[Bug rtl-optimization/44838] [4.6 regression] RTL loop unrolling causes FAIL: gcc.dg/pr39794.c

[rguenth at gcc dot gnu dot org]

------- Comment #20 from rguenth at gcc dot gnu dot org  2010-07-07 10:43 -------
(In reply to comment #19)
> Subject: Re:  [4.6 regression] RTL loop
>         unrolling causes FAIL: gcc.dg/pr39794.c
> 
> > > > > I am not sure what you mean -- I may be misunderstanding how rtl alias analysis
> > > > > works, but as far as I can tell, what unroller does (just preserving the
> > > > > MEM_ATTRs) is conservatively correct (so, potentially it may make us believe
> > > > > that there are dependences that are not really present, but it should not cause
> > > > > a wrong-code bug).
> > > > 
> > > > Consider this simplified example:
> > > > 
> > > > for (i ...)
> > > >   {
> > > > /*A*/  t = a[i];
> > > > /*B*/  a[i+1] = t;
> > > >   }
> > > > MEM_ATTRS would indicate that memory references in A and B do not alias.
> > > 
> > > but this is clearly wrong, since B in iteration X aliases with A in iteration
> > > X+1.
> > > So, not a problem in unroller.
> > 
> > It is not wrong.  You have the two identical pointers p = &a[i] and
> > q = p + 1.  *p and *q do not alias.  Never.
> 
> Well, then you have some other definition of aliasing than me.  For me, two
> memory references M1 and M2 do not alias, if on every code path, the locations
> accessed by M1 and M2 are different.  With this definition, *p and *q may
> alias,
> as the example above shows.  What is your definition?
> 

My definition is that the two statements in sequence A, B have a
true dependence if stmt B accesses memory written to by A.
Thus, in this context *p and *q do not "alias" (and this is what
the scheduler and every other optimization pass queries).


-- 


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=44838