Don't constant-fold (negateFloat# 0.0#), because the compiler's
internal representation of floating-point literals (Rational) can't
represent -0.0. This means that
main = print (-0.0)
now gives the same results with -O as it does without.
Fixes test arith005.
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-negOp (MachFloat f) = Just (mkFloatVal (-f))
-negOp (MachDouble d) = Just (mkDoubleVal (-d))
-negOp (MachInt i) = intResult (-i)
-negOp l = Nothing
+negOp (MachFloat 0.0) = Nothing -- can't represent -0.0 as a Rational
+negOp (MachFloat f) = Just (mkFloatVal (-f))
+negOp (MachDouble 0.0) = Nothing
+negOp (MachDouble d) = Just (mkDoubleVal (-d))
+negOp (MachInt i) = intResult (-i)
+negOp l = Nothing
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intOp2 op (MachInt i1) (MachInt i2) = intResult (i1 `op` i2)