为 ADT 派生显示实例不适用于更高种类的类型族

Question

我刚刚使用默认示例 gist available here 处理 Chris Done 的 ADT 并遇到了一个问题：我的 ADT 具有由更高种类的类型系列定义的字段，无法使用派生的显示实例。 GHC 告诉我我需要为 Type Family 派生一个 Show 实例，但我不知道该怎么做。到目前为止，这是我所拥有的，任何 cmets 都会有所帮助。

在以下示例中（使用 ghc 8.8.1），目标是为ShowMe 定义Show 的实例，如果可能，使用派生。

{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE PartialTypeSignatures #-}
{-# LANGUAGE ConstraintKinds #-}

data Tag = A | B deriving (Show)

type family TF (p :: Tag) a where
  TF 'A a = ()
  TF 'B a = a

data ShowMe p = ShowMe
  { a :: !(TF p String)
  , b :: String
  }

main =  connect showMeDefaults { a = "some string" }
   where
     connect :: ShowMe B -> IO ()
     connect _ = pure ()
     showMeDefaults :: ShowMe A
     showMeDefaults = ShowMe { a = (), b = "asdf" }

-- This works to define Show
{-
instance Show (ShowMe p) where
  show _ = "hello"
-}
-- This instance is the line that causes an error
deriving instance Show (ShowMe p)

随后，我从 GHC 收到一个我不熟悉的错误：

show_tf.hs:35:1: error:
    • No instance for (Show (TF p String))
        arising from a use of ‘showsPrec’
    • In the first argument of ‘(.)’, namely ‘(showsPrec 0 b1)’
      In the second argument of ‘(.)’, namely
        ‘((.)
            (showsPrec 0 b1)
            ((.)
               GHC.Show.showCommaSpace
               ((.)
                  (showString "b = ") ((.) (showsPrec 0 b2) (showString "}")))))’
      In the second argument of ‘(.)’, namely
        ‘((.)
            (showString "a = ")
            ((.)
               (showsPrec 0 b1)
               ((.)
                  GHC.Show.showCommaSpace
                  ((.)
                     (showString "b = ") ((.) (showsPrec 0 b2) (showString "}"))))))’
      When typechecking the code for ‘showsPrec’
        in a derived instance for ‘Show (ShowMe p)’:
        To see the code I am typechecking, use -ddump-deriv
   |
35 | deriving instance Show (ShowMe p)

如果我们重新编译，使用ghc -ddump-deriv，将返回以下内容：

[1 of 1] Compiling Main             ( show_tf.hs, show_tf.o )

==================== Derived instances ====================
Derived class instances:
  instance GHC.Show.Show Main.Tag where
    GHC.Show.showsPrec _ Main.A = GHC.Show.showString "A"
    GHC.Show.showsPrec _ Main.B = GHC.Show.showString "B"


Derived type family instances:



==================== Filling in method body ====================
GHC.Show.Show [Main.Tag]
  GHC.Show.show = GHC.Show.$dmshow @(Main.Tag)



==================== Filling in method body ====================
GHC.Show.Show [Main.Tag]
  GHC.Show.showList = GHC.Show.$dmshowList @(Main.Tag)


Linking show_tf ...

从概念上讲，我认为我应该能够为TF 派生一个 Show 实例，但是当我这样做时，我得到以下信息：

show_tf.hs:36:31: error:
    • Illegal type synonym family application ‘TF 'A a’ in instance:
        Show (TF 'A a)
    • In the stand-alone deriving instance for
        ‘(Show a) => Show (TF 'A a)’

   |
36 | deriving instance (Show a) => Show (TF 'A a)

如果我只是尝试自己为TF 'A a 定义 Show 实例，也会出现此错误。我搜索了“非法类型同义词”，但没有想出解决方法。

Answer 1

你需要添加

{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE UndecidableInstances #-}

然后向 GHC 建议所需的上下文：

deriving instance Show (TF p String) => Show (ShowMe p)

GHC 不会自动添加该上下文，因为它可能会让程序员感到惊讶。