如何在 Rust 中连接两个切片? [英] How do I concatenate two slices in Rust?

问题描述

我想从向量中取出 x 的第一个和最后一个元素并将它们连接起来.我有以下代码:

I want to take the x first and last elements from a vector and concatenate them. I have the following code:

fn main() {
    let v = (0u64 .. 10).collect::<Vec<_>>();
    let l = v.len();
    vec![v.iter().take(3), v.iter().skip(l-3)];
}

这给了我错误

error[E0308]: mismatched types
 --> <anon>:4:28
  |
4 |     vec![v.iter().take(3), v.iter().skip(l-3)];
  |                            ^^^^^^^^^^^^^^^^^^ expected struct `std::iter::Take`, found struct `std::iter::Skip`
<anon>:4:5: 4:48 note: in this expansion of vec! (defined in <std macros>)
  |
  = note: expected type `std::iter::Take<std::slice::Iter<'_, u64>>`
  = note:    found type `std::iter::Skip<std::slice::Iter<'_, u64>>`

如何获得 1、2、3、8、9、10 的 vec?我使用的是 Rust 1.12.

How do I get my vec of 1, 2, 3, 8, 9, 10? I am using Rust 1.12.

推荐答案

你应该 collect() take() 和 extend() 的结果 将它们与 skip() 的 collect() ed 结果:

You should collect() the results of the take() and extend() them with the collect()ed results of skip():

let mut p1 = v.iter().take(3).collect::<Vec<_>>();
let p2 = v.iter().skip(l-3);

p1.extend(p2);

println!("{:?}", p1);

编辑:正如 Neikos 所说，你甚至不需要收集 skip() 的结果，因为 extend() 接受实现 IntoIterator 的参数(Skip 这样做，因为它是一个 Iterator).

Edit: as Neikos said, you don't even need to collect the result of skip(), since extend() accepts arguments implementing IntoIterator (which Skip does, as it is an Iterator).

编辑 2:不过，您的数字有点偏差；为了得到 1, 2, 3, 8, 9, 10 你应该声明 v 如下:

Edit 2: your numbers are a bit off, though; in order to get 1, 2, 3, 8, 9, 10 you should declare v as follows:

let v = (1u64 .. 11).collect::<Vec<_>>();

由于Range 是左闭右开.

Since the Range is left-closed and right-open.

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