MATCH: Fix `(1 >> X) != 0` pattern for vector types

I had missed that integer_onep can match vector types with uniform constant of `1`.
This means the shifter could be an scalar type and then doing a comparison against `0`
would be an invalid transformation.
This fixes the problem by adding a check for the type of the integer_onep to make
sure it is a INTEGRAL_TYPE_P (which does not match a vector type).

OK? Bootstrapped and tested on x86_64-linux-gnu with no regressions.

	PR tree-optimization/111414

gcc/ChangeLog:

	* match.pd (`(1 >> X) != 0`): Check to see if
	the integer_onep was an integral type (not a vector type).

gcc/testsuite/ChangeLog:

	* gcc.c-torture/compile/pr111414-1.c: New test.

diff --git a/gcc/match.pd b/gcc/match.pd
index 07ffd831132e7a60c3f79399963b575bd941b69e..97db0eb5f25e24608fd64120ece4e262aa1fbef8 100644 (file)
--- a/gcc/match.pd
+++ b/gcc/match.pd
@@ -4206,8 +4206,9 @@ DEFINE_INT_AND_FLOAT_ROUND_FN (RINT)
  /* `(1 >> X) != 0` -> `X == 0` */
  /* `(1 >> X) == 0` -> `X != 0` */
  (simplify
-  (cmp (rshift integer_onep @0) integer_zerop)
-   (icmp @0 { build_zero_cst (TREE_TYPE (@0)); })))
+  (cmp (rshift integer_onep@1 @0) integer_zerop)
+   (if (INTEGRAL_TYPE_P (TREE_TYPE (@1)))
+    (icmp @0 { build_zero_cst (TREE_TYPE (@0)); }))))
 
 /* (CST1 << A) == CST2 -> A == ctz (CST2) - ctz (CST1)
    (CST1 << A) != CST2 -> A != ctz (CST2) - ctz (CST1)

diff --git a/gcc/testsuite/gcc.c-torture/compile/pr111414-1.c b/gcc/testsuite/gcc.c-torture/compile/pr111414-1.c
new file mode 100644 (file)
index 0000000..13fbdae
--- /dev/null
+++ b/gcc/testsuite/gcc.c-torture/compile/pr111414-1.c
@@ -0,0 +1,13 @@
+int a, b, c, d, e, f, g;
+int h(int i) { return b >= 2 ?: i >> b; }
+void j() {
+  int k;
+  int *l = &c;
+  for (; d; d++) {
+    g = h(0 != j);
+    f = g >> a;
+    k = f << 7;
+    e = k > 5 ? k : 0;
+    *l ^= e;
+  }
+}