p2 needs to be dependent on p.
> Both are not the
> case IMHO
They are indeed not dependent on each other, but I think that's not relevant.
> - that both p and p2 are dependent on "*gp" does not count.
I think what's relevant is that both pointer expressions p and p2 are based on
object P gp.
> I don't think writing
>
> **fp = 1;
> **fp2 = 2;
>
> changes any of this.
Agreed.
> If it does then this is worth a DR I think.
> I think the standard refers to 'P' as rvalue,
The standard refers to P as a pointer object.
> thus "modifying P
> to point to a copy of the array object..." would not modify the lvalue *fp
> but the rvalue *fp.
>
Modifying gp will modify pointer expressions p, p2, *fp and *fp2. The standard
does not mention any distinction between lvalue or rvalue pointer expressions.
