如何从一个范围创建一个 Vec 并将其洗牌? [英] How do I create a Vec from a range and shuffle it?

问题描述

我有以下代码:

extern crate rand;

use rand::{thread_rng, Rng};

fn main() {
    let mut vec: Vec<u32> = (0..10).collect();
    let mut slice: &[u32] = vec.as_mut_slice();

    thread_rng().shuffle(slice);
}

并得到以下错误:

error[E0308]: mismatched types
 --> src/main.rs:9:26
  |
9 |     thread_rng().shuffle(slice);
  |                          ^^^^^ types differ in mutability
  |
  = note: expected type `&mut [_]`
             found type `&[u32]`

我想我理解向量和切片的内容是不可变的，这会导致这里出现错误，但我不确定.

I think I understand that the content of vectors and slices is immutable and that causes the error here, but I'm unsure.

as_mut_slice 的签名是 pub fn as_mut_slice<'a>(&'a mut self) ->&'a mut [T]，所以切片应该是可变的，但不知何故不是.

The signature of as_mut_slice is pub fn as_mut_slice<'a>(&'a mut self) -> &'a mut [T], so the slice should be mutable, but it somehow isn't.

我知道必须有一个简单的解决方法，但我尽了最大努力却无法让它发挥作用.

I know that there must be an easy fix, but I tried my best and couldn't get it to work.

推荐答案

兰德 v0.6.0

现在不推荐使用 Rng::shuffle 方法；应该使用 rand::seq::SliceRandom 特征.它在所有切片上提供 shuffle() 方法，该方法接受 Rng 实例:

Rand v0.6.0

The Rng::shuffle method is now deprecated; rand::seq::SliceRandom trait should be used. It provides the shuffle() method on all slices, which accepts an Rng instance:

// Rust edition 2018 no longer needs extern crate

use rand::thread_rng;
use rand::seq::SliceRandom;

fn main() {
    let mut vec: Vec<u32> = (0..10).collect();
    vec.shuffle(&mut thread_rng());
    println!("{:?}", vec);
}

看到它在游乐场.

你很亲密.这应该有效:

You're very close. This should work:

extern crate rand;

use rand::{thread_rng, Rng};

fn main() {
    let mut vec: Vec<u32> = (0..10).collect();
    let slice: &mut [u32] = &mut vec;

    thread_rng().shuffle(slice);
}

&mut [T] 隐式强制转换为 &[T]，并且您使用 注释了 slice 变量&[u32]，因此切片变得不可变:&mut [u32] 被强制转换为 &[u32].变量上的 mut 在这里不相关，因为切片只是借用到其他人拥有的数据中，因此它们没有继承的可变性 - 它们的可变性被编码在它们的类型中.

&mut [T] is implicitly coercible to &[T], and you annotated the slice variable with &[u32], so the slice became immutable: &mut [u32] was coerced to &[u32]. mut on the variable is not relevant here because slices are just borrows into data owned by someone else, so they do not have inherited mutability - their mutability is encoded in their types.

实际上，您根本不需要在 slice 上添加注解.这也有效:

In fact, you don't need an annotation on slice at all. This works as well:

extern crate rand;

use rand::{thread_rng, Rng};

fn main() {
    let mut vec: Vec<u32> = (0..10).collect();
    let slice = vec.as_mut_slice();

    thread_rng().shuffle(slice);
}

你甚至不需要中间变量:

You don't even need the intermediate variable:

extern crate rand;

use rand::{thread_rng, Rng};

fn main() {
    let mut vec: Vec<u32> = (0..10).collect();
    thread_rng().shuffle(&mut vec);
}

您应该阅读 Rust 编程语言 因为它解释了所有权和借用的概念以及它们如何与可变性相互作用.

You should read The Rust Programming Language as it explains the concepts of ownership and borrowing and how they interact with mutability.

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