[Haskell-cafe] Fucntion composing
Daniel Fischer
daniel.is.fischer at googlemail.com
Mon Apr 11 12:34:26 CEST 2011
On Monday 11 April 2011 11:22:51, Adam Krauze 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?
Composition treats one argument, so
h x y = g 0 (f x y)
= (g 0) (f x y)
~>
h x = (g 0) . (f x)
~>
h = ((g 0) .) . f
for each argument of the function to be applied first, you need one
composition operator.
But (((foo .) .) .) . bar and such quickly become unreadable, so know when
to stop point-freeing such multi-compositions.
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