在haskell中的奇数/偶数函数中的相互递归 [英] Mutual recursion in odd/even functions in haskell

问题描述

在Graham Hutton的Haskell编程的第6章中，有一节叫做6.5相互递归，它包含以下示例：

  even :: Int  - > Bool 
 even 0 = True 
 even（n + 1）= odd n 
 
 odd :: Int  - > Bool 
 odd 0 = False 
 odd（n + 1）= even n 
  

我想尝试一下。我把代码放在Hof.hs文件中，运行 ghci （版本7.8.3），输入

 ：l Hof.hs 
  

并获得以下错误消息：

 
 Hof.hs：3：7：模式中的分析错误：n + 1 
失败，模块加载：无。 
 

为什么我会收到此消息？是代码在语法上是过时的还是什么？

解决方案

  even :: Int  - > Bool 
 even 0 = True 
 even n = odd（n-1）
 
 odd :: Int  - > Bool 
 odd 0 = False 
 odd n = even（n  -  1）
  

请注意，这个函数在负输入中失败了，所以你可能想用 abs 来扩充它。

In chapter 6 of "Programming in Haskell" by Graham Hutton there is a section called "6.5 Mutual recursion", that contains the following example:

even :: Int -> Bool
even 0       = True
even (n + 1) = odd n

odd :: Int -> Bool
odd 0       = False
odd (n + 1) = even n

I wanted to try it out. I put the code in Hof.hs file, ran ghci(version 7.8.3), typed

:l Hof.hs

and got the following error message

Hof.hs:3:7: Parse error in pattern: n + 1
Failed, modules loaded: none.

Why am I getting this message? Is the code syntactically out of date or something?

解决方案

n + k patterns have been removed from Haskell and are no longer available. Write it like this instead:

even :: Int -> Bool
even 0 = True
even n = odd (n - 1)

odd :: Int -> Bool
odd 0 = False
odd n = even (n - 1)

Note that this function fails horribly for negative inputs, so you probably want to augment it with abs.

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