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<main>
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<h1 id="智能指针引起的重复借用错误"><a class="header" href="#智能指针引起的重复借用错误">智能指针引起的重复借用错误</a></h1>
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<p>本文将彻底解决一个困扰广大 Rust 用户已久的常见错误： 当智能指针和结构体一起使用时导致的借用错误: <code>cannot borrow</code>mut_s<code> as mutable because it is also borrowed as immutable</code>.</p>
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<p>相信看过<a href="https://course.rs/fight-with-compiler/intro.html"><<对抗 Rust 编译检查系列>></a>的读者都知道结构体中的不同字段可以独立借用吧？</p>
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<h2 id="结构体中的字段借用"><a class="header" href="#结构体中的字段借用">结构体中的字段借用</a></h2>
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<p>不知道也没关系，我们这里再简单回顾一下:</p>
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<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
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</span><span class="boring">fn main() {
|
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</span>struct Test {
|
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a : u32,
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b : u32
|
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}
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impl Test {
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fn increase(&mut self) {
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let mut a = &mut self.a;
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let mut b = &mut self.b;
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*b += 1;
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*a += 1;
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}
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}
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<span class="boring">}</span></code></pre></pre>
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<p>这段代码看上去像是重复借用了<code>&mut self</code>,违反了 Rust 的借用规则，实际上在聪明的 Rust 编译器面前，这都不是事。它能发现我们其实借用了目标结构体的不同字段，因此完全可以将其借用权分离开来。</p>
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<p>因此，虽然我们不能同时对整个结构体进行多次可变借用，但是我们可以分别对结构体中的不同字段进行可变借用，当然，一个字段至多也只能存在一个可变借用，这个最基本的所有权规则还是不能违反的。变量<code>a</code>引用结构体字段<code>a</code>，变量<code>b</code>引用结构体字段<code>b</code>，从底层来说，这种方式也不会造成两个可变引用指向了同一块内存。</p>
|
|
|
<h2 id="refcell"><a class="header" href="#refcell">RefCell</a></h2>
|
|
|
<p>如果你还不知道 RefCell，可以看看<a href="https://course.rs/advance/smart-pointer/cell-refcell.html">这篇文章</a>，当然不看也行，简而言之，RefCell 能够实现：</p>
|
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<ul>
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|
|
<li>将借用规则从编译期推迟到运行期，但是并不会绕过借用规则，当不符合时，程序直接<code>panic</code></li>
|
|
|
<li>实现内部可变性：简单来说，对一个不可变的值进行可变借用，然后修改内部的值</li>
|
|
|
</ul>
|
|
|
<h2 id="被-refcell-包裹的结构体"><a class="header" href="#被-refcell-包裹的结构体">被 RefCell 包裹的结构体</a></h2>
|
|
|
<p>既然了解了结构体的借用规则和<code>RefCell</code>, 我们来看一段结合了两者的代码：</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::cell::RefCell;
|
|
|
use std::io::Write;
|
|
|
|
|
|
struct Data {
|
|
|
string: String,
|
|
|
}
|
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|
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|
struct S {
|
|
|
data: Data,
|
|
|
writer: Vec<u8>,
|
|
|
}
|
|
|
|
|
|
fn write(s: RefCell<S>) {
|
|
|
let mut mut_s = s.borrow_mut();
|
|
|
let str = &mut_s.data.string;
|
|
|
mut_s.writer.write(str.as_bytes());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>以上代码从<code>s</code>中可变借用出结构体<code>S</code>，随后又对结构体中的两个字段进行了分别借用，按照之前的规则这段代码应该顺利通过编译：</p>
|
|
|
<pre><code class="language-console">error[E0502]: cannot borrow `mut_s` as mutable because it is also borrowed as immutable
|
|
|
--> src/main.rs:16:5
|
|
|
|
|
|
|
15 | let str = &mut_s.data.string;
|
|
|
| ----- immutable borrow occurs here
|
|
|
16 | mut_s.writer.write(str.as_bytes());
|
|
|
| ^^^^^ --- immutable borrow later used here
|
|
|
| |
|
|
|
| mutable borrow occurs here
|
|
|
</code></pre>
|
|
|
<p>只能说，还好它报错了，否则本篇文章已经可以结束。。。错误很简单，首先对结构体<code>S</code>的<code>data</code>字段进行了不可变借用，其次又对<code>writer</code>字段进行了可变借用，这个符合之前的规则：对结构体不同字段分开借用，为何报错了？</p>
|
|
|
<h2 id="深入分析"><a class="header" href="#深入分析">深入分析</a></h2>
|
|
|
<p>第一感觉，问题是出在<code>borrow_mut</code>方法返回的类型上，先来看看:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>pub fn borrow_mut(&self) -> RefMut<'_, T>
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>可以看出，该方法并没有直接返回我们的结构体，而是一个<code>RefMut</code>类型，而要使用该类型，需要经过编译器为我们做一次隐式的<code>Deref</code>转换，编译器展开后的代码大概如下:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::cell::RefMut;
