[PATCH] Fix PR61964

[Richard Biener]

This fixes PR61964, we have to rely on structural equality when
determining equivalency of non-register LHS.

Bootstrapped and tested on x86_64-unknown-linux-gnu, applied.

Richard.

2014-07-31  Richard Biener  <rguent...@suse.de>

        PR tree-optimization/61964
        * tree-ssa-tail-merge.c (gimple_equal_p): Handle non-SSA LHS solely
        by structural equality.

        * gcc.dg/torture/pr61964.c: New testcase.

Index: gcc/tree-ssa-tail-merge.c
===================================================================
--- gcc/tree-ssa-tail-merge.c   (revision 213317)
+++ gcc/tree-ssa-tail-merge.c   (working copy)
@@ -1161,17 +1168,9 @@ gimple_equal_p (same_succ same_succ, gim
       lhs2 = gimple_get_lhs (s2);
       if (TREE_CODE (lhs1) != SSA_NAME
          && TREE_CODE (lhs2) != SSA_NAME)
-       {
-         /* If the vdef is the same, it's the same statement.  */
-         if (vn_valueize (gimple_vdef (s1))
-             == vn_valueize (gimple_vdef (s2)))
-           return true;
-
-         /* Test for structural equality.  */
-         return (operand_equal_p (lhs1, lhs2, 0)
-                 && gimple_operand_equal_value_p (gimple_assign_rhs1 (s1),
-                                                  gimple_assign_rhs1 (s2)));
-       }
+       return (operand_equal_p (lhs1, lhs2, 0)
+               && gimple_operand_equal_value_p (gimple_assign_rhs1 (s1),
+                                                gimple_assign_rhs1 (s2)));
       else if (TREE_CODE (lhs1) == SSA_NAME
               && TREE_CODE (lhs2) == SSA_NAME)
        return vn_valueize (lhs1) == vn_valueize (lhs2);
Index: gcc/testsuite/gcc.dg/torture/pr61964.c
===================================================================
--- gcc/testsuite/gcc.dg/torture/pr61964.c      (revision 0)
+++ gcc/testsuite/gcc.dg/torture/pr61964.c      (working copy)
@@ -0,0 +1,33 @@
+/* { dg-do run } */
+
+extern void abort (void);
+
+struct node { struct node *next, *prev; } node;
+struct head { struct node *first; } heads[5];
+int k = 2;
+struct head *head = &heads[2];
+
+static int __attribute__((noinline))
+foo()
+{
+  node.prev = (void *)head;
+  head->first = &node;
+
+  struct node *n = head->first;
+  struct head *h = &heads[k];
+
+  if (n->prev == (void *)h)
+    h->first = n->next;
+  else
+    n->prev->next = n->next;
+
+  n->next = h->first;
+  return n->next == &node;
+}
+
+int main()
+{
+  if (foo ())
+    abort ();
+  return 0;
+}