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by a1369209993
972 days ago
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> (I originally claimed "+0 + -0 = +0 vs. -0 + +0 = -0" which is incorrect) Specifically, (-0.0) + (+0.0) = (+0.0) + (-0.0) = +0.0 (assuming round-to-nearest-or-even). OTOH, (-0.0) + (-0.0) = -0.0. This has nothing to do with +0.0 == -0.0 for comparison, addition just is commutative outright[0]. 0: Pedantically, I'm not sure IEEE-754 requires the specific choice of which NaN you get when you do `some_nan + a_different_nan` versus `a_different_nan + some_nan` to be commutative, but it should. |
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