Haskell：可能的修正：add（Eq a）到的上下文 [英] Haskell: Possible fix: add (Eq a) to the context of

问题描述

我是Haskell的新人，我很难理解这里的代码有什么问题。

这里是我

  data BinaryTree a = Empty |节点a（BinaryTree a）（BinaryTree a）
  

考虑形成镜像的函数reflect

  reflect :: BinaryTree a  - > BinaryTree a 
反映Empty = Empty 
 reflect（Node xlr）= Node x（reflect r）（reflect l）
  

写一个函数areMirrorImages确定两个二叉树t和u是否满足
t = reflect u。该函数不应该构建新的树，因此它不应该调用reflect或Node;

这是我写的：

  areMirrorImages :: BinaryTree a  - > BinaryTree a  - > bool 
 areMirrorImages Empty Empty = True 
 areMirrorImages（Node _ _ _）Empty = False 
 areMirrorImages Empty（Node _ _ _）= False 
 areMirrorImages（Node xlr） ll rr）
 | x == y =（（镜像lrr）&&（are镜像rll））
 |否则= False 
  

当我尝试运行它时，我在第49行得到以下错误：

无法从使用'=='时产生的上下文（）中推导出（等式a）

可能的修正：将add（Eq a）添加到'areMirrorImages'类型签名的上下文中

在表达式中：x == y

我很困惑为什么我得到这个错误，我尝试在线搜索解决方案，但我到目前为止没有发现。感谢。

解决方案

目前，您的二叉树可以容纳任何类型，即使是不能使用 == 。因此，如果你在包含这种类型的树上调用 areMirrorImages ...会发生什么？

要做的是在 areMirrorImages 函数上放置约束，以便它只能接受二进制树，它可以比较的内容。

如下所示：

  areMirrorImages :: Eq a => BinaryTree a  - > BinaryTree a  - > Bool 
  

I'm kind of new to Haskell and I'm having a hard time understanding what is wrong with my code here.

Here is what I'm supposed to do:
Consider the following definition of a binary tree

data BinaryTree a = Empty | Node a (BinaryTree a) (BinaryTree a)  

Consider the function reflect that forms the mirror image of a binary tree by exchanging left and right all the way down

reflect :: BinaryTree a -> BinaryTree a  
reflect Empty = Empty  
reflect (Node x l r) = Node x (reflect r) (reflect l)  

Write a function areMirrorImages that determines whether two binary trees t and u satisfy t = reflect u. The function should not build new trees, so it should not call reflect or Node; although it may use Node in patterns.

Here is what I've written:

areMirrorImages :: BinaryTree a -> BinaryTree a -> Bool  
areMirrorImages Empty Empty = True  
areMirrorImages (Node _ _ _) Empty = False  
areMirrorImages Empty (Node _ _ _) = False  
areMirrorImages (Node x l r) (Node y ll rr)  
    | x==y = ((areMirrorImages l rr) && (areMirrorImages r ll))  
    | otherwise = False  

When I try running it I get the following error on line 49:
Could not deduce (Eq a) from the context () arising from a use of '=='
Possible fix: add (Eq a) to the context of the type signature for 'areMirrorImages'
In the expression: x==y

I am confused as to why I'm getting this error and I tried finding solutions online but I have found nothing so far. Thanks.

解决方案

At the moment, your binary trees can hold any type whatsoever - even types that can't be compared using ==. So if you called areMirrorImages on a tree containing such a type ... what should happen?

What you need to do is place a constraint on the areMirrorImages function, so that it can only accept binary trees that it can compare the contents of.

Something like the following:

areMirrorImages :: Eq a => BinaryTree a -> BinaryTree a -> Bool

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