如何将任何类型的列表作为函数参数 Haskell

Question

Hello I want to take lists of any type eg [Int] , [Char] , etc as parameters to a function.

Basically all I want to do is something along the lines of:

xyz :: [a] -> [a] -> (Int, Int)

Were [a] could be a list of any type.

Answer 1

Well, you pretty much got it already. Just write the signature like that and define your function. Except in a real-life application you will (probably; see @amalloy's comment) need to require a to belong to some type class, in order to do something meaningful with it:

xyz :: Integral a => [a] -> [a] -> (Int, Int)
xyz ls1 ls2 = (x, y) where
    x = fromInteger $ toInteger $ sum ls1
    y = fromInteger $ toInteger $ sum ls2