为什么Haskell中的Show实例（拥抱）导致堆栈溢出错误？ [英] Why does this Show instance in Haskell (Hugs) cause a stack overflow error?

问题描述

以下是Haskell中的一个多态数据类型，由Hugs解释。我试图创建一个Show for Equality的实例。
$ b 实例声明说如果在Show中有一个类型a，那么Equality a在Show中。它应该以a = b的形式将两个参数打印到构造函数Equals ab中。

 数据Equality a = Equals aa 
 
 instance（显示a）=>显示（平等a）其中
显示（等于ab）= a ++=++ b 
  

然而，在Hugs中输入（Equality 9 9）这样的结果会产生：
$ b $错误 - C堆栈溢出

所以，我尝试用几个空格缩进show（Equals ab）...行。我不确定会有什么区别，但只是玩弄，然后得到这个：

 推断类型不是一般的足够
 ***表达式：show 
 ***预期类型：Show（Equality a）=>平等 - >字符串
 ***推断类型：Show（Equality [Char]）=>平等[Char]  - >字符串
  

任何人都可以解释为什么会出现这些错误，或者建议实现此show实例的更好方法？

谢谢！
解决方案奇怪的空白敏感度。如果没有缩进，编译器无法确定以下语句属于其中。

错误因为没有约束的多态类型不能确保 a 和 b 可以与= 字符串。如果您有等于1 1 ，该怎么办？如何在不使Ints字符串第一次连接的情况下进行连接？

然而，如果显示 a和b， / code>将这些值设置为可以与字符串连接的值。

 数据Equality a = Equa aa 
 
 instance（显示a）=>显示（平等a）其中
显示（等于ab）=（显示a）++=++（显示b）
  

The following is a polymorphic data type in Haskell, interpreted by Hugs. I am trying to create an instance of Show for Equality.

The instance declaration says that if a type "a" is in Show, then Equality a is in Show. It should print the two arguments to the constructor Equals a b in the form "a = b".

data Equality a = Equals a a 

instance (Show a) => Show (Equality a) where
show (Equals a b) = a ++ " = " ++ b

Yet, typing something into Hugs like "(Equality 9 9)" yields:

ERROR - C stack overflow

So, I tried indenting the "show (Equals a b)..." line with a couple of spaces. I'm not sure what the difference would be, but was just playing around and then got this:

Inferred type is not general enough
*** Expression    : show
*** Expected type : Show (Equality a) => Equality a -> String
*** Inferred type : Show (Equality [Char]) => Equality [Char] -> String

Can anyone explain why these errors are occurring, or suggest a better way of implementing this show instance?

Thank you!

解决方案

The indenting does matter due to Haskell's at-times-strange whitespace sensitivity. Without the indent, the compiler cannot tell that the following statement belongs to the where.

The error that you are getting is because the polymorphic type, having no constraints, does not ensure that a and b can concatenated with the " = " string. What if you have Equals 1 1. How would you concatenate that without making the Ints strings first?

However, if you show a and b first, all works out because show martials the values into something that can be concatenated with a String.

data Equality a = Equals a a

instance (Show a) => Show (Equality a) where
    show (Equals a b) = (show a) ++ " = " ++ (show b)

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