[Haskell-beginners] Re: Function composition with more than 1 parameter
Daniel Fischer
daniel.is.fischer at web.de
Sat Oct 25 19:36:00 EDT 2008
Am Sonntag, 26. Oktober 2008 00:44 schrieb Glurk:
> > Sometimes pointfree style is clearer and more readable, sometimes it
> > makes things obscure.
> > Get some experience writing both, pointfree and pointful versions, to
>
> develop
>
> > a feeling when to write pointfree and when not (and when to use mixed
> > style, e.g. "hm x = length . matches x" might be clearer).
>
> I think that's good advice :)
> Having a better understanding of how it works is never a bad thing ! :)
>
> I still have trouble understanding a simple one like :-
>
> matches = filter . (==)
>
> which actually is kind of nice and clean, but when I try to understand it I
> get confused. In particular I wonder how it "knows" where the parameters go
> ?
>
> (==) is of type (==) :: Eq a => a -> a -> Bool
>
> yet I'm giving it incorrect parameters, I'm giving it a [a]
> not a a (or am I ?)
For any composition f . g, the evaluation goes
(f . g) x = f (g x), so
(filter . (==)) x = filter ((==) x)
Now if x has type a which is an instance of Eq, x is a suitable argument for
(==) and ((==) x) has type a -> Bool, hence ((==) x) is a suitable argument
for filter, so what happens is exactly what you describe below.
>
> Or, does it take the only valid parameter I give it, the a, and then form a
> function out of this (a==), and then compose this new function with filter,
> which does take a function of this new type. Then, there is one parameter
> left, [a], which is what this new function needs...and so it all works out
> !
>
> Why doesn't it try to take the 2 parameters for (==), and bomb out when it
> finds the second parameter is wrong ?
Because as soon as (==) has received its first parameter, the section (a ==)
becomes the first argument of filter, so the second argument to (==) is then
provided by filter (you remember:
filter prd (x:xs)
| prd x = x:filter prd xs
| otherwise = filter prd xs
filter _ [] = []
)
> Is this where the monomorphism
> restriction things comes into play ? Or now because I've given it a type,
> somehow it can now work things out.... ?
No, strictly speaking, every function has only one argument, it is only a
matter of convenience to say that functions which return functions take
several arguments. And it's the fact that a function has only one argument
which is the reason why it works out.
>
> Does anyone know of a good tutorial on how this all works ? :)
>
> Thanks
>
More information about the Beginners
mailing list