什么是Haskell的箭头关联性？

Question

I am currently learning haskell and struggling with the following test : Suppose we have such a type :

type Endo a = a -> a

I have to choose all types that are equivalent to Endo (Endo Int)

(Int -> Int) -> (Int -> Int)
(Int -> Int) -> Int -> Int
Int
Int -> Int
Int -> Int -> Int -> Int
(Int -> Int) -> Int
Int -> Int -> (Int -> Int)
Int -> (Int -> Int)
Int -> Int -> Int

Since the type of Endo Int is Int -> Int I understand that type I need has 4 Ints such as (Int -> Int) -> (Int -> Int) . But I do not really understand which parenthesis are unnecessary

Answer 1

It is right associative in Haskell, so following are equivalent:

a -> b -> c
a -> (b -> c)

Function that takes arguments a and b is equivalent to function that given argument a returns function that takes argument b.

It is not left associative.

So the answer is first two.

Answer 2

Your intuition is correct: (Int -> Int) -> (Int -> Int) is indeed Endo (Endo Int) .

Just recall that -> associates on the right, ie

a -> b -> c     means    a -> (b -> c)

Given that, you should now be able to solve the exercise.