Haskell 中的依赖类型

Question

以下内容在 Haskell 中不起作用-

{-# LANGUAGE GADTs, DataKinds, TypeFamilies, UndecidableInstances,
             RankNTypes, PolyKinds #-}

import Data.Kind

data Payload :: (f :: a -> Type) -> (e :: a) -> Type where
  MkPayload :: (e :: a) -> (t :: f e) -> Payload f e

payload :: Payload f e -> f e
payload (MkPayload e t) = t

    • Expecting one more argument to ‘f :: a -> Type’
      Expected a type, but ‘f :: a -> Type’ has kind ‘a -> Type’
    • In the kind ‘(f :: a -> Type) -> (e :: a) -> Type’
      In the data type declaration for ‘Payload’
  |
6 | data Payload :: (f :: a -> Type) -> (e :: a) -> Type where
  |                  ^^^^^^^^^^^^^^

还有其他方法可以在 Haskell 中定义依赖类型吗？

Answer 1

您不能像在 Agda 中习惯的那样在类型中使用_:: _符号。 相反，只留下名字，只写类型：

{-# LANGUAGE GADTs, DataKinds, TypeFamilies, UndecidableInstances,
             RankNTypes, PolyKinds #-}

import Data.Kind ( Type )

data Payload :: (a -> Type) -> a -> Type where
  MkPayload :: a -> f e -> Payload f e

payload :: Payload f e -> f e
payload (MkPayload e t) = t

不鼓励使用 CUSK（完整的用户指定种类）符号，而应该使用独立的种类签名：

{-# LANGUAGE GADTs, DataKinds, TypeFamilies, UndecidableInstances,
             RankNTypes, PolyKinds, StandaloneKindSignatures #-}

import Data.Kind ( Type )

type Payload :: (a -> Type) -> a -> Type
data Payload f e where
  MkPayload :: a -> f e -> Payload f e

payload :: Payload f e -> f e
payload (MkPayload e t) = t

省略名称确实意味着您失去了一些表现力，但这不是本示例所必需的。 有一些技术可以恢复大部分表现力，例如 David Young 提到的单例。