I'm trying to write something like this in Haskell:
length . nub . intersect
but it doesn't work.
*Main Data.List> :t intersect
intersect :: Eq a => [a] -> [a] -> [a]
*Main Data.List> :t nub
nub :: Eq a => [a] -> [a]
*Main Data.List> :t length
length :: [a] -> Int
Based on the type, my understanding is that intersect returns a type of [a] and donates to nub , which takes exactly a type of [a] , then also returns a type of [a] to length , then finally the return should be an Int. What's wrong with it?
The problem here is that intersect takes 2 arguments (in a sense)
you can provide one of the arguments explicitly:
> let f a = length . nub . intersect a
> :t f
f :: Eq a => [a] -> [a] -> Int
or you can use a fun little operator like (.:) = (.) . (.):
> let (.:) = (.) . (.)
> :t length .: (nub .: intersect)
length .: (nub .: intersect) :: Eq a => [a] -> [a] -> Int
here is a version where you don't need the parens:
import Data.List
infixr 9 .:
(.:) :: (c -> d) -> (a -> b -> c) -> a -> b -> d
(.:) = (.).(.)
f :: Eq a => [a] -> [a] -> Int
f = length .: nub .: intersect