On Tue, 2007-09-25 at 12:24 +0100, Andrew Coppin wrote:

Chaddaï Fouché wrote:

...

2007/9/25, Andrew Coppin :

...

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.

In effect, yes. As many others have pointed out, you *can* do this point-free (all of lambda calculus can be re-written using the functions s f g x = f x (g x) k x y = x (1)), but it's not a good idea. Good style in Haskell as in any other language is a matter of taste, not rules. jcc (1) http://www.madore.org/~david/programs/unlambda/

Andrew Coppin wrote:

Chaddaï Fouché wrote:

...

2007/9/25, Andrew Coppin :

...

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 (*))