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.
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