haskell - 用 `to`镜头进入 zipper

我正在努力lens和zippers。考虑下面在ghci中运行的代码

> import Control.Lens
> import Control.Zipper
>
> :t within (ix 1) $ zipper ([1,2,3] :: [Int])
> within (ix 1) $ zipper ([1,2,3] :: [Int])
   :: Control.Monad.MonadPlus m => m (Zipper Top Int [Int] :>> Int)

有了data A t = A t，如何创建类似Control.Monad.MonadPlus m => m (Zipper Top Int [Int] :>> A Int)的 zipper 类型？

我尝试了within (ix 1 . to A) $ zipper ([1,2,3] :: [Int])，但它给出了一个错误:

Could not deduce (Contravariant
                    (Bazaar (Indexed Int) (A Int) (A Int)))
  arising from a use of ‘to’
from the context (Control.Monad.MonadPlus m)
  bound by the inferred type of
           it :: Control.Monad.MonadPlus m =>
                 m (Zipper Top Int [Int] :>> A Int)
  at Top level
In the second argument of ‘(.)’, namely ‘to A’
In the first argument of ‘within’, namely ‘(ix 1 . to A)’
In the expression: within (ix 1 . to A)

最佳答案

一种方法是制作一个Iso并以此为准。在ghci中:

> import Control.Lens
> import Control.Zipper
>
> data A t = A t
> let _A = iso A (\(A a) -> a)
>
> let a = within (ix 1 . _A) $ zipper ([1,2,3] :: [Int])
> :t a
a :: MonadPlus m => m (Zipper Top Int [Int] :>> A Int)
> a ^? _Just . focus
Just (A 2)

编辑:您需要(\(A a) -> a)的原因是这样您就可以退出。

> data A t = A t
> let _A = iso A (error "Can't unA")
>
> let a = within (ix 1 . _A) $ zipper ([1,2,3] :: [Int])
> a ^? _Just . focus
Just (A 2)
> fmap upward a ^? _Just . focus
Just [1,*** Exception: Can't unA

我不认为没有一种用于提取A的函数，就没有一种有效的方法。您可以编写无效的Traversal，但仍无法正常工作:

> data A t = A t
> let _A f a = a <$ f (A a)
>
> let a = within (ix 1 . _A) $ zipper ([1,2,3] :: [Int])
> let b = a & _Just . focus .~ A 10
> b ^? _Just . focus
Just (A 10)
> fmap upward b ^? _Just . focus
Just [1,2,3] -- Should be Just [1, 10, 3]

关于haskell - 用 `to`镜头进入 zipper ，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/27842805/