8

尝试基于具有类型级别列表的 GADT 定义模式同义词时出现错误。

我设法把它归结为这个例子:

{-# LANGUAGE GADTs #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE PatternSynonyms #-}
module Example where

data L (as :: [*]) where
  L :: a -> L '[a]

pattern LUnit = L ()

给我:

Example.hs:11:17:
    Couldn't match type ‘a’ with ‘()’
      ‘a’ is a rigid type variable bound by
          the type signature for Example.$bLUnit :: (t ~ '[a]) => L t
          at Example.hs:11:17
    Expected type: L t
      Actual type: L '[()]
    In the expression: L ()
    In an equation for ‘$bLUnit’: $bLUnit = L ()

Example.hs:11:19:
    Could not deduce (a ~ ())
    from the context (t ~ '[a])
      bound by a pattern with constructor
                 L :: forall a. a -> L '[a],
               in a pattern synonym declaration
      at Example.hs:11:17-20
      ‘a’ is a rigid type variable bound by
          a pattern with constructor
            L :: forall a. a -> L '[a],
          in a pattern synonym declaration
          at Example.hs:11:17
    In the pattern: ()
    In the pattern: L ()

这是一个错误,还是我做错了什么?

4

1 回答 1

7

感谢dfeuer 的评论这张票,我能够通过在模式定义中添加类型签名来编译我的示例代码:

{-# LANGUAGE GADTs #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE PatternSynonyms #-}
module Example where

data L (as :: [*]) where
  L :: a -> L '[a]

pattern LUnit :: L '[()]
pattern LUnit = L ()

这也很好地推广到多态模式

data F (fs :: [* -> *]) a where
  F :: f a -> F '[f] a

pattern FId :: a -> F '[Identity] a
pattern FId a = F (Identity a)
于 2016-03-30T22:44:27.933 回答