通过独立派生获得

Jul*_*rch 3 haskell deriving

我不确定我在这里做错了什么:

data Vector2D u = Vector2D { 
    _x :: u, 
    _y :: u 
} deriving stock (Show, Eq, Functor, Foldable, Traversable)

{-# INLINE addVector2 #-}
addVector2 :: (Additive a) => Vector2D a -> Vector2D a -> Vector2D a 
addVector2 (Vector2D { _x = x1, _y = y1 }) (Vector2D { _x = x2, _y = y2 }) = 
    Vector2D { _x = x1 + x2, _y = y1 + y2 }

instance (Additive a) => Additive (Vector2D a) where
    (+) = addVector2

newtype Square a = Square {
    unpackSquare :: Vector2D a
} deriving stock (Show, Eq)
Run Code Online (Sandbox Code Playgroud)

到目前为止,这是正常的(添加剂在代数包中定义,但非常明显).

但是,现在我想要聪明地使用DerivingVia和StandaloneDeriving,我甚至无法获得下一行编译

deriving instance (Additive a) => Additive (Square a) via (Vector2D a)
Run Code Online (Sandbox Code Playgroud)

但这让我感到高兴

    * Expected kind `k0 -> * -> Constraint',
        but `Additive (Square a)' has kind `Constraint'
    * In the stand-alone deriving instance for
        `(Additive a) => Additive (Square a) via (Vector2D a)'

谁能告诉我我做错了什么?我正在运行GHC 8.6.2

HTN*_*TNW 12

它的

deriving via (Vector2D a) instance (Additive a) => Additive (Square a)
Run Code Online (Sandbox Code Playgroud)

你写它的方式via看起来像一个类型变量

deriving instance (Additive a) => Additive (Square a) via (Vector2D a)
-- <==>
deriving instance forall a via. (Additive a) => Additive (Square a) (via) (Vector2D a)
Run Code Online (Sandbox Code Playgroud)

这会产生一种类型不匹配的错误,因为Additive (Square a) :: Constraint已经饱和,但是你已经将它应用于另外两个参数.

这是速记形式的倒退:

data T' = ...
  deriving Class via T 
Run Code Online (Sandbox Code Playgroud)

  • [在这里](https://downloads.haskell.org/~ghc/latest/docs/html/users_guide/glasgow_exts.html#deriving-strategies)。它表示,一般来说,*所有*策略看起来都像“派生 strat 实例头”,而“via _”是策略的一个示例。 (2认同)