|
|
|
use std::ops::{Deref, DerefMut};
|
|
|
|
|
|
fn write(s: RefCell<S>) {
|
|
|
let mut mut_s: RefMut<S> = s.borrow_mut();
|
|
|
let str = &Deref::deref(&mut_s).data.string;
|
|
|
DerefMut::deref_mut(&mut mut_s).writer.write(str.as_bytes());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>可以看出，对结构体字段的调用，实际上经过一层函数，一层函数！？我相信你应该想起了什么，是的，在<a href="https://course.rs/fight-with-compiler/borrowing/ref-exist-in-out-fn.html">上一篇文章</a>中讲过类似的问题， 大意就是<strong>编译器对于函数往往只会分析签名，并不关心内部到底如何使用结构体</strong>。</p>
|
|
|
<p>而上面的<code>&Deref::deref(&mut_s)</code>和<code>DerefMut::deref_mut(&mut mut_s)</code>函数，签名全部使用的是结构体，并不是结构体中的某一个字段，因此对于编译器来说，该结构体明显是被重复借用了！</p>
|
|
|
<h2 id="解决方法"><a class="header" href="#解决方法">解决方法</a></h2>
|
|
|
<p>因此要解决这个问题，我们得把之前的展开形式中的<code>Deref::deref</code>消除掉，这样没有了函数签名，编译器也将不再懒政。</p>
|
|
|
<p>既然两次<code>Deref::deref</code>调用都是对智能指针的自动<code>Deref</code>，那么可以提前手动的把它<code>Deref</code>了，只做一次！</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn write(s: RefCell<S>) {
|
|
|
let mut mut_s = s.borrow_mut();
|
|
|
let mut tmp = &mut *mut_s; // Here
|
|
|
let str = &tmp.data.string;
|
|
|
tmp.writer.write(str.as_bytes());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>以上代码通过<code>*</code>对<code>mut_s</code>进行了解引用，获得结构体，然后又对结构体进行了可变借用<code>&mut</code>，最终赋予<code>tmp</code>变量，那么该变量就持有了我们的结构体的可变引用，而不再是持有一个智能指针。</p>
|
|
|
<p>此后对<code>tmp</code>的使用就回归到文章开头的那段代码：分别借用结构体的不同字段，成功通过编译！</p>
|
|
|
<h4 id="展开代码"><a class="header" href="#展开代码">展开代码</a></h4>
|
|
|
<p>我们再来模拟编译器对正确的代码进行一次展开:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::cell::RefMut;
|
|
|
use std::ops::DerefMut;
|
|
|
|
|
|
fn write(s: RefCell<S>) {
|
|
|
let mut mut_s: RefMut<S> = s.borrow_mut();
|
|
|
let tmp: &mut S = DerefMut::deref_mut(&mut mut_s);
|
|
|
let str = &tmp.data.string;
|
|
|
tmp.writer.write(str.as_bytes());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>可以看出，此时对结构体的使用不再有<code>DerefMut::deref</code>的身影，我们成功消除了函数边界对编译器的影响！</p>
|
|
|
<h2 id="不仅仅是-refcell"><a class="header" href="#不仅仅是-refcell">不仅仅是 RefCell</a></h2>
|
|
|
<p>事实上，除了 RefCell 外，还有不少会导致这种问题的智能指针，当然原理都是互通的，我们这里就不再进行一一深入讲解，只简单列举下：</p>
|
|
|
<ul>
|
|
|
<li><code>Box</code></li>
|
|
|
<li><code>MutexGuard</code>(来源于 Mutex)</li>
|
|
|
<li><code>PeekMut</code>(来源于 BinaryHeap)</li>
|
|
|
<li><code>RwLockWriteGuard</code>(来源于 RwLock)</li>
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<li><code>String</code></li>
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<li><code>Vec</code></li>
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<li><code>Pin</code></li>
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|
</ul>
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<h2 id="一个练习"><a class="header" href="#一个练习">一个练习</a></h2>
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<p>下面再来一个练习巩固一下，强烈建议大家按照文章的思路进行分析和解决：</p>
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<pre><pre class="playground"><code class="language-rust edition2021">use std::rc::Rc;
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use std::cell::RefCell;
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|
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pub struct Foo {
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pub foo1: Vec<bool>,
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pub foo2: Vec<i32>,
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|
}
|
|
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fn main() {
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let foo_cell = Rc::new(RefCell::new(Foo {
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|
|
foo1: vec![true, false],
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|
foo2: vec![1, 2]
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|
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}));
|
|
|
|
|
|
let borrow = foo_cell.borrow_mut();
|
|
|
let foo1 = &borrow.foo1;
|
|
|
// 下面代码会报错,因为`foo1`和`foo2`发生了重复借用
|
|
|
borrow.foo2.iter_mut().enumerate().for_each(|(idx, foo2)| {
|
|
|
if foo1[idx] {
|
|
|
*foo2 *= -1;
|
|
|
}
|
|
|
});
|
|
|
}</code></pre></pre>
|
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|
<h2 id="总结"><a class="header" href="#总结">总结</a></h2>
|
|
|
<p>当结构体的引用穿越函数边界时，我们要格外小心，因为编译器只会对函数签名进行检查，并不关心内部到底用了结构体的哪个字段，当签名都使用了结构体时，会立即报错。</p>
|
|
|
<p>而智能指针由于隐式解引用<code>Deref</code>的存在，导致了两次<code>Deref</code>时都让结构体穿越了函数边界<code>Deref::deref</code>，结果造成了重复借用的错误。</p>
|
|
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<p>解决办法就是提前对智能指针进行手动解引用，然后对内部的值进行借用后，再行使用。</p>
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