这种模式似乎详尽无遗，但我仍然收到警告 [英] This pattern seems exhaustive, but I'm still getting warnings

问题描述

我正在学习 sml，并编写了以下简单函数:

I'm learning sml, and wrote the following simple function:

(* Return a list with every other element of the input list *)
fun everyOther [] = []
  | everyOther [x] = [x]
  | everyOther x = let
        val head::head2::tail = x
    in
        head::everyOther(tail)
    end;

产生以下警告:

! Toplevel input:
!   val head::head2::tail = x
!       ^^^^^^^^^^^^^^^^^
! Warning: pattern matching is not exhaustive

我相信该函数永远不会失败，因为 val head::head2::tail 将始终适用于具有两个或多个元素的列表，并且涵盖了一个元素和零个元素的情况.据我所知，此功能按预期工作.我认为这个问题可能与 [] 的使用有关，但我真的不知道.

I believe the function can never fail, since val head::head2::tail will always work for lists with two or more elements and the case of one element and zero elements is covered. As far as I can tell, this function works as expected. I think the issue might be related to the use of [] but I really don't know.

我的问题实际上是三重的:

My question is actually three fold:

1. 为什么 sml 认为这不是详尽无遗的(我怎么会误解)?
2. 此功能是否会失败?
3. 我在用这种方式编写函数是在做蠢事吗?

推荐答案

1. SML 给你这个警告是因为它不知道 x 至少有两个元素.它只知道 x 是一个列表，它不记得 x 必须不匹配前两个模式的事实，才能进入第三种情况.

1. SML gives you that warning because it doesn't know that x has at least two elements. All it knows is that x is a list, it doesn't remember the fact that x had to have not matched the first two patterns, to go into the third case.

不，代码不能失败.

没有理由在 let 语句中执行模式匹配.您可以将模式放在 fun 语句中，这样可以减少代码并消除警告:

There is no reason to perform the pattern match in the let-statement. You can just put the pattern to the fun statement, which will result in less code and remove the warning:

fun everyOther [] = []
  | everyOther [x] = [x]
  | everyOther (head::head2::tail) = head :: everyOther tail;

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