Am Montag 21 September 2009 21:47:40 schrieb MoC:
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.)
And it's not futile :) The type of (.) is (forall a, b, c.) (b -> c) -> (a -> b) -> a -> c Now, to not get confused, choose different type variables for the type of the first (.) and the second, say the first has type (b -> c) -> (a -> b) -> a -> c and the second has type (v -> w) -> (u -> v) -> u -> w === (v -> w) -> ((u -> v) -> (u -> w)) Feeding it as an argument to the first, we must unify this type with the type of the first (.)'s (first) argument, (b -> c), thus b = (v -> w) c = (u -> v) -> u -> w and the type of (.) (.) becomes (a -> b) -> a -> c, which is (a -> (v -> w)) -> a -> ((u -> v) -> u -> w) by what we got for b and c. Now remove those parentheses which aren't necessary because (->) is right associative, to get (a -> v -> w) -> a -> (u -> v) -> u -> w and rename the type variables to get exactly lambdabot's answer.
Regards, MoC