Question

请原谅我可能滥用下面的类别理论术语。如果我觉得我有一半的线索，我会认为自己非常成功。

我发现自己编写了一系列类来处理多个arities类型构造函数的产品。像这样：

import Control.Applicative

-- | RWS monad.
newtype RWS r w s a = RWS {runRWS :: r -> s -> (a, s, w)}

-- | A class for unary type constructors that support a Cartesian
-- product operation.
class ProductObject f where
    (***) :: f a -> f b -> f (a, b)
infixr ***

-- | Example instance of 'ProductObject'.
instance ProductObject [] where
    (***) = liftA2 (,)


-- | A class for binary type constructors (read as morphisms
-- between their type parameters) that support a product morphism
-- operation.
class ProductMorphism arrow where
    (****) :: arrow a b -> arrow c d -> arrow (a, c) (b, d)
infixr ****

-- | Example instance of 'ProductMorphism'.  
instance ProductMorphism (->) where
    f **** g = \(a, c) -> (f a, g c)


-- | A class for ternary type constructors (read as two-place
-- multiarrows @a, b -> c@) with products.
class ProductMultimorphism2 arr2 where
    (*****) :: arr2 a b c -> arr2 d e f -> arr2 (a, d) (b, e) (c, f)
infixr *****


-- | A class for ternary type constructors (read as two-place
-- multiarrows @a, b -> c@) with products.
class ProductMultimorphism3 arr3 where
    (******) :: arr3 a b c d -> arr3 e f g h -> arr3 (a, e) (b, f) (c, g) (d, h)
infixr ******

-- | Let's pretend that the 'RWS' monad was not a type synonym
-- for 'RWST'.  Then an example of 'ProductMorphism3' would be:
instance ProductMultimorphism3 RWS where
    f ****** g = RWS $ \(fr, gr) (fs, gs) ->
        let (fa, fs', fw) = runRWS f fr fs
            (ga, gs', gw) = runRWS g gr gs
        in ((fa, ga), (fs', gs'), (fw, gw))

现在，由于几个原因，这很烦人。最大的一个是我必须修改我的应用程序中的一个类型以添加第三个参数，这意味着现在我必须查找该类型的****的所有用法并将其更改为*****

我可以申请一些技巧来缓解这种情况吗？我试图理解GHC中的PolyKinds是否适用于此，但（a）这种情况正在缓慢进行，（b）我听说GHC 7.4.x中的PolyKinds是错误的。

Answer 1

好吧，我自己想出来了：

{-# LANGUAGE MultiParamTypeClasses, TypeFamilies #-}

import Control.Applicative

class Product f g where
    type Prod a b :: *
    (***) :: f -> g -> Prod f g


instance Product [a] [b] where
    type Prod [a] [b] = [(a, b)]
    (***) = liftA2 (,)

instance Product (a -> b) (c -> d) where
    type Prod (a -> b) (c -> d) = (a, c) -> (b, d)
    f *** g = \(a, c) -> (f a, g c)

如何简化几个arities类型构造函数的“product”类型类？

1 个答案: