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Question

我有一个带有“隐藏”（推断）类型和具体值的数据类型。现在我尝试实现一个改变这两个但却无法通过GHC的函数。

我的示例代码是：

data T tag val = T val

data A = A
data B = B

mkIntVal :: T a b -> T Int b
mkIntVal (T x) = T x

mkCharVal :: T a b -> T Char b
mkCharVal (T x) = T x

convert :: T Int a -> T Char b
convert (T A) = mkCharVal $ T B
convert (T B) = mkCharVal $ T A

它产生的错误是：

test.hs:13:12:
    Couldn't match type `A' with `B'
    In the pattern: A
    In the pattern: T A
    In an equation for `convert': convert (T A) = mkCharVal $ T B

test.hs:13:17:
    Couldn't match type `B' with `A'
    Expected type: T Char b
      Actual type: T Char B
    In the expression: mkCharVal $ T B
    In an equation for `convert': convert (T A) = mkCharVal $ T B

要做这项工作需要做些什么？我是否必须更改数据结构？

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我正在尝试扩展Don Stewart的解决方案以处理多态数据类型。我一直在玩实例定义，但最有希望看到的是：

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE FlexibleContexts #-}

data C a = C a deriving Show

class Convertable inVal outVal outTag | outVal -> outTag where
    convert :: T Int inVal -> T outTag outVal

instance Convertable A B Char where
    convert (T A) = mkCharVal $ T B

instance Convertable B A Char where
    convert (T B) = mkCharVal $ T A

instance Convertable a b Char => Convertable (C a) (C (T Char b)) Char where
    convert (T (C val)) = mkCharVal $ T (C (convert val)) -- line 29

但是这给了我另一个错误信息：

test.hs:29:57:
    Could not deduce (a ~ T Int inVal0)
    from the context (Convertable a b Char)
      bound by the instance declaration at test.hs:28:10-70
      `a' is a rigid type variable bound by
          the instance declaration at test.hs:28:22
    In the first argument of `convert', namely `val'
    In the first argument of `C', namely `(convert val)'
    In the first argument of `T', namely `(C (convert val))'

正如唐所说，我应该有可能对如何实施这一点感兴趣。

解决方案

经过更多'玩'后，我终于想出了一些有用的东西。这看起来对你好吗？

{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE OverlappingInstances #-}


data T tag val = T val deriving Show

data A = A deriving Show
data B = B deriving Show
data C a = C a deriving Show


class Convertable inTag inVal outTag outVal | inTag -> outTag, inVal -> outVal
where
    convert :: T inTag inVal -> T outTag outVal

instance Convertable Int A Char B where
    convert (T A) = T B

instance Convertable Int B Char A where
    convert (T B) = T A

instance (Convertable Int (T Int a) Char (T Char b), Convertable Int a Char b)
    => Convertable Int (C (T Int a)) Char (C (T Char b)) where
    convert (T (C x)) = T (C (convert x))

instance Convertable Int (C (T Int A)) Char (C (T Char B)) where
    convert (T (C x)) = T (C (convert x))

instance Convertable Int (C (T Int B)) Char (C (T Char A)) where
    convert (T (C x)) = T (C (convert x))

用法：

*Main> convert $ mkIntVal $ T $ C $ mkIntVal $ T A
T (C (T B))
*Main> :t it
it :: T Char (C (T Char B))

Answer 1

convert函数的每个案例都有不同的冲突类型：

convertA :: T t A -> T Char B
convertA (T A) = mkCharVal $ T B

convertB :: T t B -> T Char A
convertB (T B) = mkCharVal $ T A

你可以通过类型类来统一这些，

{-# LANGUAGE MultiParamTypeClasses  #-}
{-# LANGUAGE FunctionalDependencies #-}


class C a b c | b -> c where
    convert :: T t a -> T c b

instance C A B Char where
    convert (T A) = mkCharVal (T B)

instance C B A Char where
    convert (T B) = mkCharVal (T A)

如果你真的希望在不同类型的单一功能，转换到不同的方向。请注意如何将T与标记一起使用，将其丢弃，并使用由值类型确定的新标记替换标记和值。

变换参数化类型（带推断参数）

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解决方案

1 个答案: