在Haskell获取玫瑰树的根

本文介绍了在Haskell获取玫瑰树的根的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 编写函数root :: Rose a - > a和children :: Rose a - > [Rose a] ，它们分别返回存储在玫瑰树根部的值，即玫瑰树的根部的子元素。 他们给我下面的基本代码开始： data Rose a = a：[Rose a] derived（Eq，Show） 吨知道什么（：>）的意思。我尝试通过输入ghci来理解它： 输入：：t（：>）输出：a - > [玫瑰a] - > Rose a 所以我认为这意味着你有一个 a value，它将用于从列表中查找 Rose a 并返回 Rose a ，但我仍然困惑下一步该做什么。 如果我看一下 root的签名：Rose a - >一个，函数看起来像这样： root（Rose a）= a children的功能:: Rose a - > [Rose a] ： children（Rose a）=（Rose a）：[] 这是不正确的，我不知道如何使它正确。 解决方案声明 数据Rose a = a：> [玫瑰a] 基本上相当于 换句话说， Node 是一个包含一个数据和一个子节点列表的数据结构。但是在上面的定义中，不是称它为 Node ，而是叫做：> 。这只是一个虚构的名字; Haskell允许您创建像这样的用户定义的操作符。 如果使用名称 Node ，您可以编写 $ b $ root（Node datum children）= datum 或者，更简单地说， root（Node a rs）= a 由于给定的名称实际上是：> ，你必须把它写成 root（a：> rs）= a 特别是，您似乎试图使用 Rose ，但那是类型构造函数，而不是值构造函数。同样，你似乎试图使用：运算符，但这是列表而不是玫瑰树。 Recently I started to learn about Haskell, and am struggling with the following exercise: Write functions root :: Rose a -> a and children :: Rose a -> [Rose a]that return the value stored at the root of a rose tree, respectively the children of theroot of a rose tree.They gave me the following basic code to start:data Rose a = a :> [Rose a] deriving (Eq, Show)I don't know what (:>) means. I tried to understand it by typing in the ghci Input: :t (:>)Output: a -> [Rose a] -> Rose aSo I think it means that you have an a value, which will be used to lookup Rose a out of a list and returns Rose a, but I'm still confused what to do next. If I look at the signature of root :: Rose a -> a, the function would look like this: root (Rose a) = a And the function of children :: Rose a -> [Rose a]:children (Rose a) = (Rose a):[]This is not correct, and I don't how to make it correct. 解决方案 The declarationdata Rose a = a :> [Rose a]is basically equivalent todata Rose a = Node a [Rose a]In other words, Node is a data structure containing a datum and a list of child nodes. But in the definition above, rather than calling it Node, it's called :>. It's just a made-up name; Haskell allows you to create user-defined operators like this.If the name Node was used, you could writeroot (Node datum children) = datumOr, more briefly,root (Node a rs) = aSince the name given is actually :>, you'd have to write it asroot (a :> rs) = aIn particular, you seem to be trying to use Rose, but that's the type constructor, not the value constructor. Similarly, you seem to be trying to use the ":" operator, but that's for lists, not rose trees.Hopefully that clears up some things for you. 这篇关于在Haskell获取玫瑰树的根的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！