Re: [tree-ssa, RFC] new, very experimental ssa-vn

[Pop Sébastian <pop at gauvain dot u-strasbg dot fr>]

Hi Steven, 

On Mon, Jan 05, 2004 at 02:24:23AM +0100, Steven Bosscher wrote:
> 
> I still have a bug that I don't quite understand yet.  Consider:
> 
> foo (void)
> {
>   int i, j;
> 
>   for (i = 0; i < 10; i++)
>     for (j = 0; j < 10; j++)
>       { }
> }
> 
> We would find that i and j are congruent, but clearly they are independent.
> I am not sure if this is a bug in my implementation, or just a mess-up in
> the removal algorithm.  Morgan gets it all wrong, and Simpson's thesis
> doesn't help much either.  Looking at my dumps, I seem to be doing the
> Right Thing, but it gives a useless result.  Worked around by requiring
> that PHIs can only be congruent if they are in the same BB (by hashing
> bb_for_stmt), I think it does not pessimize the algorithm (at least, not
> too much ;-), all that happens is that we don't find congruences that would
> not give opportunities for redundancy elimination anyway.
> 

I would check the loop_num (loop_of_stmt (phi-node)) instead of
bb_for_stmt (phi-node).  If the loops differ, then the considered
evolutions belong to two different dimensions.  In terms of chains of
recurrences, the above example is translated to: 
i  ->  {0, +, 1}_loop1
j  ->  {0, +, 1}_loop2
Since (loop1 != loop2) you don't have the same evolution for i and j.

Could you try the following condition:

if (loop_num (loop_of_stmt (phi1)) == loop_num (loop_of_stmt (phi2)))
  /* then "valid value representative for `phi1' is `phi2'"  */

> 
> Sébastian, is it OK if I put this on the lno branch as an experimental pass
> (i.e. disabled by default) once I have it bootstrapping, hopefully soon?
> 

Rule: Everybody is allowed to commit into lno-branch WITHOUT asking
for permission, from the moment that the commit does not broke others
work.

Since you've guarded your pass with a flag that is off by default, you
can check it in without worries.

Thanks.