Andrew Coppin wrote: > Chaddaï Fouché wrote: >> 2007/9/25, Andrew Coppin <[EMAIL PROTECTED]>: >> >>> This is why I found it so surprising - and annoying - that you can't >>> use >>> a 2-argument function in a point-free expression. >>> >>> For example, "zipWith (*)" expects two arguments, and yet >>> >>> sum . zipWith (*) >>> >>> fails to type-check. You just instead write >>> >>> \xs ys -> sum $ zipWith(*) xs ys >>> >>> >> >> (sum . zipWith (*)) xs ys >> == (sum (zipWith (*) xs)) ys >> >> so you try to apply sum :: [a] -> Int to a function (zipWith (*) xs) >> :: [a] -> [b], it can't work ! >> >> (sum.) . zipWith (*) >> works, but isn't the most pretty expression I have seen. >> > > I'm still puzzled as to why this breaks with my example, but works > perfectly with other people's examples... > > So you're saying that > > (f3 . f2 . f1) x y z ==> f3 (f2 (f1 x) y) z > > ? In that case, that would mean that > > (map . map) f xss ==> map (map f) xss > > which *just happens* to be what we want. But in the general case where > you want > > f3 (f2 (f1 x y z)) > > there's nothing you can do except leave point-free. Well, there's one thing. You can change your three argument function into a one argument function of a 3-tuple, and then change the composed function back again:
let uncurry3 = \f (x,y,z) -> f x y z curry3 = \f x y z -> f (x,y,z) in curry3 $ f3 . f2 . uncurry3 f1 In your earlier example, this would have been: curry $ sum . uncurry (zipWith (*)) _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe