在Haskell获取玫瑰树的根 [英] Get the root of a Rose tree in Haskell

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

< pre $ 数据Rose a = Node a [Rose a]


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。

<希望能为你清除一些东西。

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 the
root 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 a

So 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 declaration

data Rose a = a :> [Rose a]

is basically equivalent to

data 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 write

root (Node datum children) = datum

Or, more briefly,

root (Node a rs) = a

Since the name given is actually :>, you'd have to write it as

root (a :> rs) = a

In 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IT屋！