折叠Haskell中的多态列表

Question

我有一个记录集合,分布在一个相互引用的大型Haskell应用程序中的多种类型中.涉及的所有类型都实现了一个共同的类型类.类型类包含对变量及其所有子项起作用的para函数,非常类似于uniplate的函数.

这是我想要构建的简化代码示例.是否有可能(并且合理)获得通用功能来折叠在GHC中实现给定类型类的记录字段...

{-# LANGUAGE RankNTypes #-}

myPara :: forall a r . (Data a, Foo a)
       => (forall b . Foo b => b -> [r] -> r)
       -> a
       -> r

-- or as a fold
myFold :: forall a r . (Data a, Foo a)
       => (forall b . Foo b => r -> b -> r)
       -> r
       -> b
       -> r

但通用到足以与任意类型类一起使用？

{-# LANGUAGE ExistentialQuantification #-}
{-# LANGUAGE DeriveDataTypeable #-}

import Data.Data
import Data.Generics.Uniplate.Data

class Foo a where 
  fooConst :: a -> Int

data Bar = Bar {barBaz :: Baz} deriving (Typeable, Data)

instance Foo Bar where
  fooConst _ = 2

data Baz = Baz {barBar :: Bar} deriving (Typeable, Data)

instance Foo Baz where
  fooConst _ = 3

func :: Int
func = foldl (\ x y -> x + fooConst y) 0 instances where
  instances :: forall a . (Data a, Foo a) => [a]
  instances = universeBi bar
  bar = Bar{barBaz = baz}
  baz = Baz{barBar = bar}

使用GHC 7.2.1进行编译(显然)失败:

Repro.hs:21:42:
    Ambiguous type variable `a0' in the constraints:
      (Data a0) arising from a use of `instances' at Repro.hs:21:42-50
      (Foo a0) arising from a use of `instances' at Repro.hs:21:42-50
    Probable fix: add a type signature that fixes these type variable(s)
    In the third argument of `foldl', namely `instances'
    In the expression: foldl (\ x y -> x + fooConst y) 0 instances
    In an equation for `func':
        func
          = foldl (\ x y -> x + fooConst y) 0 instances
          where
              instances :: forall a. (Data a, Foo a) => [a]
              instances = universeBi bar
              bar = Bar {barBaz = baz}
              baz = Baz {barBar = bar}

Answer 1

已经有一段时间..

您是否尝试过存在量化的数据构造器？

data Foo = forall a. MyTypeClass a => Bar [a]

func (Bar l) = map typeClassMember a

现在，func将适用于 Foo 类型的任何内容，它隐藏了内部类型。