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我在 Haskell 中实现了如下的转换器：

{-# LANGUAGE RankNTypes #-}

import Prelude hiding (foldr)
import Data.Foldable

type Reducer b a = a -> b -> b
type Transducer a b = forall t. Reducer t b -> Reducer t a

class Foldable c => Collection c where
    insert :: a -> c a -> c a
    empty  :: c a

reduce :: Collection c => Transducer a b -> c a -> c b
reduce f = foldr (f insert) empty

mapping :: (a -> b) -> Transducer a b
mapping f g x = g (f x)

现在我想定义一个通用map函数。因此，我将上述代码加载到 GHCi 中：

Prelude> :load Transducer
[1 of 1] Compiling Main             ( Transducer.hs, interpreted )
Ok, modules loaded: Main.
*Main> let map = reduce . mapping

<interactive>:3:20:
    Couldn't match type ‘Reducer t0 b1 -> Reducer t0 a1’
                  with ‘forall t. Reducer t b -> Reducer t a’
    Expected type: (a1 -> b1) -> Transducer a b
      Actual type: (a1 -> b1) -> Reducer t0 b1 -> Reducer t0 a1
    Relevant bindings include
      map :: (a1 -> b1) -> c a -> c b (bound at <interactive>:3:5)
    In the second argument of ‘(.)’, namely ‘mapping’
    In the expression: reduce . mapping
*Main> let map f = reduce (mapping f)
*Main> :t map
map :: Collection c => (a -> b) -> c a -> c b

所以我无法定义map = reduce . mapping。但是，我可以定义map f = reduce (mapping f).

我相信这个问题是由单态限制引起的。我真的很想写map = reduce . mapping而不是map f = reduce (mapping f)。因此，我有两个问题：

  1. 是什么导致了这个问题？确实是单态限制吗？
  2. 我该如何解决这个问题？
4

1 回答 1

5

如果你做Transducer一个newtype, 比 GHC 会更好地计算类型。存在类型变量不会逃脱范围 - 转换器将保持多态性。

换句话说，具有以下定义的map = reduce . mapping作品

{-# LANGUAGE RankNTypes #-}

import Prelude hiding (foldr, map, (.), id)
import Control.Category
import Data.Foldable

type Reducer b a = a -> b -> b
newtype Transducer a b = MkTrans { unTrans :: forall t. Reducer t b -> Reducer t a }

class Foldable c => Collection c where
    insert :: a -> c a -> c a
    empty  :: c a

instance Collection [] where
  insert = (:)
  empty = []

reduce :: Collection c => Transducer a b -> c a -> c b
reduce f = foldr (unTrans f insert) empty

mapping :: (a -> b) -> Transducer a b
mapping f = MkTrans $ \g x -> g (f x)

filtering :: (a -> Bool) -> Transducer a a
filtering f = MkTrans $ \g x y -> if f x then g x y else y

map :: Collection c => (a -> b) -> c a -> c b
map = reduce . mapping

filter :: Collection c => (a -> Bool) -> c a -> c a
filter = reduce . filtering

instance Category Transducer where
  id = MkTrans id
  MkTrans f . MkTrans g = MkTrans $ \x -> g (f x)

dub :: Num a => a -> a
dub x = x + x

test1 :: [Int]
test1 = reduce (filtering even . mapping dub) [1..10]
-- [2,4,6,8,10,12,14,16,18,20]

test2 :: [Int]
test2 = reduce (mapping dub . filtering even) [1..10]
-- [4,8,12,16,20]

*Main> :t reduce . mapping
reduce . mapping :: Collection c => (a -> b) -> c a -> c b

您还可以查看http://www.reddit.com/r/haskell/comments/2cv6l4/clojures_transducers_are_perverse_lenses/ 其中定义type Transducer a b =:: (a -> Constant (Endo x) a) -> (b -> Constant (Endo x) b)和其他各种。还有其他有趣的讨论。

于 2015-01-12T10:04:58.727 回答