When the divisor is bool or has [0, 1] range, as division by
0 is UB, the only remaining option in valid programs is division by 1,
so we can optimize X / bool_range_Y into X.
2020-12-15 Jakub Jelinek <jakub@redhat.com>
PR tree-optimization/96094
* match.pd (X / bool_range_Y -> X): New simplification.
* gcc.dg/tree-ssa/pr96094.c: New test.
/* But not for 0 / 0 so that we can get the proper warnings and errors. */
(if (!integer_zerop (@1))
@0))
- /* X / -1 is -X. */
+ /* X / -1 is -X. */
(simplify
- (div @0 integer_minus_onep@1)
- (if (!TYPE_UNSIGNED (type))
- (negate @0)))
+ (div @0 integer_minus_onep@1)
+ (if (!TYPE_UNSIGNED (type))
+ (negate @0)))
+ /* X / bool_range_Y is X. */
+ (simplify
+ (div @0 SSA_NAME@1)
+ (if (INTEGRAL_TYPE_P (type) && ssa_name_has_boolean_range (@1))
+ @0))
/* X / X is one. */
(simplify
(div @0 @0)
--- /dev/null
+/* PR tree-optimization/96094 */
+/* { dg-do compile } */
+/* { dg-options "-O2 -fdump-tree-optimized" } */
+/* { dg-final { scan-tree-dump-times "return 34;" 2 "optimized" } } */
+/* { dg-final { scan-tree-dump-times "return y_\[0-9]*\\\(D\\\);" 2 "optimized" } } */
+/* { dg-final { scan-tree-dump-times "return \[^\n\r;]*;" 4 "optimized" } } */
+
+int
+foo (int x)
+{
+ if (x >= 2U)
+ return 34;
+ return 34 / x;
+}
+
+int
+bar (int x, int y)
+{
+ if (x >= 2U)
+ return y;
+ return y / x;
+}
+
+int
+baz (_Bool x)
+{
+ return 34 / x;
+}
+
+int
+qux (_Bool x, int y)
+{
+ return y / x;
+}