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<main>
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<h1 id="引用与借用"><a class="header" href="#引用与借用">引用与借用</a></h1>
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<p>上节中提到，如果仅仅支持通过转移所有权的方式获取一个值，那会让程序变得复杂。 Rust 能否像其它编程语言一样，使用某个变量的指针或者引用呢？答案是可以。</p>
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<p>Rust 通过 <code>借用(Borrowing)</code> 这个概念来达成上述的目的，<strong>获取变量的引用，称之为借用(borrowing)</strong>。正如现实生活中，如果一个人拥有某样东西，你可以从他那里借来，当使用完毕后，也必须要物归原主。</p>
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<h2 id="引用与解引用"><a class="header" href="#引用与解引用">引用与解引用</a></h2>
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<p>常规引用是一个指针类型，指向了对象存储的内存地址。在下面代码中，我们创建一个 <code>i32</code> 值的引用 <code>y</code>，然后使用解引用运算符来解出 <code>y</code> 所使用的值:</p>
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<pre class="playground"><code class="language-rust edition2021">fn main() {
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let x = 5;
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let y = &x;
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assert_eq!(5, x);
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assert_eq!(5, *y);
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}</code></pre>
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<p>变量 <code>x</code> 存放了一个 <code>i32</code> 值 <code>5</code>。<code>y</code> 是 <code>x</code> 的一个引用。可以断言 <code>x</code> 等于 <code>5</code>。然而，如果希望对 <code>y</code> 的值做出断言，必须使用 <code>*y</code> 来解出引用所指向的值（也就是<strong>解引用</strong>）。一旦解引用了 <code>y</code>，就可以访问 <code>y</code> 所指向的整型值并可以与 <code>5</code> 做比较。</p>
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<p>相反如果尝试编写 <code>assert_eq!(5, y);</code>，则会得到如下编译错误：</p>
|
|
|
<pre><code class="language-text">error[E0277]: can't compare `{integer}` with `&{integer}`
|
|
|
--> src/main.rs:6:5
|
|
|
|
|
|
|
6 | assert_eq!(5, y);
|
|
|
| ^^^^^^^^^^^^^^^^^ no implementation for `{integer} == &{integer}` // 无法比较整数类型和引用类型
|
|
|
|
|
|
|
= help: the trait `std::cmp::PartialEq<&{integer}>` is not implemented for
|
|
|
`{integer}`
|
|
|
</code></pre>
|
|
|
<p>不允许比较整数与引用，因为它们是不同的类型。必须使用解引用运算符解出引用所指向的值。</p>
|
|
|
<h2 id="不可变引用"><a class="header" href="#不可变引用">不可变引用</a></h2>
|
|
|
<p>下面的代码，我们用 <code>s1</code> 的引用作为参数传递给 <code>calculate_length</code> 函数，而不是把 <code>s1</code> 的所有权转移给该函数：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let s1 = String::from("hello");
|
|
|
|
|
|
let len = calculate_length(&s1);
|
|
|
|
|
|
println!("The length of '{}' is {}.", s1, len);
|
|
|
}
|
|
|
|
|
|
fn calculate_length(s: &String) -> usize {
|
|
|
s.len()
|
|
|
}</code></pre>
|
|
|
<p>能注意到两点：</p>
|
|
|
<ol>
|
|
|
<li>无需像上章一样：先通过函数参数传入所有权，然后再通过函数返回来传出所有权，代码更加简洁</li>
|
|
|
<li><code>calculate_length</code> 的参数 <code>s</code> 类型从 <code>String</code> 变为 <code>&String</code></li>
|
|
|
</ol>
|
|
|
<p>这里，<code>&</code> 符号即是引用，它们允许你使用值，但是不获取所有权，如图所示：
|
|
|
