haskell - 创建完全依赖的串联

Question

关于串联的一个很好的真实事实是，如果我知道方程中的任何两个变量:

a ++ b = c

然后我知道第三个。

我想在我自己的 concat 中捕捉到这个想法，所以我使用了函数依赖。

{-# Language DataKinds, GADTs, FlexibleContexts, FlexibleInstances, FunctionalDependencies, KindSignatures, PolyKinds, TypeOperators, UndecidableInstances #-}
import Data.Kind (Type)

class Concatable
   (m  :: k -> Type)
   (as :: k)
   (bs :: k)
   (cs :: k)
   | as bs -> cs
   , as cs -> bs
   , bs cs -> as
   where
     concat' :: m as -> m bs -> m cs

现在我想出这样的异构列表:

data HList ( as :: [ Type ] ) where
  HEmpty :: HList '[]
  HCons  :: a -> HList as -> HList (a ': as)

但是当我尝试将这些声明为 Concatable我有一个问题

instance Concatable HList '[] bs bs where
  concat' HEmpty bs = bs
instance
  ( Concatable HList as bs cs
  )
    => Concatable HList (a ': as) bs (a ': cs)
  where
    concat' (HCons head tail) bs = HCons head (concat' tail bs)

我不满足我的第三个功能依赖。或者更确切地说，编译器认为我们不相信。
这是因为编译器认为在我们的第二个实例中， bs ~ (a ': cs) 可能是这种情况。 .如果 Concatable as (a ': cs) cs 可能是这种情况.

如何调整我的实例以使所有三个依赖项都得到满足？

Answer 1

所以，正如评论所暗示的，GHC 不会自己解决它，但我们可以通过一些类型级编程来帮助它。介绍一些TypeFamilies .所有这些函数都是将列表操作提升到类型级别的非常简单的翻译:

-- | This will produce the suffix of `cs` without `as`
type family DropPrefix (as :: [Type]) (cs :: [Type]) where
  DropPrefix '[] cs = cs
  DropPrefix (a ': as) (a ': cs) = DropPrefix as cs

-- Similar to the logic in the question itself: list concatenation. 
type family Concat (as :: [Type]) (bs :: [Type]) where
  Concat '[] bs = bs
  Concat (head ': tail) bs = head ': Concat tail bs

-- | Naive list reversal with help of concatenation.
type family Reverse (xs :: [Type]) where
  Reverse '[] = '[]
  Reverse (x ': xs) = Concat (Reverse xs) '[x]

-- | This will produce the prefix of `cs` without `bs`
type family DropSuffix (bs :: [Type]) (cs :: [Type]) where
  DropSuffix bs cs = Reverse (DropPrefix (Reverse bs) (Reverse cs))

-- | Same definition of `HList` as in the question
data HList (as :: [Type]) where
  HEmpty :: HList '[]
  HCons :: a -> HList as -> HList (a ': as)

-- | Definition of concatenation at the value level
concatHList :: (cs ~ Concat as bs) => HList as -> HList bs -> HList cs
concatHList HEmpty bs = bs
concatHList (HCons head tail) bs = HCons head (concatHList tail bs)

使用我们可以使用的这些工具，我们实际上可以达到小时目标，但首先让我们定义一个具有所需属性的函数:

推理能力cs来自 as和 bs

推理能力as来自 bs和 cs

推理能力bs来自 as和 cs

瞧:

concatH ::
     (cs ~ Concat as bs, bs ~ DropPrefix as cs, as ~ DropSuffix bs cs)
  => HList as
  -> HList bs
  -> HList cs
concatH = concatHList

让我们测试一下:

foo :: HList '[Char, Bool]
foo = HCons 'a' (HCons True HEmpty)

bar :: HList '[Int]
bar = HCons (1 :: Int) HEmpty

λ> :t concatH foo bar
concatH foo bar :: HList '[Char, Bool, Int]
λ> :t concatH bar foo
concatH bar foo :: HList '[Int, Char, Bool]

最后是最终目标:

class Concatable (m :: k -> Type) (as :: k) (bs :: k) (cs :: k)
  | as bs -> cs
  , as cs -> bs
  , bs cs -> as
  where
  concat' :: m as -> m bs -> m cs

instance (cs ~ Concat as bs, bs ~ DropPrefix as cs, as ~ DropSuffix bs cs) =>
         Concatable HList as bs cs where
  concat' = concatH

λ> :t concat' HEmpty bar
concat' HEmpty bar :: HList '[Int]
λ> :t concat' foo bar
concat' foo bar :: HList '[Char, Bool, Int]
λ> :t concat' bar foo
concat' bar foo :: HList '[Int, Char, Bool]