为什么如果我使用“[xs]”，该功能不起作用。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 = 
（取n [xs]，drop n [xs]）
  

如果我将变量设为 xs 而不是 [xs] ，签名在两种情况下都是相同的。使用 [xs] 会给出该模式不详尽的错误。测试输入：我知道它告诉我输入的内容不被我的代码覆盖，但不清楚底下发生了什么。

测试输入： [1,2,3] 2 。

解决方案

不知何故，很多人认为 [xs] as模式意味着您< xs 来统一列表。但是这是不正确的，因为函数签名（隐式派生或者明确声明）已经会阻止你在你用非列表项调用函数的地方编写代码。

<一个列表有两个构造函数：

• 空列表 [] ;和 h 的$ consc （h：t）
  
• head （第一个元素）和 t tail （包含其余元素的列表）。

然而，Haskell也引入了一些语法糖。例如 [1] 是（1：[]）和 [1 ，4，2] for （1：（4：（2：[]）））。

这意味着如果你写了 [xs] ，在窗帘后面定义了一个（xs：[]）这意味着你完全匹配所有列表 一个元素，而那个单独的元素（不是整个列表）则是 xs 。

无论如何，解决方案是使用：

 分割 xs  n =（取n  xs ，放下n  xs ） 

既然 take :: Int - > [a] - > [a] 和 drop :: Int - > [a] - > [code> xs 应该是一个列表，Haskell会自动派生出 code> n 应该是 Int ，并且 xs an [a] 。

请注意，您可以使用 splitAt :: Int - > [a] - > （[a]，[a]） 。

  split = flip splitAt 
 我们可以将签名等同于您的目标。 c> 




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

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.

Test input: [1,2,3] 2.

解决方案

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.

A list has two constructors:

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

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

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.

Anyway, the solution is to use:

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

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].

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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