haskell - 我可以派生 `Data` 的新类型实例吗？

Question

我有一个新类型T:

newtype T = T Text

我想为其导出数据。因此，使用 -XGeneralizedNewtypeDeriving -XDerivingStrategies 我这样做

deriving newtype instance Data T

我希望这对 GHC 来说是一个简单的推导，但我收到了一条令人讨厌的错误消息(见下文)。该错误消息似乎来自于 newtype 派生所产生的 coerce 应用。

如果我正确理解角色注释，它们将需要在 Data 的实例方法上进行类型声明，但实际上没有。我需要滚动自己的实例吗？

    • Couldn't match representation of type ‘c1 Text’
                               with that of ‘c1 T’
        arising from a use of ‘ghc-prim-0.5.2.0:GHC.Prim.coerce’
      NB: We cannot know what roles the parameters to ‘c1’ have;
        we must assume that the role is nominal
    • In the expression:
        ghc-prim-0.5.2.0:GHC.Prim.coerce
          @(forall (c :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep
                         -> TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
            forall (d :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep)
                   (b :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
            Data d => c (d -> b) -> d -> c b
            -> forall (g :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
               g -> c g
               -> Text -> c Text)
          @(forall (c :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep
                         -> TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
            forall (d :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep)
                   (b :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
            Data d => c (d -> b) -> d -> c b
            -> forall (g :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
               g -> c g
               -> T -> c T)
          gfoldl
      In an equation for ‘gfoldl’:
          gfoldl
            = ghc-prim-0.5.2.0:GHC.Prim.coerce
                @(forall (c :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep
                               -> TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
                  forall (d :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep)
                         (b :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
                  Data d => c (d -> b) -> d -> c b
                  -> forall (g :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
                     g -> c g
                     -> Text -> c Text)
                @(forall (c :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep
                               -> TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
                  forall (d :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep)
                         (b :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
                  Data d => c (d -> b) -> d -> c b
                  -> forall (g :: TYPE ghc-prim-0.5.2.0:GHC.Types.LiftedRep).
                     g -> c g
                     -> T -> c T)
                gfoldl
      When typechecking the code for ‘gfoldl’
        in a derived instance for ‘Data T’:
        To see the code I am typechecking, use -ddump-deriv
      In the instance declaration for ‘Data T’

可能相关:GeneralizedNewtypeDeriving fails for PersistFieldSql

编辑:也许我应该使用 -XDeriveDataTypeable？

Answer 1

来自docs :

A derived instance is derived only for declarations of these forms (after expansion of any type synonyms):

newtype T v1..vn                   = MkT (t vk+1..vn) deriving (C t1..tj)
newtype instance T s1..sk vk+1..vn = MkT (t vk+1..vn) deriving (C t1..tj)

where [...]

• C is not Read, Show, Typeable, or Data. These classes should not “look through” the type or its constructor. You can still derive these classes for a newtype, but it happens in the usual way, not via this new mechanism. Confer with Default deriving strategy.

要点是:newtype 的 Data 实例应该根据 newtype 本身的构造函数来制作，而不是根据(几个)底层类型的构造函数。

广义的 newtype 实例只会将基础类型的 Data 实例“强制”为 newtype，但这是错误的。

(不过，您收到的错误消息可能会更有帮助。)

结论:尝试使用 DeriveDataTypeable 代替。这应该派生出正确的实例。

newtype T = T Text deriving (Data)

<小时/>

更准确地说，我们应该在适当的实例中看到这一点:

> data U = A | B deriving Data
> newtype T = T U deriving Data
> toConstr (T A)
T

相反，“newtype 派生”实例将在此处生成 A，从而公开底层构造函数而不是预期的 T 构造函数。