There exists a representation of a negative double whose negation is also negative, i.e. a < 0 && -a < 0 holds true? I'm completely out of my depth.
My answer to "implement fabs(), incorrectly". Seven and one third internet points to anyone who can say why it's incorrect, without looking up the spec. Hint: It's implementation defined. (Using C99 semantics because that's the only one I know reasonably well. :)https://twitter.com/oe1cxw/status/1144274172253822976 …
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No. That's only possible with twos-complement integers. But you are very close.
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Does it treat NaNs incorrectly in some way? Alternately, might it return -0 sometimes?
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Only one of those two things is true. ;)
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With -NaN as input, your function returns -NaN instead of +NaN.
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On some implementations yes, but I _think_ that's within allowed behavior for fabs(), so it's technically not incorrect.
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fmax on zeros is required to return one of them but it is implementation defined which so fabs(0) = fmax(+0, -0) = -0 is possible and that is not fabs!
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Seven and one third internet points for you! :D
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I don't see the problem with this one. Either the compiler doesn't promise IEEE 754 and -x doesn't even need to be exact, or it does but (see other discussion) the sign bit of NaN is meaningless (and even getting the correct payload is a stretch tbh). Is this about 0. and -0.?
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Yes, it's about negative zero. fmax(+0.0, -0.0) is allowed to return -0.0, and fabs(-0.0) is required to return +0.0.
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