Hi!
The following testcase ICEs during regimplificatgion since the addition of
(convert (eqne zero_one_valued_p@0 INTEGER_CST@1))
simplification. That simplification is novel in the sense that in
gimplify_expr it can turn an expression (comparison in particular) into
a SSA_NAME. Normally when gimplify_expr sees originally a SSA_NAME, it does
case SSA_NAME:
/* Allow callbacks into the gimplifier during optimization. */
ret = GS_ALL_DONE;
break;
and doesn't try to recalculate side effects because of that, but in this
case gimplify_expr normally enters the:
default:
switch (TREE_CODE_CLASS (TREE_CODE (*expr_p)))
{
case tcc_comparison:
then does
*expr_p = gimple_boolify (*expr_p);
and then
*expr_p = fold_convert_loc (input_location,
org_type, *expr_p);
with this new match.pd simplification turns that tcc_comparison class
into SSA_NAME. Unlike the outer SSA_NAME handling though, this falls
through into
recalculate_side_effects (*expr_p);
dont_recalculate:
break;
but unfortunately recalculate_side_effects doesn't handle SSA_NAME and ICEs
on it.
SSA_NAMEs don't ever have TREE_SIDE_EFFECTS set on those, so the following
patch fixes it by handling it similarly to the tcc_constant case.
Bootstrapped/regtested on x86_64-linux and i686-linux, ok for trunk?
2024-01-06 Jakub Jelinek <ja...@redhat.com>
PR tree-optimization/113228
* gimplify.cc (recalculate_side_effects): Do nothing for SSA_NAMEs.
* gcc.c-torture/compile/pr113228.c: New test.
--- gcc/gimplify.cc.jj 2024-01-03 11:51:40.744603324 +0100
+++ gcc/gimplify.cc 2024-01-05 13:32:34.351336320 +0100
@@ -3344,6 +3344,9 @@ recalculate_side_effects (tree t)
return;
default:
+ if (code == SSA_NAME)
+ /* No side-effects. */
+ return;
gcc_unreachable ();
}
}
--- gcc/testsuite/gcc.c-torture/compile/pr113228.c.jj 2024-01-05
13:27:42.876330301 +0100
+++ gcc/testsuite/gcc.c-torture/compile/pr113228.c 2024-01-05
13:27:22.503609458 +0100
@@ -0,0 +1,17 @@
+/* PR tree-optimization/113228 */
+
+int a, b, c, d, i;
+
+void
+foo (void)
+{
+ int k[3] = {};
+ int *l = &a;
+ for (d = 0; c; c--)
+ for (i = 0; i <= 9; i++)
+ {
+ for (b = 1; b <= 4; b++)
+ k[0] = k[0] == 0;
+ *l |= k[d];
+ }
+}
Jakub
