我不能让我的基于GADT的玩具动态类型与参数类型一起使用

Question

因此,为了帮助我理解一些更高级的Haskell/GHC特性和概念,我决定采用基于GADT的动态类型数据实现,并将其扩展到参数类型.(我为这个例子的长度道歉.)

{-# LANGUAGE GADTs #-}

module Dyn ( Dynamic(..), 
             toDynamic,
             fromDynamic
           ) where

import Control.Applicative

----------------------------------------------------------------
----------------------------------------------------------------
--
-- Equality proofs
--

-- | The type of equality proofs.
data Equal a b where
    Reflexivity :: Equal a a
    -- | Inductive case for parametric types
    Induction   :: Equal a b -> Equal (f a) (f b)

instance Show (Equal a b) where
    show Reflexivity = "Reflexivity"
    show (Induction proof) = "Induction (" ++ show proof ++ ")"

----------------------------------------------------------------
----------------------------------------------------------------
--
-- Type representations
--

-- | Type representations.  If @x :: TypeRep a@, then @x@ is a singleton
-- value that stands in for type @a@.
data TypeRep a where 
    Integer :: TypeRep Integer
    Char :: TypeRep Char
    Maybe :: TypeRep a -> TypeRep (Maybe a)
    List :: TypeRep a -> TypeRep [a]

-- | Typeclass for types that have a TypeRep
class Representable a where
    typeRep :: TypeRep a

instance Representable Integer where typeRep = Integer
instance Representable Char where typeRep = Char

instance Representable a => Representable (Maybe a) where 
    typeRep = Maybe typeRep

instance Representable a => Representable [a] where 
    typeRep = List typeRep


-- | Match two types and return @Just@ an equality proof if they are
-- equal, @Nothing@ if they are not.
matchTypes :: TypeRep a -> TypeRep b -> Maybe (Equal a b)
matchTypes Integer Integer = Just Reflexivity
matchTypes Char Char = Just Reflexivity
matchTypes (List a) (List b) = Induction <$> (matchTypes a b)
matchTypes (Maybe a) (Maybe b) = Induction <$> (matchTypes a b)
matchTypes _ _ = Nothing


instance Show (TypeRep a) where
    show Integer = "Integer"
    show Char = "Char"
    show (List a) = "[" ++ show a ++ "]"
    show (Maybe a) = "Maybe (" ++ show a ++ ")"


----------------------------------------------------------------
----------------------------------------------------------------
--
-- Dynamic data
--

data Dynamic where
    Dyn :: TypeRep a -> a -> Dynamic

instance Show Dynamic where
    show (Dyn typ val) = "Dyn " ++ show typ

-- | Inject a value of a @Representable@ type into @Dynamic@.
toDynamic :: Representable a => a -> Dynamic
toDynamic = Dyn typeRep

-- | Cast a @Dynamic@ into a @Representable@ type.
fromDynamic :: Representable a => Dynamic -> Maybe a
fromDynamic = fromDynamic' typeRep

fromDynamic' :: TypeRep a -> Dynamic -> Maybe a
fromDynamic' target (Dyn source value) = 
    case matchTypes source target of
      Just Reflexivity -> Just value
      Nothing -> Nothing
      -- The following pattern causes compilation to fail.
      Just (Induction _) -> Just value

但是,在最后一行(我的行号与示例不匹配)的编译失败了:

../src/Dyn.hs:105:34:
    Could not deduce (a2 ~ b)
    from the context (a1 ~ f a2, a ~ f b)
      bound by a pattern with constructor
                 Induction :: forall a b (f :: * -> *).
                              Equal a b -> Equal (f a) (f b),
               in a case alternative
      at ../src/Dyn.hs:105:13-23
      `a2' is a rigid type variable bound by
           a pattern with constructor
             Induction :: forall a b (f :: * -> *).
                          Equal a b -> Equal (f a) (f b),
           in a case alternative
           at ../src/Dyn.hs:105:13
      `b' is a rigid type variable bound by
          a pattern with constructor
            Induction :: forall a b (f :: * -> *).
                         Equal a b -> Equal (f a) (f b),
          in a case alternative
          at ../src/Dyn.hs:105:13
    Expected type: a1
      Actual type: a
    In the first argument of `Just', namely `value'
    In the expression: Just value
    In a case alternative: Just (Induction _) -> Just value

我读了这个问题的方法,编译器无法弄清楚,在Inductive :: Equal a b -> Equal (f a) (f b),a和b必须等于对非底部值.所以我尝试过Inductive :: Equal a a -> Equal (f a) (f a),但是在定义中也失败了matchTypes :: TypeRep a -> TypeRep b -> Maybe (Equal a b):

../src/Dyn.hs:66:60:
    Could not deduce (a2 ~ a1)
    from the context (a ~ [a1])
      bound by a pattern with constructor
                 List :: forall a. TypeRep a -> TypeRep [a],
               in an equation for `matchTypes'
      at ../src/Dyn.hs:66:13-18
    or from (b ~ [a2])
      bound by a pattern with constructor
                 List :: forall a. TypeRep a -> TypeRep [a],
               in an equation for `matchTypes'
      at ../src/Dyn.hs:66:22-27
      `a2' is a rigid type variable bound by
           a pattern with constructor
             List :: forall a. TypeRep a -> TypeRep [a],
           in an equation for `matchTypes'
           at ../src/Dyn.hs:66:22
      `a1' is a rigid type variable bound by
           a pattern with constructor
             List :: forall a. TypeRep a -> TypeRep [a],
           in an equation for `matchTypes'
           at ../src/Dyn.hs:66:13
    Expected type: TypeRep a1
      Actual type: TypeRep a
    In the second argument of `matchTypes', namely `b'
    In the second argument of `(<$>)', namely `(matchTypes a b)'

更改matchTypes :: TypeRep a -> TypeRep b -> Maybe (Equal a b)要生成的类型matchTypes :: TypeRep a -> TypeRep b -> Maybe (Equal a a)不起作用(只需将其作为命题阅读).也不对matchTypes :: TypeRep a -> TypeRep a -> Maybe (Equal a a)(另一个平凡的命题,并将此作为我的理解它需要的用户fromDynamic' to know the一个in theTypeRep一个contained in theDynamic`).

所以,我很难过.有关如何向前推进的任何指示？

Answer 1

问题是您的模式的通配符模式丢失了相等的信息.如果以这种方式编码归纳,则无法编写涵盖所有情况的(有限)模式集合.解决方案是将感应从您的数据类型转移到定义的值.相关更改如下所示:

data Equal a b where
    Reflexivity :: Equal a a

induction :: Equal a b -> Equal (f a) (f b)
induction Reflexivity = Reflexivity

matchTypes (List a) (List b) = induction <$> matchTypes a b
matchTypes (Maybe a) (Maybe b) = induction <$> matchTypes a b

fromDynamic' :: TypeRep a -> Dynamic -> Maybe a
fromDynamic' target (Dyn source value) = 
    case matchTypes source target of
      Just Reflexivity -> Just value
      Nothing -> Nothing

这样,模式fromDynamic'是详尽的,但没有任何信息丢失的通配符.