What you've actually defined is function application. Function composition has a different type: (.) :: (b -> c) -> (a -> b) -> a -> c (I'd read this as: compose takes a function from b to c, a function from a to b, and something of type a, and produces a c.) You ran into an order of operations issue that masked this bug: p . q 4 /= (p . q) 4 Since you're working through this I don't want to give away the actual definition of function composition, but be sure to follow the type. On 2012-01-15 16.17.24 +0100, TP wrote:
Hi,
I have a basic question concerning function composition. I have used http://www.haskell.org/tutorial/functions.html to write a composition function:
Prelude> let f�g = f g Prelude> let p = (*2) Prelude> let q = (+3) Prelude> p�q 4 14 Prelude> :t (�) (�) :: (t1 -> t) -> t1 -> t
If I understand well, this means that the infix operator "�" takes a function of type t1, i.e. g in f�g, and applies f on it, which takes a type t1 as input and returns f(g) which is of type t. The final result is of type t. So the first argument is represented above by "(t1->t)", and the second by "t1", the final result being of type t.
However, I am not able to get the type of p�q
Prelude> :t p�q
<interactive>:1:3: Couldn't match expected type `Integer' with actual type `Integer -> Integer' In the second argument of `(�)', namely `q' In the expression: p � q Prelude>
What's the problem here? How can I obtain the type of p�q?
Thanks in advance,
TP
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