>> pointfree "h x y = (g 0 (f x y))"

h = (g 0 .) . f

[1] http://hackage.haskell.org/package/pointfree

On 11 April 2011 10:22, Adam Krauze <ajschylos@mac.com> wrote:

Hello,
as I am newbie to Haskell and my introductory question is:

given functions say f and g with type signatures

f :: (Num a) => [a] -> [a] -> [(a,a)] // f takes two lists and zips them into one in some special way
g :: (Num a) => a -> [(a,a)] -> [a] // g using some Num value calculates list of singletons from list of pairs

of course g 0 :: (Num a) => [(a,a)] ->[a]

now I want to create function h :: (Num a) => [a] -> [a] -> [a] in such way

that (g 0) consumes output of f.

But when I try

Prelude> :t (g 0).f

I get an error:

<interactive>:1:9:
Couldn't match expected type `[(a0, a0)]'
with actual type `[a1] -> [(a1, a1)]'
Expected type: [a1] -> [(a0, a0)]
Actual type: [a1] -> [a1] -> [(a1, a1)]
In the second argument of `(.)', namely `f'
In the expression: (g 0) . f

In pointfull representation it works well

Prelude> let h x y = (g 0 (f x y))

How to do pointfree definition of h?

Ajschylos.

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Ozgur Akgun