如何修复Haskell中的“函数中的非穷举模式"错误? [英] How to fix 'Non-exhaustive patterns in function' error in haskell?

问题描述

我正在尝试创建一个函数，该函数将元组列表作为参数并按第二个元素进行排序.它不会打印其他内容，仅显示错误"***异常:main.hs:20:1-76:函数sortWords中的非穷尽模式" 这是代码:

I am trying to create a function which takes a list of tuples as argument and sort the by the second element. It doesn't print anytning else, just the error '*** Exception: main.hs:20:1-76: Non-exhaustive patterns in function sortWords' Here is the code:

sortWords :: [(String, Int)] -> [(String, Int)]

sortWords [(str,num)] = sortBy (\x y -> compare (snd x) (snd y)) [(str,num)]`

这是我调用函数的方式

main = do
    putStrLn $ show $ sortWords [("friend",1),("she",2)]

我不得不说我在 http://Repl.it 网站

谢谢！

推荐答案

sortWords [(str,num)] = 

(上面的)函数定义模式匹配包含单个元素的列表，该元素是一个元组，具有两个值的变量.

Your function definition (above) pattern matches for a list containing a single element which is a tuple with variables for each of the two values.

您似乎只想要一个变量，而不想要任何模式匹配:

You seem to want just a variable and not a pattern match at all:

sortWords xs  = sortBy (\x y -> compare (snd x) (snd y)) xs

或eta减少:

sortWords = sortBy (\x y -> compare (snd x) (snd y))

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