Avoid redundant GORI calcuations.

When GORI evaluates a statement, if operand 1 and 2 are both in the
dependency chain, GORI evaluates the name through both operands sequentially
and combines the results.

If either operand is in the dependency chain of the other, this
evaluation will do the same work twice, for questionable gain.
Instead, simple evaluate only the operand which depends on the other
and keep the evaluation linear in time.

	* gimple-range-gori.cc (compute_operand1_and_operand2_range):
	Check for interdependence between operands 1 and 2.

1 files changed, 17 insertions, 4 deletions

diff --git a/gcc/gimple-range-gori.cc b/gcc/gimple-range-gori.cc
index abc70cd..4ee0ae3 100644
--- a/gcc/gimple-range-gori.cc
+++ b/gcc/gimple-range-gori.cc
@@ -1291,13 +1291,26 @@ gori_compute::compute_operand1_and_operand2_range (vrange &r,
 {
   Value_Range op_range (TREE_TYPE (name));
 
-  // Calculate a good a range for op2.  Since op1 == op2, this will
-  // have already included whatever the actual range of name is.
-  if (!compute_operand2_range (op_range, handler, lhs, name, src, rel))
+  // If op1 is in the def chain of op2, we'll do the work twice to evalaute
+  // op1.  This can result in an exponential time calculation.
+  // Instead just evaluate op2, which will eventualy get to op1.
+  if (in_chain_p (handler.operand1 (), handler.operand2 ()))
+    return compute_operand2_range (r, handler, lhs, name, src, rel);
+
+  // Likewise if op2 is in the def chain of op1.
+  if (in_chain_p (handler.operand2 (), handler.operand1 ()))
+    return compute_operand1_range (r, handler, lhs, name, src, rel);
+
+  // Calculate a good a range through op2.
+  if (!compute_operand2_range (r, handler, lhs, name, src, rel))
     return false;
 
+  // If op1 == op2 there is again no need to go further.
+  if (handler.operand1 () == handler.operand2 ())
+    return true;
+
   // Now get the range thru op1.
-  if (!compute_operand1_range (r, handler, lhs, name, src, rel))
+  if (!compute_operand1_range (op_range, handler, lhs, name, src, rel))
     return false;
 
   // Both operands have to be simultaneously true, so perform an intersection.