GHCi中功能的非穷举模式 [英] Non exhaustive pattern in function in GHCi

问题描述

我想创建一个显示列表的最后一个元素的函数。
这是我的代码：

  ghci> let myLast :: [a]  - > a 
 ghci> let myLast [] =错误
 ghci> let myLast [x] = x 
 ghci>让myLast（x：xs）= myLast xs 
  

我得到以下错误：

$ b

  ***例外：函数myLast中的非详尽模式
  

我明白，如果您错过了某个案例，您会收到此错误消息，但我想我已经包含了所有可能性。任何想法？

解决方案

如果您在每一行中使用 let 每个定义都会创建一个名为 myLast 的新函数，并映射所有以前的定义。所以你最终得到的结果是相当于

GHCi> let myLast（x：xs）= myLast xs

单独。

您可能需要的是制作一个haskell文件，比如 MyLast.hs ，包含

 模块MyLast其中
 
 myLast： ：[a]  - > a 
 myLast [] =错误
 myLast [x] = x 
 myLast（x：xs）= myLast xs 
  

然后，您可以使用 ghci MyLast.hs 将该文件加载到GHCi中。

关键字 let 只有在您已经在GHCi中时才需要（或者在 IO 或者在另一个函数中），并且想要创建一个本地定义。但是你只能使用 let 一次，例如



GHCi> let myLast :: [a] - > a; myLast [] =错误; myLast [x] = x; myLast（x：xs）= myLast xs

或

  twiceLast :: [Int]  - > [Int] 
 twiceLast = let myLast [] =错误
 myLast [x] = x 
 myLast（x：xs）= myLast xs $ b $ in \xs  - > 2 *最后xs 
  

然而，我会喜欢写成



  twiceLast =（2 *）。 myLast 
其中myLast [] =错误
 myLast [x] = x 
 myLast（x：xs）= myLast xs 
  




I want to make a function that displays the last element of a list. This is my code:

ghci> let myLast :: [a] -> a
ghci> let myLast [] = error 
ghci> let myLast [x] = x
ghci> let myLast (x:xs) = myLast xs

And I get the following error:

***Exception: Non-exhaustive patterns in function myLast

I understood that you get this error when you're missing a case, but I think I have included all the possibilities. Any ideas?

解决方案

If you use a let in each line, each definition will make a new function named myLast, shadowing all the previous definitions. So what you end up with is equivalent to

GHCi> let myLast (x:xs) = myLast xs

alone.

What you probably want is to make a haskell file, say MyLast.hs, containing

module MyLast where

myLast :: [a] -> a
myLast [] = error 
myLast [x] = x
myLast (x:xs) = myLast xs

you can then load that file into GHCi with ghci MyLast.hs.

The keyword let is only needed when you're already in GHCi (or, in some monad like IO, or in another function) and want to make a local definition. But then you must only use the let once, e.g.

GHCi> let myLast :: [a]->a; myLast [] = error; myLast [x] = x; myLast (x:xs) = myLast xs

or

twiceLast :: [Int] -> [Int]
twiceLast = let myLast [] = error 
                myLast [x] = x
                myLast (x:xs) = myLast xs
            in \xs -> 2 * last xs

which I would, however, prefer to write as

twiceLast = (2*) . myLast
 where myLast [] = error 
       myLast [x] = x
       myLast (x:xs) = myLast xs

这篇关于GHCi中功能的非穷举模式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！