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 _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners