如何获得子切片?

问题描述

我有一个 &[T] 类型的变量 a;如何获得对 a 子切片的引用?

I have a variable a of type &[T]; how can I get a reference to a subslice of a?

举个具体的例子，我想得到 a 的前半部分和后半部分，前提是 a.len() 是偶数.

As a concrete example, I'd like to get the first and second halves of a, provided a.len() is even.

推荐答案

你使用切片语法:

fn main() {
    let data: &[u8] = b"12345678";
    println!("{:?} - {:?}", &data[..data.len()/2], &data[data.len()/2..]);
}

(试试这里)

一般语法是

&slice[start_idx..end_idx]

给出从 slice 派生的切片，从 start_idx 开始，到 end_idx-1 结束(即右侧索引处的项不包括在内).可以省略任何一个索引(甚至两个)，这意味着相应的零或切片长度.

which gives a slice derived from slice, starting at start_idx and ending at end_idx-1 (that is, item at the right index is not included). Either index could be omitted (even both), which would mean zero or slice length, correspondingly.

注意，如果你想将某个位置的切片分割成两个切片，通常使用split_at()方法会更好:

Note that if you want to split a slice at some position into two slices, it is often better to use split_at() method:

let data = b"12345678";
let (left, right): (&[u8], &[u8]) = data.split_at(4);

此外，这是从另一个可变切片中获取两个可变切片的唯一方法:

Moreover, this is the only way to obtain two mutable slices out of another mutable slice:

let mut data: Vec<u8> = vec![1, 2, 3, 4, 5, 6, 7, 8];
let data_slice: &mut [u8] = &mut data[..];
let (left, right): (&mut [u8], &mut [u8]) = data_slice.split_at_mut(4);

然而，这些基本的东西在关于 Rust 的官方书籍中有解释.如果你想学习 Rust，你应该从那里开始.

However, these basic things are explained in the official book on Rust. You should start from there if you want to learn Rust.

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