[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  <rguenther@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;
+}