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<h1 id="同时在函数内外使用引用导致的重复借用错误"><a class="header" href="#同时在函数内外使用引用导致的重复借用错误">同时在函数内外使用引用导致的重复借用错误</a></h1>
<p>本文将彻底解决一个困扰广大 Rust 用户已久的常见错误：因为在函数内外同时借用一个引用，导致了重复借用错误<code>cannot borrow *self as mutable because it is also borrowed as immutable</code>.</p>
<blockquote>
<p>本文大部分内容节选自<a href="https://course.rs/pitfalls/index.html">Rust 常见陷阱</a>专题，由于借用是新手绕不过去的坎，因此将其提取出来形成一个新的系列</p>
</blockquote>
<h2 id="正确的代码"><a class="header" href="#正确的代码">正确的代码</a></h2>
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>struct Test {
a : u32,
b : u32
}
impl Test {
fn increase(&amp;mut self) {
let mut a = &amp;mut self.a;
let mut b = &amp;mut self.b;
*b += 1;
*a += 1;
}
}
<span class="boring">}</span></code></pre>
<p>这段代码是可以正常编译的，也许有读者会有疑问，<code>self</code>在这里被两个变量以可变的方式借用了，明明违反了 Rust 的所有权规则，为何它不会报错？</p>
<p>答案要从很久很久之前开始(啊哒~~~由于我太啰嗦，被正义群众来了一下，那咱现在开始长话短说，直接进入主题)。</p>
<h4 id="正确代码为何不报错"><a class="header" href="#正确代码为何不报错">正确代码为何不报错？</a></h4>
<p>虽然从表面来看，<code>a</code>和<code>b</code>都可变引用了<code>self</code>，但是 Rust 的编译器在很多时候都足够聪明，它发现我们其实仅仅引用了同一个结构体中的不同字段，因此完全可以将其的借用权分离开来。</p>
<p>因此，虽然我们不能同时对整个结构体进行可变引用，但是我们可以分别对结构体中的不同字段进行可变引用，当然，一个字段至多也只能存在一个可变引用，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>
<p>至此，正确代码我们已经挖掘完毕，再来看看重构后的错误代码。</p>
<h2 id="重构后的错误代码"><a class="header" href="#重构后的错误代码">重构后的错误代码</a></h2>
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>struct Test {
a : u32,
b : u32
}
impl Test {
fn increase_a (&amp;mut self) {
self.a += 1;
}
fn increase(&amp;mut self) {
let b = &amp;mut self.b;
self.increase_a();
*b += 1;
}
}
<span class="boring">}</span></code></pre>
<p>果然不正义的代码就是不好看，但是邪恶的它更强了吗？</p>
<pre><code class="language-console">error[E0499]: cannot borrow `*self` as mutable more than once at a time
--&gt; src/main.rs:14:9
|
13 | let b = &amp;mut self.b;
| ----------- first mutable borrow occurs here
14 | self.increase_a();
| ^^^^ second mutable borrow occurs here
15 | *b += 1;
| ------- first borrow later used here
</code></pre>
<p>嗯，最开始提到的错误，它终于出现了。</p>
<h2 id="大聪明编译器"><a class="header" href="#大聪明编译器">大聪明编译器</a></h2>
<p>为什么？明明之前还是正确的代码，就因为放入函数中就报错了？我们先从一个简单的理解谈起，当然这个理解也是浮于表面的，等会会深入分析真实的原因。</p>
<p>之前讲到 Rust 编译器挺聪明，可以识别到引用到不同的结构体字段，因此不会报错。但是现在这种情况下，编译器又不够聪明了，一旦放入函数中，编译器将无法理解我们对<code>self</code>的使用：它仅仅用到了一个字段，而不是整个结构体。</p>
<p>因此它会简单的认为，这个结构体作为一个整体被可变借用了，产生两个可变引用，一个引用整个结构体，一个引用了结构体字段<code>b</code>，这两个引用存在重叠的部分，最终导致编译错误。</p>
<h2 id="被冤枉的编译器"><a class="header" href="#被冤枉的编译器">被冤枉的编译器</a></h2>
<p>在工作生活中，我们无法理解甚至错误的理解一件事，有时是因为层次不够导致的。同样，对于本文来说，也是因为我们对编译器的所知不够，才冤枉了它，还给它起了一个屈辱的“大聪明”外号。</p>
<h4 id="深入分析"><a class="header" href="#深入分析">深入分析</a></h4>
<blockquote>
<p>如果只改变相关函数的实现而不改变它的签名，那么不会影响编译的结果</p>
</blockquote>
<p>何为相关函数？当函数<code>a</code>调用了函数<code>b</code>，那么<code>b</code>就是<code>a</code>的相关函数。</p>
<p>上面这句是一条非常重要的编译准则，意思是，对于编译器来说，只要函数签名没有变，那么任何函数实现的修改都不会影响已有的编译结果(前提是函数实现没有错误- , -)。</p>
