[Bug tree-optimization/33512] Simple bitwise simplification missed

["ktietz at gcc dot gnu.org" <gcc-bugzilla at gcc dot gnu dot org>]

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=33512

--- Comment #14 from Kai Tietz <ktietz at gcc dot gnu.org> 2012-04-24 08:30:49 UTC ---
Hmm, I have right now in my tree

Index: tree-ssa-forwprop.c
===================================================================
--- tree-ssa-forwprop.c (revision 186753)
+++ tree-ssa-forwprop.c (working copy)
@@ -1912,17 +1912,25 @@

    /* Simplify (A & B) OP0 (C & B) to (A OP0 C) & B. */
    if (def1_code == def2_code
-       && def1_code == BIT_AND_EXPR
-       && operand_equal_for_phi_arg_p (gimple_assign_rhs2 (def1),
-                                      gimple_assign_rhs2 (def2)))
+       && def1_code == BIT_AND_EXPR)
     {
-      tree b = gimple_assign_rhs2 (def1);
-      tree a = def1_arg1;
-      tree c = def2_arg1;
-      tree inner = fold_build2 (code, TREE_TYPE (arg2), a, c);
+      tree inner, a = NULL_TREE, b = NULL_TREE, c = NULL_TREE;
+
+      if (operand_equal_for_phi_arg_p (def1_arg2, def2_arg2))
+       { b = def1_arg2; a = def1_arg1; c = def2_arg1; }
+      else if (operand_equal_for_phi_arg_p (def1_arg1, def2_arg1))
+       { b = def1_arg1; a = def1_arg2; c = def2_arg2; }
+      else if (operand_equal_for_phi_arg_p (def1_arg1, def2_arg2))
+       { b = def1_arg1; a = def1_arg2; c = def2_arg1; }
+      else if (operand_equal_for_phi_arg_p (def1_arg2, def2_arg1))
+       { b = def1_arg2; a = def1_arg1; c = def2_arg2; }
+      if (a && c)
+       inner = fold_build2 (code, TREE_TYPE (arg2), a, c);
+      if (!a || !b || !c)
+       ;
       /* If A OP0 C (this usually means C is the same as A) is 0
         then fold it down correctly. */
-      if (integer_zerop (inner))
+      else if (integer_zerop (inner))
        {
          gimple_assign_set_rhs_from_tree (gsi, inner);
          update_stmt (stmt);

It has the advantage of handling also cases for (A & B) OP0 (C & B) to (A OP0
C) & B, (B & A) OP0 (C & B) to (A OP0 C) & B, (A & B) OP0 (B & C) to (A OP0 C)
& B, and (B & A) OP0 (B & C) to (A OP0 C) & B.

It bootstraps fine.