Haskell:可能的解决方法:将 (Eq a) 添加到 [英] Haskell: Possible fix: add (Eq a) to the context of
问题描述
我是 Haskell 的新手，我很难理解我的代码有什么问题.
I'm kind of new to Haskell and I'm having a hard time understanding what is wrong with my code here.
这是我应该做的:
考虑以下二叉树的定义
data BinaryTree a = Empty | Node a (BinaryTree a) (BinaryTree a)
考虑函数reflect，通过一直向下交换左右来形成二叉树的镜像
reflect :: BinaryTree a -> BinaryTree a
reflect Empty = Empty
reflect (Node x l r) = Node x (reflect r) (reflect l)
编写一个函数 areMirrorImages 判断两棵二叉树 t 和 u 是否满足t = 反映你.该函数不应构建新树，因此不应调用反射或节点；尽管它可能会在模式中使用 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 x l r) (Node y ll rr)
| x==y = ((areMirrorImages l rr) && (areMirrorImages r ll))
| otherwise = False
当我尝试运行它时，在第 49 行出现以下错误:
无法从使用=="的上下文 () 推导出 (Eq a)
可能的解决方法:将 (Eq a) 添加到areMirrorImages"的类型签名的上下文中
表达式中:x==y
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
我很困惑为什么我会收到这个错误，我尝试在网上寻找解决方案，但到目前为止我什么也没找到.谢谢.
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.
推荐答案
目前，您的二叉树可以容纳任何类型——甚至是无法使用 == 进行比较的类型.所以如果你在包含这种类型的树上调用 areMirrorImages ......会发生什么?
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?
你需要做的是在areMirrorImages函数上放置一个约束，这样它就只能接受它可以比较内容的二叉树.
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.
类似于以下内容:
areMirrorImages :: Eq a => BinaryTree a -> BinaryTree a -> Bool
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