[Haskell-beginners] Question regarding infering the type of a function

[Joe Fredette]

The type of `(.)` is

     (.) :: (b->c) -> (a->b) -> a -> c

it can be defined as:

     (.) f g x = f (g x)

So, let f = (.), then we can plug that in and make it infix in the  
defn to get:

     (.) (.) g x q y = g x . q $ y

I had to add a few variables there to make it pointed, as opposed to  
the more succint semipointfree version:

     (.) (.) g x = (.) g x

So we can see that this function, `(.) (.)` takes four arguments, a  
function on 2 inputs (g, because (.) is going to feed it another one,  
and it's taking one already, in the form of x). One of the variables  
of appropriate some arbitrary type `t` (since g is the only thing that  
will ever see it, you can think of it as a kind of index to an indexed  
family of functions of type `b->c`.) giving `(.) (.)` just `x` will  
reduce the type to something you should recognize, that is if:

     foo x = something in (\ g q y -> (.) (.) g x' q y)

then foo has type:

     foo :: (b -> c) -> (a -> b) -> a -> c

eg, foo = (.)

So, (.) (.) effectively says, "Given a indexed family of functions  
from `b -> c`, indexed by some type `t` compose the `t`th function  
with some function `q` from `a -> b`, and return a function of type `a  
-> c`.

You might use it for something like mapping over a list at a given  
starting point, or something.

HTH.

/Joe


On Sep 21, 2009, at 3:47 PM, MoC wrote:

> Hi everybody,
>
> today somebody on IRC (#haskell on Freenode) jokingly submitted the  
> following to lambdabot:
> > (.) (.)
> It turned out that (.) (.) :: (a1 -> b -> c) -> a1 -> (a -> b) -> a - 
> > c. I got interested in why that is so, but unfortunately I have  
> not been able to figure this out by myself yet. I know that partial  
> application is used, but don't know to what it would expand to and  
> why.
> Could someone please explain?
> (I asked on the channel, but nobody did answer so I thought I'd try  
> my luck here.)
>
> Regards,
>   MoC
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