Making (a, a) a Functor

问题

How can I make (a, a) a Functor without resorting to a newtype?

Basically I want it to work like this:

instance Functor (a, a) where
  fmap f (x, y) = (f x, f y)

But of course that's not a legal way to express it:

Kind mis-match
The first argument of `Functor' should have kind `* -> *',
but `(a, a)' has kind `*'
In the instance declaration for `Functor (a, a)'

What I really want is a type-level function like this: \a -> (a, a) (invalid syntax). So a type alias, perhaps?

type V2 a = (a, a)
instance Functor V2 where
    fmap f (x, y) = (f x, f y)

I would think this would work, but it doesn't. First I get this complaint:

Illegal instance declaration for `Functor V2'
(All instance types must be of the form (T t1 ... tn)
 where T is not a synonym.
 Use -XTypeSynonymInstances if you want to disable this.)
In the instance declaration for `Functor V2'

If I follow the advice and add the TypeSynonymInstances extension, I get a new error:

Type synonym `V2' should have 1 argument, but has been given 0
In the instance declaration for `Functor V2'

Well, duh, that's the point! V2 has kind * -> * which is what is required of a Functor instance. Well, ok, I can use a newtype like this:

newtype V2 a = V2 (a, a)
instance Functor V2 where
  fmap f (V2 (x, y)) = V2 (f x, f y)

But now I've got to sprinkle V2s liberally throughout my code instead of just being able to deal with simple tuples, which kind of defeats the point of making it a Functor; at that point I might as well make my own function vmap :: (a -> b) -> (a, a) -> (b, b).

So is there any way to do this nicely, i.e. without a newtype?

回答1:

As others have stated, there's no way to do this without resorting to newtypes or data declarations. However, have you looked at Control.Arrow? Many of those functions are very useful with tuples, for example:

vmap :: (a -> b) -> (a,a) -> (b,b)
vmap f = f *** f

回答2:

You can declare

instance Functor ((,) a) where
  ...

However that doesn't constrain the first element of your pair, and fmap would only act on the second element.

The issue is that a tuple doesn't enforce a relationship between the types of the two elements.

If you don't want a newtype decorator you can make your own fresh type:

data Pair a = P a a

instance Functor Pair where
  ...

which will be easier to work with than a newtype around a tuple.

回答3:

With singletons you can define a Functor type class for defunctionalized symbols (Type ~> Type instead of Type -> Type)

{-# Language ExplicitNamespaces, TypeApplications, TypeOperators, KindSignatures, ScopedTypeVariables, DataKinds, TypeInType, TypeFamilies, AllowAmbiguousTypes, InstanceSigs #-}

import Data.Kind (Type)
import Data.Singletons (type (~>), Apply)

class Functor' (f :: Type ~> Type) where
  fmap' :: (a -> a') -> (Apply f a -> Apply f a')

data Dup :: Type ~> Type

type instance Dup `Apply` a = (a, a)

instance Functor' Dup where
  fmap' :: (a -> a') -> ((a, a) -> (a', a'))
  fmap' f (a1, a2) = (f a1, f a2)

This gives you a Prelude.Functor instance automatically

newtype f $ a = App (Apply f a)

instance Functor' f => Functor (($) f) where
  fmap :: (a -> a') -> (f $ a -> f $ a')
  fmap f (App fa) = App (fmap' @f f fa)

来源：https://stackoverflow.com/questions/4812633/making-a-a-a-functor

标签

haskell

types

functor

typeclass