具有默認值的相互遞歸定義的類型類方法

Question

我想定義一個具有兩種方法的類型類，其中實現任何一種方法就足夠了（但如果需要，您可以獨立實現這兩種方法）。 這種情況與Eq中的情況相同，其中x == y = not (x /= y)和x /= y = not (x == y) 。 到目前為止一切順利，我可以做同樣的事情：

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，我還想提供bdistribute的通用默認實現，如果bdistribute沒有定義，我可以這樣做：

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，只要我想兩者都做，它就會中斷：

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

帶有以下錯誤消息：

bdistribute的定義沖突

現在，我可以看到這個錯誤是如何產生的； 而且，我認為我想要的也是合理且定義明確的：如果您手寫DistributiveB實例，則可以覆蓋bdistribute和/或bshape ，但是如果您只編寫instance DistributiveB MyB ，則bshape定義為從bdistribute定義的bdistribute和gdistributeDefault 。

Answer 1

一個折衷方案是放棄第一個默認定義：當用戶手動實現bshape時，要求添加一行以從中獲取bdistribute的另一個“默認”實現並不過分。

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

所以實例看起來像這樣：

-- Generic default
instance DistributiveB MyB

-- Manual bshape
instance DistributiveB MyB where
  distribute = distributeDefault  -- one extra line of boilerplate
  bshape = ...  -- custom definition

-- Manual distribute
instance DistributiveB MyB where
  distribute = ...
  -- bshape has a default implementation