Haskell:为 zipper 创建类型类

Question

所以我一直在阅读一些关于 Haskell(以及其他函数式语言，我想)中的 Zipper 模式来遍历和修改数据结构的内容，我认为这对我来说是一个磨练创建类型技能的好机会Haskell 中的类(class)，因为
该类可以为我提供一个通用的遍历接口(interface)来编写代码，而与遍历的数据结构无关。

我想我可能需要两个类——一个用于根数据结构，一个用于创建的特殊数据结构
遍历第一个:

module Zipper where

class Zipper z where
  go'up :: z -> Maybe z
  go'down :: z -> Maybe z
  go'left :: z -> Maybe z
  go'right :: z -> Maybe z

class Zippable t where
  zipper :: (Zipper z) => t -> z
  get :: (Zipper z) => z -> t
  put :: (Zipper z) => z -> t -> z

但是当我用一些简单的数据结构(如列表)尝试这些时:

-- store a path through a list, with preceding elements stored in reverse
data ListZipper a = ListZipper { preceding :: [a], following :: [a] }

instance Zipper (ListZipper a) where
  go'up ListZipper { preceding = [] } = Nothing
  go'up ListZipper { preceding = a:ps, following = fs } = 
      Just $ ListZipper { preceding = ps, following = a:fs }
  go'down ListZipper { following = [] } = Nothing
  go'down ListZipper { preceding = ps, following = a:fs } = 
      Just $ ListZipper { preceding = a:ps, following = fs }
  go'left _ = Nothing
  go'right _ = Nothing

instance Zippable ([a]) where
  zipper as = ListZipper { preceding = [], following = as }
  get = following
  put z as = z { following = as }

或者二叉树:

-- binary tree that only stores values at the leaves
data Tree a = Node { left'child :: Tree a, right'child :: Tree a } | Leaf a
-- store a path down a Tree, with branches not taken stored in reverse
data TreeZipper a = TreeZipper { branches :: [Either (Tree a) (Tree a)], subtree :: Tree a }

instance Zipper (TreeZipper a) where
  go'up TreeZipper { branches = [] } = Nothing
  go'up TreeZipper { branches = (Left l):bs, subtree = r } =  
      Just $ TreeZipper { branches = bs, subtree = Node { left'child = l, right'child = r } }
  go'up TreeZipper { branches = (Right r):bs, subtree = l } =  
      Just $ TreeZipper { branches = bs, subtree = Node { left'child = l, right'child = r } }
  go'down TreeZipper { subtree = Leaf a } = Nothing
  go'down TreeZipper { branches = bs, subtree = Node { left'child = l, right'child = r } } =
      Just $ TreeZipper { branches = (Right r):bs, subtree = l }
  go'left TreeZipper { branches = [] } = Nothing
  go'left TreeZipper { branches = (Right r):bs } = Nothing
  go'left TreeZipper { branches = (Left l):bs, subtree = r } =
      Just $ TreeZipper { branches = (Right r):bs, subtree = l }
  go'right TreeZipper { branches = [] } = Nothing
  go'right TreeZipper { branches = (Left l):bs } = Nothing
  go'right TreeZipper { branches = (Right r):bs, subtree = l } =
      Just $ TreeZipper { branches = (Left l):bs, subtree = r }

instance Zippable (Tree a) where
  zipper t = TreeZipper { branches = [], subtree = t }
  get TreeZipper { subtree = s } = s
  put z s = z { subtree = s }

我无法编译它，我的每个 Zippable 都会出现很多这样的错误实例定义:

zipper .hs:28:14:
无法匹配预期的类型“z”
针对推断类型“ListZipper a”
`z' 是一个刚性类型变量，由
Zipper.hs:10:20 中“zipper”的类型签名
在表达式中:ListZipper {preceding = [], following = as}
在“ zipper ”的定义中:
zipper as = ListZipper {preceding = [], following = as}
在方法“ zipper ”的定义中

所以我不确定从这里去哪里。我怀疑我的问题是我正在尝试绑定(bind)这两个实例
一起，当 (Zipper z) =>声明只想 z任意 Zipper .

Answer 1

您还可以使用类型同义词族代替多参数类型类和功能依赖项。在这样的情况下，它们提供了一个更清晰、更易于理解的解决方案。在这种情况下，类和实例将变为:

class Zippable t where
  type ZipperType t :: *
  enter :: t -> ZipperType t
  focus :: ZipperType t -> t

instance Zippable [a] where
  type ZipperType [a] = ListZipper a
  enter = ...
  focus = ...

Fun with type functions对于已经熟悉 Haskell 的人来说，这是对类型同义词族的极好介绍。我还写了 an article关于如何经常使用类型同义词系列而不是功能依赖关系。

希望这可以帮助!