[PATCH 1/2] Auto-generate maybe_fold_and/or_comparisons from match.pd

[Richard Biener]

On Mon, 9 Sep 2019, Martin Liška wrote:

> Hi.
> 
> I'm sending slightly updated version of the patch where we
> need to properly select type in maybe_fold_comparisons_from_match_pd
> function for the created SSA_NAMEs. We can be called for a VECTOR_TYPE
> and so that we can't return a boolean_type_node.
> 
> Patch can bootstrap on x86_64-linux-gnu and survives regression tests.
> 
> Ready to be installed?

2019-07-16  Li Jia He  <helijia@linux.ibm.com>
            Martin Liska  <mliska@suse.cz>

        * gimple.h (gimple_init): Declare.
        (gimple_size): Likewise.
        * gimple.c (gimple_init): Remove static and inline restrictions.
        (gimple_alloc): Only allocate memory and call gimple_init.
        (gimple_size): Likewise.

Likewise?

        * tree-ssanames.c (init_ssa_name_imm_use): Use make_ssa_name_fn.
        (make_ssa_name_fn): New.

You didn't touch make_ssa_name_fn.

Since we're needing another iteration:

+  /* Allocate gimple stmt1 on the stack.  */
+  gimple *stmt1 = (gimple *) XALLOCAVEC (char, gimple_size 
(GIMPLE_ASSIGN, 2));

You can use gassign *stmt1 here so all the gimple_assign_ fns below
get cheaper.

+  if (op.resimplify (NULL, follow_all_ssa_edges))
+    {
+      if (gimple_simplified_result_is_gimple_val (&op))
+       {
+         tree res = op.ops[0];
+         switch (TREE_CODE (res))
+           {
+           case SSA_NAME:
+               {
+                 gimple *def = SSA_NAME_DEF_STMT (res);

you shouldn't expand SSA names here unless that SSA name is
exactly lhs1 or lhs2 from above.  So

         if (res == lhs1)
           return build2 (...);
         else if (res == lhs2)
           return build2 (..);
         else
           return res;

plus you miss the case where 'op' became a simplified comparison
in itself.  So,

     if (op.code.is_tree_code ()
         && TREE_CODE_CLASS ((enum tree_code)op.code) == tcc_comparison)
       {
          tree op0 = op.ops[0];
          tree op1 = op.ops[1];
          if (op0 == lhs1 || op0 == lhs2 || op1 == lhs1 || op1 == lhs2)
            return NULL_TREE;  /* not simple */
          return build2 ((enum tree_code)op.code, op.type,
                         op0, op1);
       }

note you need not fold_ again.  It's of course ugly that we
need to build a GENERIC tree here but that's the current interface
and thus OK at the moment.

Thanks,
Richard.