Mitar wrote: > > Why is 0/0 (which is NaN) > 1 == False and at the same time 0/0 < 1 == > > False. This means that 0/0 == 1? No, because also 0/0 == 1 == False. > > I understand that proper mathematical behavior would be that as 0/0 is > > mathematically undefined that 0/0 cannot be even compared to 1. > > There is probably an implementation reason behind it, but do we really > > want such "hidden" behavior? Would not it be better to throw some kind > > of an error?
Like nearly all programming languages, Haskell implements the standard IEEE behavior for floating point numbers. That leads to some mathematical infelicities that are especially irking to us in Haskell, but the consensus was that it is best to follow the standard. That is why NaN==NaN, NaN<NaN, and NaN>Nan are all False, The special case of 1/0 is less clear, though. One might decide that it should be an error rather than NaN, as some languages have. Cristian Baboi wrote: > I think it's a bug. > Here is why: > > let f = (\x -> x/0) in f 0 == f 0 > > Referential transparency say that f 0 must equal to f 0, but in this case > it is not. :-) This does not violate referential transparency. f 0 is always the same value. (==) is a function like any other; in this case, it does not satisfy the mathematical laws we would like it to in order to conform to standard floating point behavior. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
