On Mon, 26 Sep 2011, Jakub Jelinek wrote:

> Hi!
> 
> Adding Joseph and Jason to CC.
> 
> On Mon, Sep 26, 2011 at 04:56:20PM +0200, Richard Guenther wrote:
> > Let's see what kind of fallout we get ;)  For example, if the
> > following is valid C code I expect we will vectorize the second
> > loop (disambiguating p[i] and q[i]) bogously:
> > 
> > void foo (int *p)
> > {    
> >   int * __restrict p1 = p;
> >   int * __restrict p2 = p + 32;
> >   int *q;
> >   int i;
> >   for (i = 0; i < 32; ++i)
> >     p1[i] = p2[i];
> >   p = p1;
> >   q = p2 - 31;
> >   for (i = 0; i < 32; ++i)
> >     p[i] = q[i];
> > }
> > 
> > because p and q base on different restrict qualified pointers
> > (p1 and p2 respective).  At the moment we are safe from this
> > because of the TYPE_RESTRICT checks.
> > 
> > Any opinion on the above?  Is it valid to base non-restrict
> > pointers on restrict ones?  It would be sort-of weird at least,
> > but at least I don't think the first loop use is bogus (even
> > though the pointed-to objects are the same).
> 
> If the last loop was
>   for (i = 0; i < 32; i++)
>     q[i] = p[i];
> then I believe the above would be clearly invalid C99, because
> an object X (say incoming p[4]) would be modified in the same block
> using a pointer based on p1 and using a pointer not based on p1
> (q), which would violate the requirements that if the object is
> modified through lvalue whose address is based on p1, all modifications
> to B in that block should be done through lvalues whose address is
> based on p1.  In the above testcase all modifications are made through
> lvalues whose addresses are p1 based though, so it is less clear.
> Joseph?

If an object that is accessed by a restricted pointer is also modified, 
then all accesses (not just all modifications) must be through pointers 
based on the restricted pointer.  So in the original loop with p[i] = 
q[i], q[i] for i from 0 to 30 is an object that was previously modified 
through p1 and is now being accessed through p2.  So this code appears 
invalid to me.

-- 
Joseph S. Myers
jos...@codesourcery.com

Reply via email to