Haskell:函数中的非穷举模式(简单函数) [英] Haskell: Non-exhaustive patterns in function (simple functions)

问题描述

对于该函数的第一个和第三个版本为什么会出现此错误，而第二个定义却能正常工作，我感到困惑.

I am confused as to why the 1st and 3rd versions of this functions give this error whereas the second definition works fine.

-- head and tail
third :: [a] -> a
third [a] = head (tail (tail[a]))

-- Pattern matching
third2 :: [a] -> a
third2 (_:_:x:_) = x

-- List indexing
third3 :: [a] -> a
third3 [a] = [a]!!2

预先感谢

推荐答案

奇怪的是，第二个没有抱怨非详尽无遗模式，因为 third2 将不匹配长度为零，一或两个的列表. third 和 third3 函数抱怨，因为 [a] 不是变量，这是一种模式.将 [a] 还原为(a:[])，所以您可以将它们写为

That is odd that the second one does not complain about non-exhaustive patterns, since third2 will not match lists of length zero, one, or two. The third and third3 functions complain because [a] is not a variable, it is a pattern. [a] desugars to (a:[]), so you could have written them as

third (a:[]) = head (tail (a:[]))

third3 (a:[]) = (a:[]) !! 2

这两个都不起作用，因为它们是单个元素列表.我怀疑你想要的是什么

Neither of which will work, as those are single element lists. I suspect what you want is

third a = head (tail a)

third3 a = a !! 2

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