<img alt="&String s pointing at String s1" src="https://pic1.zhimg.com/80/v2-fc68ea4a1fe2e3fe4c5bb523a0a8247c_1440w.jpg" class="center" /></p>
|
|
|
<p>通过 <code>&s1</code> 语法，我们创建了一个<strong>指向 <code>s1</code> 的引用</strong>，但是并不拥有它。因为并不拥有这个值，当引用离开作用域后，其指向的值也不会被丢弃。</p>
|
|
|
<p>同理，函数 <code>calculate_length</code> 使用 <code>&</code> 来表明参数 <code>s</code> 的类型是一个引用：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn calculate_length(s: &String) -> usize { // s 是对 String 的引用
|
|
|
s.len()
|
|
|
} // 这里，s 离开了作用域。但因为它并不拥有引用值的所有权，
|
|
|
// 所以什么也不会发生
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>人总是贪心的，可以拉女孩小手了，就想着抱抱柔软的身子（读者中的某老司机表示，这个流程完全不对），因此光借用已经满足不了我们了，如果尝试修改借用的变量呢？</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let s = String::from("hello");
|
|
|
|
|
|
change(&s);
|
|
|
}
|
|
|
|
|
|
fn change(some_string: &String) {
|
|
|
some_string.push_str(", world");
|
|
|
}</code></pre>
|
|
|
<p>很不幸，妹子你没抱到，哦口误，你修改错了：</p>
|
|
|
<pre><code class="language-console">error[E0596]: cannot borrow `*some_string` as mutable, as it is behind a `&` reference
|
|
|
--> src/main.rs:8:5
|
|
|
|
|
|
|
7 | fn change(some_string: &String) {
|
|
|
| ------- help: consider changing this to be a mutable reference: `&mut String`
|
|
|
------- 帮助：考虑将该参数类型修改为可变的引用: `&mut String`
|
|
|
8 | some_string.push_str(", world");
|
|
|
| ^^^^^^^^^^^ `some_string` is a `&` reference, so the data it refers to cannot be borrowed as mutable
|
|
|
`some_string`是一个`&`类型的引用，因此它指向的数据无法进行修改
|
|
|
</code></pre>
|
|
|
<p>正如变量默认不可变一样，引用指向的值默认也是不可变的，没事，来一起看看如何解决这个问题。</p>
|
|
|
<h2 id="可变引用"><a class="header" href="#可变引用">可变引用</a></h2>
|
|
|
<p>只需要一个小调整，即可修复上面代码的错误：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let mut s = String::from("hello");
|
|
|
|
|
|
change(&mut s);
|
|
|
}
|
|
|
|
|
|
fn change(some_string: &mut String) {
|
|
|
some_string.push_str(", world");
|
|
|
}</code></pre>
|
|
|
<p>首先，声明 <code>s</code> 是可变类型，其次创建一个可变的引用 <code>&mut s</code> 和接受可变引用参数 <code>some_string: &mut String</code> 的函数。</p>
|
|
|
<h4 id="可变引用同时只能存在一个"><a class="header" href="#可变引用同时只能存在一个">可变引用同时只能存在一个</a></h4>
|
|
|
<p>不过可变引用并不是随心所欲、想用就用的，它有一个很大的限制： <strong>同一作用域，特定数据只能有一个可变引用</strong>：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut s = String::from("hello");
|
|
|
|
|
|
let r1 = &mut s;
|
|
|
let r2 = &mut s;
|
|
|
|
|
|
println!("{}, {}", r1, r2);
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>以上代码会报错：</p>
|
|
|
<pre><code class="language-console">error[E0499]: cannot borrow `s` as mutable more than once at a time 同一时间无法对 `s` 进行两次可变借用
|
|
|
--> src/main.rs:5:14
|
|
|
|
|
|
|
4 | let r1 = &mut s;
|
|
|
| ------ first mutable borrow occurs here 首个可变引用在这里借用
|
|
|
5 | let r2 = &mut s;
|
|
|
| ^^^^^^ second mutable borrow occurs here 第二个可变引用在这里借用
|
|
|
6 |
|
|
|
7 | println!("{}, {}", r1, r2);
|
|
|
| -- first borrow later used here 第一个借用在这里使用
|
|
|
</code></pre>
|
|
|
