如果我使用“[xs]"，为什么这个功能不起作用?而不是“xs"? [英] Why doesn't this function work if I use "[xs]" instead of "xs"?

问题描述

split :: [a] -> Int -> ([a], [a])
split [xs] n = 
    (take n [xs], drop n [xs])

如果我将变量指定为 xs 而不是 [xs]，则相同的代码有效，两种情况下的签名都相同.使用 [xs] 会给出模式不详尽的错误.我明白这说明我提供的输入没有包含在我的代码中，但不清楚幕后发生了什么.

The same code works if I give the variable as xs instead of [xs], signatures are same in both cases. Using [xs] gives the error that pattern is non-exhaustive. I understand it's telling that the input I gave is not covered by my code, but not clear what is happening under the hood.

测试输入:[1,2,3] 2.

推荐答案

不知何故，很多人认为 [xs] 作为模式意味着你统一一个列表xs.但这是不正确的，因为函数签名(隐式派生或显式声明)已经会阻止您编写代码，在其中使用非列表项调用函数.

Somehow a lot of people think that [xs] as pattern means that you unify a list with xs. But this is incorrect, since the function signature (either derived implicitly, or stated explicitly) already will prevent you to write code where you call the function with a non-list item.

一个列表有两个构造函数:

• 空列表[];和
• 缺点" (h : t) 与 h head(第一个元素)和 t tail(包含剩余元素的列表).

• the empty list []; and
• the "cons" (h : t) with h the head (first element), and t the tail (a list with the remaining elements).

不过 Haskell 也引入了一些语法糖.例如 [1] 是 (1:[]) 的缩写，[1, 4, 2] 是 (1:(4:(2:[]))).

Haskell however introduces some syntactical sugar as well. For example [1] is short for (1:[]), and [1, 4, 2] for (1:(4:(2:[]))).

所以这意味着如果你写了 [xs]，在幕后你定义了一个模式 (xs: []) 这意味着你匹配所有列表与 正好一个元素，而那个元素(不是整个列表)就是xs.

So that means that if you write [xs], behind the curtains you defined a pattern (xs: []) which thus means you match all lists with exactly one element, and that single element (not the entire list) is then xs.

反正解决方法是使用:

split xs n = (take n xs, drop n xs)

由于 采取 :: Int ->[a] ->[a] 和 drop :: Int ->[a] ->[a] 在签名中有 xs 应该是一个列表，Haskell 会自动推导出 n 应该是成为Int，xs成为[a].

Since both take :: Int -> [a] -> [a] and drop :: Int -> [a] -> [a] have in the signature that xs is supposed to be a list, Haskell will derive automatically that n is supposed to be an Int, and xs an [a].

请注意，您可以使用 splitAt :: Int ->[a] ->([a], [a]) 也是如此.我们可以使签名与您的目标签名相同:

Note that you can use splitAt :: Int -> [a] -> ([a], [a]) as well. We can make the signature equivalent to the one you target with:

split = flip splitAt

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