<p>以前面的代码为例：</p>
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>fn increase_a (&amp;mut self) {
self.a += 1;
}
fn increase(&amp;mut self) {
let b = &amp;mut self.b;
self.increase_a();
*b += 1;
}
<span class="boring">}</span></code></pre>
<p>虽然<code>increase_a</code>在函数实现中没有访问<code>self.b</code>字段，但是它的签名允许它访问<code>b</code>，因此违背了借用规则。事实上，该函数有没有访问<code>b</code>不重要，<strong>因为编译器在这里只关心签名，签名存在可能性，那么就立刻报出错误</strong>。</p>
<p>为何会有这种编译器行为，主要有两个原因：</p>
<ol>
<li>一般来说，我们希望编译器有能力独立的编译每个函数，而无需深入到相关函数的内部实现，因为这样做会带来快得多的编译速度。</li>
<li>如果没有这种保证，那么在实际项目开发中，我们会特别容易遇到各种错误。 假设我们要求编译器不仅仅关注相关函数的签名，还要深入其内部关注实现，那么由于 Rust 严苛的编译规则，当你修改了某个函数内部实现的代码后，可能会引起使用该函数的其它函数的各种错误！对于大型项目来说，这几乎是不可接受的！</li>
</ol>
<p>然后，我们的借用类型这么简单，编译器有没有可能针对这种场景，在现有的借用规则之外增加特殊规则？答案是否定的，由于 Rust 语言的设计哲学：特殊规则的加入需要慎之又慎，而我们的这种情况其实还蛮好解决的，因此编译器不会为此新增规则。</p>
<h2 id="解决办法"><a class="header" href="#解决办法">解决办法</a></h2>
<p>在深入分析中，我们提到一条重要的规则，要影响编译行为，就需要更改相关函数的签名，因此可以修改<code>increase_a</code>的签名:</p>
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>fn increase_a (a :&amp;mut u32) {
*a += 1;
}
fn increase(&amp;mut self) {
let b = &amp;mut self.b;
Test::increase_a(&amp;mut self.a);
*b += 1;
}
<span class="boring">}</span></code></pre>
<p>此时，<code>increase_a</code>这个相关函数，不再使用<code>&amp;mut self</code>作为签名，而是获取了结构体中的字段<code>a</code>，此时编译器又可以清晰的知道：函数<code>increase_a</code>和变量<code>b</code>分别引用了结构体中的不同字段，因此可以编译通过。</p>
<p>当然，除了修改相关函数的签名，你还可以修改调用者的实现：</p>
<pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>fn increase(&amp;mut self) {
self.increase_a();
self.b += 1;
}
<span class="boring">}</span></code></pre>
<p>在这里，我们不再单独声明变量<code>b</code>，而是直接调用<code>self.b+=1</code>进行递增，根据借用生命周期<a href="https://course.rs/advance/lifetime/advance.html#nllnon-lexical-lifetime">NLL</a>的规则，第一个可变借用<code>self.increase_a()</code>的生命周期随着方法调用的结束而结束，那么就不会影响<code>self.b += 1</code>中的借用。</p>
<h2 id="cpu-模拟例子"><a class="header" href="#cpu-模拟例子">CPU 模拟例子</a></h2>
<p>我们再来看一个例子:</p>
<pre class="playground"><code class="language-rust edition2021">use std::collections::HashMap;
struct Cpu {
pc: u16,
cycles: u32,
opcodes: HashMap&lt;u8, Opcode&gt;,
}
struct Opcode {
size: u16,
cycles: u32,
}
impl Cpu {
fn new() -&gt; Cpu {
Cpu {
pc: 0,
cycles: 0,
opcodes: HashMap::from([
(0x00, Opcode::new(1, 7)),
(0x01, Opcode::new(2, 6))
]),
}
}
fn tick(&amp;mut self) {
let address = self.pc as u8;
let opcode = &amp;self.opcodes[&amp;address];
step(&amp;mut self, opcode);
}
}
fn step(cpu : &amp;mut Cpu, opcode: &amp;Opcode) {
}
impl Opcode {
fn new(size: u16, cycles: u32) -&gt; Opcode {
Opcode { size, cycles }
}
}
fn main() {
let mut cpu = Cpu::new();
cpu.tick();
}</code></pre>
<h2 id="总结"><a class="header" href="#总结">总结</a></h2>
<p>知其然知其所以然，要彻底解决借用导致的编译错误，我们就必须深入了解其原理，心中有剑则手中无“贱“。</p>
<p>上面的例子就留给读者朋友自己去解决，相信你以后在遇到这种常见问题时，会更加游刃有余。</p>
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