haskell - 在类实例中需要更高种类的类型

Question

我有一个看起来像这样的类型。

data State = Unsanitized | Sanitized

type family BarTy (s :: State) :: Type where
  BarTy 'Unsanitized = String
  BarTy 'Sanitized = Int

data Foo (a :: State) = Foo (Bar a)
data Bar (a :: State) = Bar (BarTy a)

我使用类型参数来更改某些字段的类型。例如，当通过验证管道演变数据结构时，这非常有用。

现在我想编写一个类，可以从该数据结构中提取一些内容，但它是哪个状态并不重要。完整示例:

data State = Unsanitized | Sanitized

type family BarTy (s :: State) :: Type where
  BarTy 'Unsanitized = String
  BarTy 'Sanitized = Int

data Foo (a :: State) = Foo (Bar a)
data Bar (a :: State) = Bar (BarTy a)

class Extractor a where
  extract :: a -> (BarTy b)

instance Extractor (Foo a) where
  extract (Foo x) = extract x

instance Extractor (Bar a) where
  extract (Bar x) = x

这显然不起作用，因为 b 类型变量在实例中始终未知，因此会发生类型错误。我可以创建另一个类型系列，从给定类型中提取 State 类型，但这对于较大的类型来说确实很笨重。

data State = Unsanitized | Sanitized

type family BarTy (s :: State) :: Type where
  BarTy 'Unsanitized = String
  BarTy 'Sanitized = Int

type family ExtractState (t :: Type) :: State where
  ExtractState (Foo s) = s
  ExtractState (Bar s) = s
  ExtractState _ = TypeError ('Text "ExtractState: Passed type is not valid.")

data Foo (a :: State) = Foo (Bar a)
data Bar (a :: State) = Bar (BarTy a)

class Extractor a where
  extract :: a -> (BarTy (ExtractState a))

instance Extractor (Foo a) where
  extract (Foo x) = extract x

instance Extractor (Bar a) where
  extract (Bar x) = x

bar1 :: Bar 'Unsanitized
bar1 = Bar "Test"
foo1 = Foo bar1

bar2 :: Bar 'Sanitized
bar2 = Bar 1337
foo2 = Foo bar2

a = extract bar1
b = extract foo1
c = extract bar2
d = extract foo2

我理想中想做的是在类声明中进行模式匹配，但我不知道如何做到这一点，或者是否可能:

data State = Unsanitized | Sanitized

type family BarTy (s :: State) :: Type where
  BarTy 'Unsanitized = String
  BarTy 'Sanitized = Int

data Foo (a :: State) = Foo (Bar a)
data Bar (a :: State) = Bar (BarTy a)

class Extractor (t a) where
  extract :: (t a) -> (BarTy a)

instance Extractor (Foo a) where
  extract (Foo x) = extract x

instance Extractor (Bar a) where
  extract (Bar x) = x

Haskell 可以做这样的事情吗？或者还有其他方法可以实现我想要的吗？

感谢任何帮助。

Answer 1

请记住，定义类型类时可以部分应用类型:

class Extractor t where
    extract :: t a -> BarTy a

(同样的技术也用在 Functor、Applicative、Monad 的定义中，并且一般来说，任何用于与更高级的类型。)