https://gcc.gnu.org/bugzilla/show_bug.cgi?id=48885
--- Comment #15 from Richard Biener <rguenth at gcc dot gnu.org> --- (In reply to vries from comment #14) > (In reply to rguent...@suse.de from comment #13) > > On Wed, 23 Sep 2015, vries at gcc dot gnu.org wrote: > > > > > https://gcc.gnu.org/bugzilla/show_bug.cgi?id=48885 > > > > > > --- Comment #12 from vries at gcc dot gnu.org --- > > > (In reply to Richard Biener from comment #11) > > > > I'm testing the above simple fix and amend the comment. > > > > > > Consider the example with functions f and g I gave in comment 10. Using > > > the > > > patch from comment 11, I get at ealias: > > > ... > > > void f(int* __restrict__&, int*) (intD.9 * restrict & restrict pD.2252, > > > intD.9 > > > * p2D.2253) > > > { > > > intD.9 * _3; > > > > > > # VUSE <.MEM_1(D)> > > > # PT = { D.2265 } (nonlocal) > > > _3 = MEM[(intD.9 * restrict &)p_2(D) clique 1 base 1]; > > > > > > # .MEM_4 = VDEF <.MEM_1(D)> > > > MEM[(intD.9 *)_3 clique 1 base 2] = 1; > > > > > > # .MEM_6 = VDEF <.MEM_4> > > > MEM[(intD.9 *)p2_5(D) clique 1 base 0] = 2; > > > ... > > > > > > AFAIU, this is incorrect. The two stores can be now disambiguated based > > > on same > > > clique/different base, but in fact the stores can alias (in fact they do, > > > in > > > the "f (gp, gp)" call from g). > > > > How is this a valid testcase? > > You are accessing g()s *gp through > > p and p2 even though p is marked as restrict. > > To be exact, p is a restrict reference to a restrict pointer. > And AFAIU it's a valid test-case. > > > Did you mean to write > > > > void > > f (int *&__restrict__ p, int *p2) > > > > ? > > No. I'll try explain, renaming variables to help clarification, and adding a > call to g for completeness: > ... > void > f (int *__restrict__ &__restrict__ fp, int *fp2) > { > *fp = 1; > *fp2 = 2; > } > > void > g (int *__restrict__ gp) > { > f (gp, gp); > } > > void > h (void) > { > int ha; > g (&ha); > } > ... > > Let's look at the three restricts in the example. > > First, there's the second restrict in "int *__restrict &__restrict fp", > which is a reference to object gp. Since object gp is not modified during f, > the restrict has no consequence. > > Then there's the restrict in "int *__restrict__ gp". The object pointed to > is ha, and it's modified during g. So all accesses to ha during g need to be > based on gp. And that is the case. The '*fp2 = 1' is based on gp. And the > '*fp2 = 2' is based on gp. No, *fp2 is _not_ based on gp. Otherwise even simple cases like int foo (int * __restrict p, int * __restrict q) { *p = 1; *q = 0; return *p; } could not be optimized because calling foo like int bar () { int i; int *r = &i; foo (r, r); return i; } would make that invalid. With your reading both p and q are based on r (in the context of bar). The standard has some interesting wording to define "based-on". IIRC it goes like a pointer is based on 'p' if the value of the pointer changes when you modify 'p'. I think that only allows for expressions based on p, like (p + 1) or &p[2]. It does _not_ allow for new temporaries, like q = p + 1; as if you modify p q doesn't change. IMHO the restrict qualifications on a function signature are to be seen as constraints on the set of valid parameters it can be called with and f (gp, gp) is not amongst that set. > Finally, there's the first restrict in "int *__restrict &__restrict fp". > That's a copy of the type of gp.