<p>这段代码出错的原因在于，第一个可变借用 <code>r1</code> 必须要持续到最后一次使用的位置 <code>println!</code>，在 <code>r1</code> 创建和最后一次使用之间，我们又尝试创建第二个可变借用 <code>r2</code>。</p>
|
|
|
<p>对于新手来说，这个特性绝对是一大拦路虎，也是新人们谈之色变的编译器 <code>borrow checker</code> 特性之一，不过各行各业都一样，限制往往是出于安全的考虑，Rust 也一样。</p>
|
|
|
<p>这种限制的好处就是使 Rust 在编译期就避免数据竞争，数据竞争可由以下行为造成：</p>
|
|
|
<ul>
|
|
|
<li>两个或更多的指针同时访问同一数据</li>
|
|
|
<li>至少有一个指针被用来写入数据</li>
|
|
|
<li>没有同步数据访问的机制</li>
|
|
|
</ul>
|
|
|
<p>数据竞争会导致未定义行为，这种行为很可能超出我们的预期，难以在运行时追踪，并且难以诊断和修复。而 Rust 避免了这种情况的发生，因为它甚至不会编译存在数据竞争的代码！</p>
|
|
|
<p>很多时候，大括号可以帮我们解决一些编译不通过的问题，通过手动限制变量的作用域：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut s = String::from("hello");
|
|
|
|
|
|
{
|
|
|
let r1 = &mut s;
|
|
|
|
|
|
} // r1 在这里离开了作用域，所以我们完全可以创建一个新的引用
|
|
|
|
|
|
let r2 = &mut s;
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<h4 id="可变引用与不可变引用不能同时存在"><a class="header" href="#可变引用与不可变引用不能同时存在">可变引用与不可变引用不能同时存在</a></h4>
|
|
|
<p>下面的代码会导致一个错误：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut s = String::from("hello");
|
|
|
|
|
|
let r1 = &s; // 没问题
|
|
|
let r2 = &s; // 没问题
|
|
|
let r3 = &mut s; // 大问题
|
|
|
|
|
|
println!("{}, {}, and {}", r1, r2, r3);
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>错误如下：</p>
|
|
|
<pre><code class="language-console">error[E0502]: cannot borrow `s` as mutable because it is also borrowed as immutable
|
|
|
// 无法借用可变 `s` 因为它已经被借用了不可变
|
|
|
--> src/main.rs:6:14
|
|
|
|
|
|
|
4 | let r1 = &s; // 没问题
|
|
|
| -- immutable borrow occurs here 不可变借用发生在这里
|
|
|
5 | let r2 = &s; // 没问题
|
|
|
6 | let r3 = &mut s; // 大问题
|
|
|
| ^^^^^^ mutable borrow occurs here 可变借用发生在这里
|
|
|
7 |
|
|
|
8 | println!("{}, {}, and {}", r1, r2, r3);
|
|
|
| -- immutable borrow later used here 不可变借用在这里使用
|
|
|
</code></pre>
|
|
|
<p>其实这个也很好理解，正在借用不可变引用的用户，肯定不希望他借用的东西，被另外一个人莫名其妙改变了。多个不可变借用被允许是因为没有人会去试图修改数据，每个人都只读这一份数据而不做修改，因此不用担心数据被污染。</p>
|
|
|
<blockquote>
|
|
|
<p>注意，引用 <code>r1</code>,<code>r2</code>,<code>r3</code> 的作用域从创建开始，一直持续到它最后一次使用的地方 <code>println!(....)</code>，这个跟变量的作用域有所不同，变量的作用域从创建持续到某一个花括号 <code>}</code></p>
|
|
|
</blockquote>
|
|
|
<p>Rust 的编译器一直在优化，早期的时候，引用的作用域跟变量作用域是一致的，这对日常使用带来了很大的困扰，你必须非常小心的去安排可变、不可变变量的借用，免得无法通过编译，例如以下代码：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let mut s = String::from("hello");
|
|
|
|
|
|
let r1 = &s;
|
|
|
let r2 = &s;
|
|
|
println!("{} and {}", r1, r2);
|
|
|
// 新编译器中，r1,r2作用域在这里结束
|
|
|
|
|
|
let r3 = &mut s;
|
|
|
println!("{}", r3);
|
|
|
} // 老编译器中，r1、r2、r3作用域在这里结束
|
|
|
// 新编译器中，r3作用域在这里结束</code></pre>
|
|
|
<p>在老版本的编译器中（Rust 1.31 前），将会报错，因为 <code>r1</code> 和 <code>r2</code> 的作用域在花括号 <code>}</code> 处结束，那么 <code>r3</code> 的借用就会触发 <strong>无法同时借用可变和不可变</strong> 的规则。</p>
|
|
|
<p>但是在新的编译器中，该代码将顺利通过，因为 <strong>引用作用域的结束位置从花括号变成最后一次使用的位置</strong>，因此 <code>r1</code> 借用和 <code>r2</code> 借用在 <code>println!</code> 后，就结束了，此时 <code>r3</code> 可以顺利借用到可变引用。</p>
|
|
|
<h4 id="nll"><a class="header" href="#nll">NLL</a></h4>
|
|
|
<p>对于这种编译器优化行为，Rust 专门起了一个名字 —— <strong>Non-Lexical Lifetimes(NLL)</strong>，专门用于找到某个引用在作用域(<code>}</code>)结束前就不再被使用的代码位置。</p>
|
|
|
<p>虽然这种借用错误有的时候会让我们很郁闷，但是你只要想想这是 Rust 提前帮你发现了潜在的 BUG，其实就开心了，虽然减慢了开发速度，但是从长期来看，大幅减少了后续开发和运维成本。</p>
|
|
|
<h3 id="悬垂引用dangling-references"><a class="header" href="#悬垂引用dangling-references">悬垂引用(Dangling References)</a></h3>
|
|
|
<p>悬垂引用也叫做悬垂指针，意思为指针指向某个值后，这个值被释放掉了，而指针仍然存在，其指向的内存可能不存在任何值或已被其它变量重新使用。在 Rust 中编译器可以确保引用永远也不会变成悬垂状态：当你获取数据的引用后，编译器可以确保数据不会在引用结束前被释放，要想释放数据，必须先停止其引用的使用。</p>
|
|
|
<p>让我们尝试创建一个悬垂引用，Rust 会抛出一个编译时错误：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021">fn main() {
|
|
|
let reference_to_nothing = dangle();
|
|
|
}
|
|
|
|
|
|
fn dangle() -> &String {
|
|
|
let s = String::from("hello");
|
|
|
|
|
|
&s
|
|
|
}</code></pre>
|
|
|
<p>这里是错误：</p>
|
|
|
<pre><code class="language-text">error[E0106]: missing lifetime specifier
|
|
|
--> src/main.rs:5:16
|
|
|
|
|
|
|
5 | fn dangle() -> &String {
|
|
|
| ^ expected named lifetime parameter
|
|
|
|
|
|
|
= help: this function's return type contains a borrowed value, but there is no value for it to be borrowed from
|
|
|
help: consider using the `'static` lifetime
|
|
|
|
|
|
|
5 | fn dangle() -> &'static String {
|
|
|
| ~~~~~~~~
|
|
|
|
|
|
</code></pre>
|
|
|
<p>错误信息引用了一个我们还未介绍的功能：<a href="https://course.rs/basic/lifetime.html">生命周期(lifetimes)</a>。不过，即使你不理解生命周期，也可以通过错误信息知道这段代码错误的关键信息：</p>
|
|
|
<pre><code class="language-text">this function's return type contains a borrowed value, but there is no value for it to be borrowed from.
|
|
|
该函数返回了一个借用的值，但是已经找不到它所借用值的来源
|
|
|
</code></pre>
|
|
|
<p>仔细看看 <code>dangle</code> 代码的每一步到底发生了什么：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn dangle() -> &String { // dangle 返回一个字符串的引用
|
|
|
|
|
|
let s = String::from("hello"); // s 是一个新字符串
|
|
|
|
|
|
&s // 返回字符串 s 的引用
|
|
|
} // 这里 s 离开作用域并被丢弃。其内存被释放。
|
|
|
// 危险！
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>因为 <code>s</code> 是在 <code>dangle</code> 函数内创建的，当 <code>dangle</code> 的代码执行完毕后，<code>s</code> 将被释放，但是此时我们又尝试去返回它的引用。这意味着这个引用会指向一个无效的 <code>String</code>，这可不对！</p>
|
|
|
<p>其中一个很好的解决方法是直接返回 <code>String</code>：</p>
|
|
|
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn no_dangle() -> String {
|
|
|
let s = String::from("hello");
|
|
|
|
|
|
s
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre>
|
|
|
<p>这样就没有任何错误了，最终 <code>String</code> 的 <strong>所有权被转移给外面的调用者</strong>。</p>
|
|
|
<h2 id="借用规则总结"><a class="header" href="#借用规则总结">借用规则总结</a></h2>
|
|
|
<p>总的来说，借用规则如下：</p>
|
|
|
<ul>
|
|
|
<li>同一时刻，你只能拥有要么一个可变引用，要么任意多个不可变引用</li>
|
|
|
<li>引用必须总是有效的</li>
|
|
|
</ul>
|
|
|
<h2 id="课后练习"><a class="header" href="#课后练习">课后练习</a></h2>
|
|
|
<blockquote>
|
|
|
<p><a href="https://practice-zh.course.rs/ownership/borrowing.html">Rust By Practice</a>，支持代码在线编辑和运行，并提供详细的<a href="https://github.com/sunface/rust-by-practice/blob/master/solutions/ownership/borrowing.md">习题解答</a>。</p>
|
|
|
</blockquote>
|
|
|
|
|
|
</main>
|
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