You're right, producing +/- 0 depends on the rounding mode and it is commutative, I forgot about that completely. I edited my comment to fix that claim.
Also, yes, +0 == -0, but they can produce different results when used in the same expression, so the distinction does matter (although this doesn't affect commutativity, which is the larger point). For example, let f(x) = 1 / x. Then f(+0) = +inf, f(-0) = -inf.
I also agree with you about NaN, that's why I mentioned having to go outside floating point numbers (bit-casting).
Also, yes, +0 == -0, but they can produce different results when used in the same expression, so the distinction does matter (although this doesn't affect commutativity, which is the larger point). For example, let f(x) = 1 / x. Then f(+0) = +inf, f(-0) = -inf.
I also agree with you about NaN, that's why I mentioned having to go outside floating point numbers (bit-casting).