[Haskell-beginners] help with types and composition

Thomas Davie tom.davie at gmail.com
Sat Jul 4 06:33:24 EDT 2009


Note here that where liftM2 and ap are used, we can be more general  
than monads -- we need only Applicatives instead, substituting liftA2  
and <*> instead.

Also, compose12 can be made more general using (fmap . fmap), and  
becomes the two level fmap -- instead of applying the function inside  
one layer of functor it goes two layers in.  Similarly with hook, we  
can use liftA2 fmap, and be generalised over all functors, not just  
functions.

Bob

On 4 Jul 2009, at 11:15, Geoffrey Marchant wrote:

> Functions in Haskell aren't distinguished as 1-arg of 2-arg  
> functions. Usually we don't even think of them as such -- rather we  
> think of the type of the function or operator.
>
> But there is a wide array of combinators available:
>
> K = const
> I = id
> B = (.)
> C = flip
> S = Monad.ap  -- instance Monad (->) a   - the Reader Monad
> W = Monad.join -- also for the Reader Monad
> Y = Control.Monad.Fix.fix
>
> and the list goes on and on... Using only S and K, a great many  
> functions can be expressed, though it does get rather ugly.
>
> Perhaps the most significant reason that Haskell doesn't have more  
> composition operators is that composing functions is the least part  
> of what we compose. Using some standard class functions often yields  
> exactly what you're looking for:
>
> hook = liftM2 (.)
> mhook = ap
> fork = liftM2
> dyfork = liftM2 . liftM2
> compose12 = (.) (.) (.)  -- as previously shown
>
> Written in this form, the first four become far more general:
>
> "hook" is function composition within a monad;
> "mhook" is function application within a monad;
> "fork" raises a function for application on monads; and
> "dyfork" -- absolutely beautiful, BTW -- raises a function for  
> application upon nested monads.
>
> BTW, if anyone has a reasonably concise form for a point-free  
> compose21, I'd like to see it.
>
>
> On Fri, Jul 3, 2009 at 7:15 PM, Troy Pracy <troyp7 at gmail.com> wrote:
> I've just started learning Haskell and I've been wondering about  
> this issue as well. I can usually work out a point-free version by  
> carefullty deriving it step-by-step, but I was wondering if Haskell  
> had composition operators/functions for dealing with the various  
> forms of composition where a 2-arg function is involved.
>
> I've played around with J (APL's successor) a little and noticed  
> that J has various options for composing two functions (Ponit-free,  
> or "implicit" style is very important in J). Some of the  
> distinctions have to do with J's  native array operations and aren't  
> relevant here, but many are. Here are Haskell versions...
> (note: "monadic" below isn't used in the Haskell/CT sense -  
> "monadic" and "dyadic" in J jsut refer to how many arguments an  
> operator acts on)
>
> -- hook is the J dyadic hook as a function.
> hook :: (a->b->c) -> (d->b) -> a -> d -> c
> hook f g = \x y -> f x (g y)
> -- J's monadic hook
> mhook :: (a->b->c) -> (a->b) -> a -> c
> mhook f g = \x -> (hook f g) x x
> -- J's monadic fork
> fork :: (a->b->c) -> (d->a) -> (d->b) -> d  -> c
> fork f g h = \x -> f (g x) (h x)
> -- J's dyadic fork
> dyfork :: (a->b->c) -> (d->e->a) -> (d->e->b) -> d -> e -> c
> dyfork f g h = \x y -> f (g x y) (h x y)
> -- J's dyadic @ or @: - composition of 1-arg fn with 2-arg fn
> compose12 :: (a->b) -> (c->d->a) -> c -> d -> b
> compose12 f g = \x y -> f (g x y)
> (@:) = compose12
> {-  J's dyadic & or &: - composition of 2-arg fn with 1-arg fn,  
> resulting in a
>   2-arg fn (f&:g) which applies g to *both* args before passing them  
> to f.
>   Haskell's composition operator and partial application allow a  
> composition
>   of such fns (f . g) where g is applied only to the first arg.    -}
> compose21 :: (a->a->b) -> (c->a) -> c -> c -> b
> compose21 f g = \x y -> f (g x) (g y)
> (&:) = compose21
>
>
> / /I know a lot of Haskell is written in point-free style and I  
> would have thought Haskell would have operators for some of this,  
> but judging from the previous responses, it looks like it might not.  
> That surprises me, since some of this seems to crop up a lot, but as  
> I said I've just started learning Haskell, so I guess I'll have to  
> give myself some time to absorb the Haskell way of doing things.
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