用“to”镜头走进拉链

Question

我正在努力 lens and zippers。考虑以下在 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，我如何创建像这样的拉链类型：Control.Monad.MonadPlus m => m (Zipper Top Int [Int] :>> A Int)?

我试过 within (ix 1 . to A) $ zipper ([1,2,3] :: [Int]) 但出现错误：

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)

Answer 1

一种方法是制作一个 Iso 并与之组合。在 ghci:

> 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) 的原因是您可以退出。

> 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

我认为没有提取函数 A 的有效方法。你可以写一个无效的 Traversal 但它仍然不能正常工作：

> 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]

用“to”镜头走进拉链

stepping into zipper with `to` lens

haskell

types

haskell-lens

zipper