[Haskell-cafe] Fucntion composing

[Daniel Fischer]

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.