[ranger] x == -0.0 does not mean we can replace x with -0.0

On the true side of x == -0.0, we can't just blindly value propagate
the -0.0 into every use of x because x could be +0.0.

With this change, we only allow the transformation if
!HONOR_SIGNED_ZEROS or if the range is known not to contain 0.

gcc/ChangeLog:

	* range-op-float.cc (foperator_equal::op1_range): Do not blindly
	copy op2 range when honoring signed zeros.

diff --git a/gcc/range-op-float.cc b/gcc/range-op-float.cc
index ad2fae5..ff9fe31 100644 (file)
--- a/gcc/range-op-float.cc
+++ b/gcc/range-op-float.cc
@@ -252,8 +252,21 @@ foperator_equal::op1_range (frange &r, tree type,
   switch (get_bool_state (r, lhs, type))
     {
     case BRS_TRUE:
-      // If it's true, the result is the same as OP2.
-      r = op2;
+      if (HONOR_SIGNED_ZEROS (type)
+         && op2.contains_p (build_zero_cst (type)))
+       {
+         // With signed zeros, x == -0.0 does not mean we can replace
+         // x with -0.0, because x may be either +0.0 or -0.0.
+         r.set_varying (type);
+       }
+      else
+       {
+         // If it's true, the result is the same as OP2.
+         //
+         // If the range does not actually contain zeros, this should
+         // always be OK.
+         r = op2;
+       }
       // The TRUE side of op1 == op2 implies op1 is !NAN.
       r.set_nan (fp_prop::NO);
       break;