Haskell - 形式A的所有函数 - > A - > ... - > A.

Question

I have a type (call it A) and I want to make a typeclass of the functions of type A -> A, A -> A -> A, A -> A -> A -> ... etc. This doesn't work:

{-# LANGUAGE FlexibleInstances #-}

data A = A

class AsToA a where
  takeA :: AsToA b => a -> A -> Either A b

instance AsToA (A -> A) where
  takeA f a = Left (f a)

instance AsToA b => AsToA (A -> b) where
  takeA f a = Right (f a)

I get the following error message:

AsToA.hs:12:22:
    Couldn't match expected type ‘b1’ with actual type ‘b’
      ‘b’ is a rigid type variable bound by
          the instance declaration at AsToA.hs:11:10
      ‘b1’ is a rigid type variable bound by
          the type signature for

             takeA :: AsToA b1 => (A -> b) -> A -> Either A b1
          at AsToA.hs:12:3
    Relevant bindings include
      f :: A -> b (bound at AsToA.hs:12:9)
      takeA :: (A -> b) -> A -> Either A b1 (bound at AsToA.hs:12:3)
    In the first argument of ‘Right’, namely ‘(f a)’
    In the expression: Right (f a)

Any ideas? Thanks very much for any advice.

Answer 1

There is some confusion between the two b s:

class AsToA a where
  takeA :: AsToA b => a -> A -> Either A b

instance AsToA b => AsToA (A -> b) where
  takeA f a = Right (f a)

These are not the same. Let's rename the first one to c

class AsToA a where
  takeA :: AsToA c => a -> A -> Either A c

instance AsToA b => AsToA (A -> b) where
  takeA f a = Right (f a)

Now, Right (fa) has type Either A b but should have type Either A c for any c such that AsToA c holds. This does not type check.

The issue here is that the signature

  takeA :: AsToA c => a -> A -> Either A c

promises that takeA can return Either A c for any c , caller's choice. This is not what you want, I guess.

---

I'm still not sure about what the actual intended result is, but I guess the problem is similar to the following one.

Given a function f of type A->A->...->A return a function \\x -> fxx ... , with one application of x for each -> in the type (hence of type A->A ).

A possible solution is

{-# LANGUAGE FlexibleInstances, OverlappingInstances #-}
data A = A -- could be anything

class C f where
   takeA :: f -> A -> A

instance C (A -> A) where
   takeA f = f

instance C b => C (A -> b) where
    takeA f = \x -> takeA (f x) x

Note that this requires OverlappingInstances to be used, which is quite evil . I'd recommend to avoid it.

To avoid it, in this case it's enough to define an instance even for the type A .

{-# LANGUAGE FlexibleInstances #-}
data A = A -- could be anything

class C f where
   takeA :: f -> A -> A

instance C A where
   takeA a = \_ -> a

instance C b => C (A -> b) where
    takeA f = \x -> takeA (f x) x

Answer 2

As mentioned in the comments to the other answer, you might not really need the Either , and takeA is then basically always id , just with a type restriction. If so you can make this a method-less class:

{-# LANGUAGE FlexibleInstances, FlexibleContexts #-}

data A = A

class AsToA a

takeA :: AsToA a => a -> a
takeA = id

instance AsToA (A -> A)

instance AsToA (A -> b) => AsToA (A -> (A -> b))

Alternatively, you might want to convert the functions to a common type that allows you to pass in A s dynamically. If so Either won't be enough, but you can define your own:

{-# LANGUAGE FlexibleInstances, FlexibleContexts #-}

data A = A

data R = Result A | MoreArgs (A -> R)

class AsToA a where
    takeA :: a -> A -> R

instance AsToA (A -> A) where
    takeA f a = Result (f a)

instance AsToA (A -> b) => AsToA (A -> (A -> b)) where
    takeA f a = MoreArgs (takeA $ f a)