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我正在尝试为Linear.V类型创建一个新类型包装器并派生有用的实例。我正在尝试这个：

{-# LANGUAGE DataKinds, PolyKinds, ScopedTypeVariables, 
StandaloneDeriving, FlexibleContexts, UndecidableInstances, 
GeneralizedNewtypeDeriving, PartialTypeSignatures, TypeFamilies #-}

import Linear.V 
import Control.Lens.At

data Foo = Foo1 | Foo2 deriving (Show, Eq)

尝试 1 - 我认为 GeneralizedNewtypeDeriving 可以，但不：

newtype Bar n = Bar {
  getBar :: V n Foo
} deriving (Show, Eq, Ixed)

我收到此错误：

    • Couldn't match representation of type ‘f (V n Foo)’
                               with that of ‘f (Bar n)’
        arising from the coercion of the method ‘ix’
          from type ‘Index (V n Foo)
                     -> Control.Lens.Type.Traversal' (V n Foo) (IxValue (V n Foo))’
            to type ‘Index (Bar n)
                     -> Control.Lens.Type.Traversal' (Bar n) (IxValue (Bar n))’
      NB: We cannot know what roles the parameters to ‘f’ have;
        we must assume that the role is nominal
    • When deriving the instance for (Ixed (Bar n))

我使用这样的独立派生进行了尝试2：

newtype Bar n = Bar {
  getBar :: V n Foo
} deriving (Show, Eq)
type instance Index (Bar n) = Int
type instance IxValue (Bar n) = Foo

deriving instance Ixed (V n Foo) => Ixed (Bar n)

但后来我得到了一个不同的错误：

• Couldn't match representation of type ‘f1 (V n Foo)’
                           with that of ‘f1 (Bar n)’
    arising from a use of ‘GHC.Prim.coerce’
  NB: We cannot know what roles the parameters to ‘f1’ have;
    we must assume that the role is nominal
• In the expression:
    GHC.Prim.coerce
      @(Index (V n Foo)
        -> Control.Lens.Type.Traversal' (V n Foo) (IxValue (V n Foo)))
      @(Index (Bar n)
        -> Control.Lens.Type.Traversal' (Bar n) (IxValue (Bar n)))
      ix
  In an equation for ‘ix’:
      ix
        = GHC.Prim.coerce
            @(Index (V n Foo)
              -> Control.Lens.Type.Traversal' (V n Foo) (IxValue (V n Foo)))
            @(Index (Bar n)
              -> Control.Lens.Type.Traversal' (Bar n) (IxValue (Bar n)))
            ix
  When typechecking the code for ‘ix’
    in a derived instance for ‘Ixed (Bar n)’:
    To see the code I am typechecking, use -ddump-deriv
  In the instance declaration for ‘Ixed (Bar n)’
• Relevant bindings include
    ix :: Index (Bar n)
          -> Control.Lens.Type.Traversal' (Bar n) (IxValue (Bar n))
      (bound at a.hs:12:1)

我不确定为什么实际上会发生这两个错误。这可以以某种方式完成吗？我对高级类型级别的功能没有那么丰富的经验，到目前为止我也无法真正手动编写这个特定的实例定义，所以我也认为这是一个解决方案。但我更愿意以deriving某种方式使用该机制，因为它似乎更可重用。

编辑：我已经尝试过这个手动实例声明：

type instance Index (Bar n) = Int
type instance IxValue (Bar n) = Foo

instance Ixed (Bar n) where
  ix i f (Bar v) = ix i f v

但这会产生以下错误：

• Couldn't match type ‘V n Foo’ with ‘Bar n’
  Expected type: f (Bar n)
    Actual type: f (V n Foo)
• In the expression: ix i f v
  In an equation for ‘ix’: ix i f (Bar v) = ix i f v
  In the instance declaration for ‘Ixed (Bar n)’
• Relevant bindings include
    v :: V n Foo (bound at a.hs:14:15)
    f :: IxValue (Bar n) -> f (IxValue (Bar n)) (bound at a.hs:14:8)
    i :: Index (Bar n) (bound at a.hs:14:6)
    ix :: Index (Bar n)
          -> Control.Lens.Type.Traversal' (Bar n) (IxValue (Bar n))
      (bound at a.hs:14:3)

在我看来，编译器无法弄清楚Index两者的V n Fooand Bar nis Int。但我不确定。

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1 回答 1

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你快到了。剩下的问题是将底层的ix 遍历V n Foo（最终返回一个函数V n Foo -> f (V n Foo)）转换ix为对包装器类型的遍历Bar n，它最终应该返回一个函数Bar n -> f (Bar n)。我们必须“解包” 的定义Traversal'才能知道这一点。

在您的代码中，ix i f v具有 type f (V n Foo)，因此fmap使用Bar构造函数就足够了：

type instance Index (Bar n) = Int
type instance IxValue (Bar n) = Foo

instance Ixed (Bar n) where
  ix i f (Bar v) = fmap Bar (ix i f v)
于 2017-07-20T10:33:02.773 回答