gpt4 book ai didi

haskell - 我无法让基于 GADT 的玩具动态类型与参数类型一起使用

转载 作者:行者123 更新时间:2023-12-04 14:32:05 25 4
gpt4 key购买 nike

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为涵盖参数类型。 (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,对此的编译在最后一行失败(我的行号与示例不匹配):
../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

按照我的阅读方式,编译器无法确定 Inductive :: Equal a b -> Equal (f a) (f b) 中的内容, ab对于非底部值必须相等。所以我试过 Inductive :: Equal a a -> Equal (f a) (f a) ,但这也失败了,在 matchTypes :: TypeRep a -> TypeRep b -> Maybe (Equal a b) 的定义中:
../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) 的类型生产 matchTypes :: TypeRep a -> TypeRep b -> Maybe (Equal a a)不起作用(只是把它当作一个命题来读)。 matchTypes :: TypeRep a -> TypeRep a -> Maybe (Equal a a) 也没有(另一个微不足道的命题,据我了解,这需要 fromDynamic' to know the 一个 in the TypeRep 一个 contained in the 动态`的用户)。

所以,我很难过。关于如何在这里前进的任何指示?

最佳答案

问题是您的模式的通配符模式丢失了相等信息。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。解决方案是将归纳从您的数据类型移出到定义的值。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 fromDynamic' 中的模式是详尽的,但没有任何丢失信息的通配符。

关于haskell - 我无法让基于 GADT 的玩具动态类型与参数类型一起使用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10985298/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com