为什么我会得到“功能非穷举模式...”。当我调用我的Haskell子字符串函数？ [英] Why am I getting "Non-exhaustive patterns in function..." when I invoke my Haskell substring function?

问题描述

Haskell逻辑，数学和编程之路

一>。 （我只是通过第一章的中途，但我很享受它，并打算继续。）我已阅读了第1.5节玩Haskell游戏，其中包含许多其他示例让你熟悉[Haskell]。到目前为止，我已经了解了函数，类型声明，守护方程，关于列表模式匹配的一些信息，以及在哪里&

我坚持练习1.17，它要求我们编写一个函数substring :: String - > String - > Bool其中：

1. 如果xs是ys的前缀，则xs是ys的子字符串
如果ys等于y：ys'和xs，则

2. 是ys'的子字符串，xs是ys的子字符串


3. 其他都不是ys的子字符串

我使用前面示例中提供的前缀函数：

  prefix :: String  - >字符串 - > Bool 
 prefix [] ys = True 
前缀（x：xs）[] = False 
前缀（x：xs）（y：ys）=（x == y）&& amp ;前缀xs ys 
  

然后尝试：

  substring :: String  - >字符串 - > Bool 
 subsstring xs [] = False 
 substring xs（y：ys）|前缀xs（y：ys）= True 
 | substring xs ys = True 
 |否则= False 
  

...并且可能会对此进行其他排列。

当我运行 substringabcxxxabcyyy时，我得到 True ，但是当我运行 substringabcxxxabyyy我得到 * 异常：substring.hs：（3,0） - （5,45）：函数子串中的非穷举模式。我无法弄清楚为什么。我不明白当我使用其他时，如何有非穷举的模式。

顺便说一句，这本书还没有涵盖if-then-else。

解决方案

在函数名称中有一个拼写错误：

  subsstring xs [] = False 
  

由于错字，它声明了一个新函数 subsstring ，而不是子字符串函数。

substring 函数本身没有任何匹配第二个参数 [] 。

I'm working my way through the book The Haskell Road to Logic, Maths and Programming. (I'm only mid-way through chapter 1, but I'm enjoying it so far and intend to continue.) I've read through the section 1.5 "Playing the Haskell Game" which "consists of a number of further examples to get you acquainted with [Haskell]". So far I've learned about functions, type declarations, guarded equations, a little about list pattern matching, and where & let.

I'm stuck on exercise 1.17, which asks us to write a function substring :: String -> String -> Bool where:

1. if xs is a prefix of ys, xs is a substring of ys
2. if ys equals y:ys' and xs is a substring of ys', xs is a substring of ys
3. nothing else is a substring of ys

I used the prefix function provided in a previous example:

prefix :: String -> String -> Bool
prefix [] ys = True
prefix (x:xs) [] = False
prefix (x:xs) (y:ys) = (x==y) && prefix xs ys

And then tried:

substring :: String -> String -> Bool
subsstring xs [] = False
substring xs (y:ys) | prefix xs (y:ys) = True
                    | substring xs ys  = True
                    | otherwise        = False

...and may other permutations of this.

When I run substring "abc" "xxxabcyyy" I get True, but when I run substring "abc" "xxxabyyy" I get "* Exception: substring.hs:(3,0)-(5,45): Non-exhaustive patterns in function substring". I can't figure out why. I don't understand how there could be non-exhaustive patterns when I use "otherwise".

BTW, the book has not yet covered if-then-else. I'd prefer to keep that out of my solution, for now.

解决方案

You have a typo in the function name:

subsstring xs [] = False

Because of the typo this declares a new function subsstring, not a case of the substring function.

The substring function itself then doesn't have any case that would match a second parameter of [].

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