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给定以下代码

{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE PolyKinds #-}

type family Tagged (m :: * -> *) :: k

class Example (t :: k) (a :: *) where
  type Return t a
  a :: (Monad m, Tagged m ~ t) => a -> m (Return t a)

data A
data A' a
data B = B

instance Example A B where
  type Return A B = ()
  a B = return ()

-- This is why I want a PolyKinded 't'
instance Example A' B where
  type Return A' B = ()
  a B = return ()

我收到类型错误（指向行a :: (Monad m ...）

• Could not deduce: Return (Tagged m) a ~ Return t a
  from the context: (Example t a, Monad m, Tagged m ~ t)
    bound by the type signature for:
               a :: (Example t a, Monad m, Tagged m ~ t) =>
                    a -> m (Return t a)
...
  Expected type: a -> m (Return t a)
    Actual type: a -> m (Return (Tagged m) a)
  NB: ‘Return’ is a type function, and may not be injective
  The type variable ‘k0’ is ambiguous
• In the ambiguity check for ‘a’
  To defer the ambiguity check to use sites, enable AllowAmbiguousTypes
  When checking the class method:
    a :: forall k (t :: k) a.
         Example t a =>
         forall (m :: * -> *).
         (Monad m, Tagged m ~ t) =>
         a -> m (Return t a)
  In the class declaration for ‘Example’

a我可以向with引入一个参数，Proxy t只要我在调用站点上签名，这将起作用：test = a (Proxy :: Proxy A) B但这是我想要避免的。我想要的是

newtype Test t m a = Test
  { runTest :: m a
  } deriving (Functor, Applicative, Monad)

type instance Tagged (Test t m) = t

test :: Monad m => Test A m ()
test = a B

我想使用类型实例t从上下文中找到。Test A m ()考虑到模块将在删除种类注释后编译，似乎应该是可能的PolyKinds， 和A'. k0来自哪里？

我想解决方法是放弃 PolyKinds 并使用额外的数据类型等data ATag; data A'Tag; data BTag。

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1 回答 1

2

这只是部分答案。

我试图明确说明。

type family Tagged k (m :: * -> *) :: k

class Example k (t :: k) (a :: *) where
  type Return k (t :: k) (a :: *)
  a :: forall m . (Monad m, Tagged k m ~ t) => a -> m (Return k t a)

并且，在启用了许多扩展之后，观察到：

> :t a
a :: (Example k (Tagged k m) a, Monad m) =>
     a -> m (Return k (Tagged k m) a)

因此，编译器抱怨，因为实例Example k (Tagged k m) a不能单独确定a,m。也就是说，我们不知道如何选择k。

我想，从技术上讲，我们可能有不同的Example k (Tagged k m) a实例，例如一个 fork=*和另一个 for k=(*->*)。

直觉上，知道t应该让我们找到k，但Return非内射会阻止我们找到t。

于 2017-11-09T13:44:05.903 回答