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haskell - 具有默认值的相互递归定义的类型类方法

问题描述

我想定义一个具有两种方法的类型类，其中实现任何一种方法就足够了（但如果需要，您可以独立实现这两种方法）。Eq这种情况与, wherex == y = not (x /= y)和中的情况相同x /= y = not (x == y)。到目前为止一切顺利，我可以做同样的事情：

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，我还想提供一个通用的默认实现，如果没有定义bdistribute，我可以这样做：bdistribute

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

带有以下错误消息：

冲突的定义bdistribute

现在，我可以看到这个错误是如何产生的；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，但如果您只是编写，instance DistributiveB MyB那么您将bshape根据.bdistributebdistributegdistributeDefault

标签： haskellgenericstypeclass

解决方案

一个折衷方案是放弃第一个默认定义：当用户手动实现bshape时，要求添加一行以从中获取另一个“默认”实现并不过分bdistribute。

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

所以实例看起来像这样：

-- 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

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