在 lets-lens 教程中,您如何重构对遍历的调用以实现结束?

In the lets-lens tutorial, how do you refactor out the call to traverse in order to implement over?

the exercises中我实现了fmapT:

-- Let's remind ourselves of Traversable, noting Foldable and Functor.
--
-- class (Foldable t, Functor t) => Traversable t where
--   traverse ::
--     Applicative f => 
--     (a -> f b)
--     -> t a
--     -> f (t b)

-- | Observe that @fmap@ can be recovered from @traverse@ using @Identity@.
--
-- /Reminder:/ fmap :: Functor t => (a -> b) -> t a -> t b
fmapT ::
  Traversable t =>
  (a -> b)
  -> t a
  -> t b
fmapT =
  error "todo: fmapT"

现在我该如何实施 over

-- | Let's refactor out the call to @traverse@ as an argument to @fmapT@.

over ::
  ((a -> Identity b) -> s -> Identity t)
  -> (a -> b)
  -> s
  -> t
over = error "undefined"

您可以使用 traverse 实现 fmapT 作为:

fmapT f s = runIdentity (traverse (Identity . f) s)

现在下一个练习是通过在定义中提供 traverse 作为参数而不是 hard-coding 来重构此函数。如果您选择 Identity 作为应用类型构造函数,那么 traverse 的类型是:

(Traversable t) => (a -> Identity b) -> t a -> Identity (t b)

如果您将此作为参数提供给 fmapT,您最终会得到类似 over:

的结果
over' :: Traversable t => ((a -> Identity b) -> t a -> Identity (t b)) -> (a -> b) -> t a -> t b
over' l f s = runIdentity (l (Identity . f) s)

因为 Traversable 约束存在于 traverse 中,over' 不需要 over',它具有练习中给出的更通用的 over 类型,即 [=26] =]

over :: ((a -> Identity b) -> (s -> Identity t)) -> (a -> b) -> s -> t

over' =  over traverse