Question

我想定义一个具有两个方法的类型类，在其中实现任何一个方法就足够了（但是如果需要，您可以独立地实现两个方法）。这种情况与Eq和x == y = not (x /= y)中的x /= y = not (x == y)相同。到目前为止，我可以做的完全一样：

class (FunctorB b) => DistributiveB (b :: (Type -> Type) -> Type) where
  bdistribute :: (Distributive f) => f (b g) -> b (Compose f g)
  bdistribute x = bmap (\f -> Compose $ fmap f . bsequence' <$> x) bshape

  bshape :: b ((->) (b Identity))
  bshape = bdistribute' id

bdistribute' :: (DistributiveB b, Distributive f) => f (b Identity) -> b f
bdistribute' = bmap (fmap runIdentity . getCompose) . bdistribute

但是，我还想提供bdistribute的通用 default 实现，如果bdistribute没有定义，我可以这样做：

class (FunctorB b) => DistributiveB (b :: (Type -> Type) -> Type) where
  bdistribute :: (Distributive f) => f (b g) -> b (Compose f g)

  default bdistribute
    :: forall f g
    .  CanDeriveDistributiveB b f g
    => (Distributive f) => f (b g) -> b (Compose f g)
  bdistribute = gbdistributeDefault

  bshape :: b ((->) (b Identity))
  bshape = bdistribute' id

但是，一旦我想两者都做，它就会中断：

class (FunctorB b) => DistributiveB (b :: (Type -> Type) -> Type) where
  bdistribute :: (Distributive f) => f (b g) -> b (Compose f g)
  bdistribute x = bmap (\f -> Compose $ fmap f . bsequence' <$> x) bshape

  default bdistribute
    :: forall f g
    .  CanDeriveDistributiveB b f g
    => (Distributive f) => f (b g) -> b (Compose f g)
  bdistribute = gbdistributeDefault

  bshape :: b ((->) (b Identity))
  bshape = bdistribute' id

，并显示以下错误消息：

bdistribute的定义冲突

现在，我可以看到此错误的含义了；而且，我认为我想要的也是合理且定义明确的：如果您手写DistributiveB实例，则可以覆盖bdistribute和/或bshape，但是如果您只需编写instance DistributiveB MyB，就可以根据bshape和bdistribute定义的bdistribute和gdistributeDefault来定义1。

Answer 1

一个折衷方案是删除第一个默认定义：在用户手动实现bshape时，要求添加一行以获取{{1}的另一个“默认”实现并不过分}。

bdistribute

所以实例看起来像这样：

class FunctorB b => DistributiveB b where
  bdistribute :: Distributive f => f (b g) -> b (Compose f g)

  default bdistribute :: CanDeriveDistributiveB b f g => ...
  bdistribute = ...

  bshape :: b ((->) (b Identity))
  bshape = ...

-- Default implementation of bdistribute with an explicitly defined bshape
bdistributeDefault :: DistributiveB b => f (b g) -> b (Compose f g)
bdistributeDefault x = bmap (\f -> Compose $ fmap f . bsequence' <$> x) bshape

具有默认值的相互递归定义的类型类方法

1 个答案: