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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 >
< 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 >
< 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 >
< 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 >
< 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 >
< pre class = "playground" > < code class = "language-rust edition2021" > fn main() {
let x = 5;
let y = & x;
assert_eq!(5, x);
assert_eq!(5, *y);
}< / code > < / pre >
< 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 >
< 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，
// 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， ， ， < / 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，
let r3 = & mut s;
println!("{}", r3);
} // 老编译器中，
// 新编译器中， < / code > < / pre >
< p > 在老版本的编译器中（ < 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， < 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， ， ， ， < / 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， < / 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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