[gcc r13-6074] reassoc: Fix up (ab) handling in eliminate_redundant_comparison [PR108783]

[gcc r13-6074] reassoc: Fix up (ab) handling in eliminate_redundant_comparison [PR108783]

[Jakub Jelinek]

https://gcc.gnu.org/g:55db240d28d29aac90a2d2af3768283ba6288752

commit r13-6074-g55db240d28d29aac90a2d2af3768283ba6288752
Author: Jakub Jelinek <jakub@redhat.com>
Date:   Thu Feb 16 10:41:18 2023 +0100

    reassoc: Fix up (ab) handling in eliminate_redundant_comparison [PR108783]
    
    The following testcase ICEs because eliminate_redundant_comparison sees
    redundant comparisons in &&/|| where the comparison has (ab) SSA_NAME,
    maybe_fold_{and,or}_comparisons optimizes them into a single comparison
    and build_and_add_sum emits a new comparison close to the definition
    operands, which in this case is before a returns_twice call (which is
    invalid).  Generally reassoc just punts on (ab) SSA_NAMEs, declares them
    non-reassociable etc., so the second half of this patch does that.
    
    Though we can do better in this case; the function has special code
    when maybe_fold_{and,or}_comparisons returns INTEGER_CST (false/true)
    or when what it returns is the same as curr->op (the first of the
    comparisons we are considering) - in that case we just remove the
    second one and keep the first one.  The reason it doesn't match is that
    curr->op is a SSA_NAME whose SSA_NAME_DEF_STMT is checked to be a
    comparison, in this case _42 = a_1(ab) != 0 and the other comparison
    is also like that.  maybe_fold_{and,or}_comparisons looks through the
    definitions though and so returns a_1(ab) != 0 as tree.
    So the first part of the patch checks whether that returned comparison
    isn't the same as the curr->op comparison and if yes, it just overrides
    t back to curr->op so that its SSA_NAME is reused.  In that case we can
    handle even (ab) in {,new}op{1,2} because we don't create a new comparison
    of that, just keep using the existing one.  And t can't be (ab) because
    otherwise it wouldn't be considered a reassociable operand.
    
    The (ab) checks are needed say when we have a_1(ab) == 42 || a_1(ab) > 42
    kind of comparisons where maybe_fold_{and,or}_comparisons returns a new
    comparison not existing in the IL yet.
    
    2023-02-16  Jakub Jelinek  <jakub@redhat.com>
    
            PR tree-optimization/108783
            * tree-ssa-reassoc.cc (eliminate_redundant_comparison): If lcode
            is equal to TREE_CODE (t), op1 to newop1 and op2 to newop2, set
            t to curr->op.  Otherwise, punt if either newop1 or newop2 are
            SSA_NAME_OCCURS_IN_ABNORMAL_PHI SSA_NAMEs.
    
            * gcc.c-torture/compile/pr108783.c: New test.

Diff:
---
 gcc/testsuite/gcc.c-torture/compile/pr108783.c | 42 ++++++++++++++++++++++++++
 gcc/tree-ssa-reassoc.cc                        |  9 ++++++
 2 files changed, 51 insertions(+)

diff --git a/gcc/testsuite/gcc.c-torture/compile/pr108783.c b/gcc/testsuite/gcc.c-torture/compile/pr108783.c
new file mode 100644
index 00000000000..eae87db36d7
--- /dev/null
+++ b/gcc/testsuite/gcc.c-torture/compile/pr108783.c
@@ -0,0 +1,42 @@
+/* PR tree-optimization/108783 */
+
+__attribute__((returns_twice)) int baz (int, int);
+
+int
+bar (int x)
+{
+  return x;
+}
+
+int
+foo (int x, int y)
+{
+  int a;
+
+  a = bar (x);
+  baz (x, y);
+
+  return y && a && a;
+}
+
+int
+qux (int x, int y)
+{
+  int a;
+
+  a = bar (x);
+  baz (x, y);
+
+  return y && a != 42 && a >= 42;
+}
+
+int
+corge (int x, int y)
+{
+  int a;
+
+  a = bar (x);
+  baz (x, y);
+
+  return y || a == 42 || a > 42;
+}
diff --git a/gcc/tree-ssa-reassoc.cc b/gcc/tree-ssa-reassoc.cc
index 5522a3ada8e..f163612f140 100644
--- a/gcc/tree-ssa-reassoc.cc
+++ b/gcc/tree-ssa-reassoc.cc
@@ -2272,6 +2272,15 @@ eliminate_redundant_comparison (enum tree_code opcode,
 	  STRIP_USELESS_TYPE_CONVERSION (newop2);
 	  if (!is_gimple_val (newop1) || !is_gimple_val (newop2))
 	    continue;
+	  if (lcode == TREE_CODE (t)
+	      && operand_equal_p (op1, newop1, 0)
+	      && operand_equal_p (op2, newop2, 0))
+	    t = curr->op;
+	  else if ((TREE_CODE (newop1) == SSA_NAME
+		    && SSA_NAME_OCCURS_IN_ABNORMAL_PHI (newop1))
+		   || (TREE_CODE (newop2) == SSA_NAME
+		       && SSA_NAME_OCCURS_IN_ABNORMAL_PHI (newop2)))
+	    continue;
 	}
 
       if (dump_file && (dump_flags & TDF_DETAILS))