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<h1 id="代码重构导致的可变借用错误"><a class="header" href="#代码重构导致的可变借用错误">代码重构导致的可变借用错误</a></h1>
<p>相信大家都听说过<strong>重构一时爽,一直重构一直爽</strong>的说法,私以为这种说法是很有道理的,不然技术团队绩效从何而来?但是,在 Rust 中,重构可能就不是那么爽快的事了,不信?咱们来看看。</p>
<h2 id="欣赏下报错"><a class="header" href="#欣赏下报错">欣赏下报错</a></h2>
<p>很多时候,错误也是一种美,但是当这种错误每天都能见到时(呕):</p>
<pre><code class="language-css">error[e0499]: cannot borrow ` * self` as mutable more than once at a time;
</code></pre>
<p>虽然这一类错误长得一样,但是我这里的错误可能并不是大家常遇到的那些妖艳错误,废话不多说,一起来看看。</p>
<h2 id="重构前的正确代码"><a class="header" href="#重构前的正确代码">重构前的正确代码</a></h2>
<pre><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></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><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>
<p>由于领导说我们这个函数没办法复用,那就敷衍一下呗:</p>
<pre><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></pre>
<p>既然领导说了,咱照做,反正他也没说怎么个复用法,咱就来个简单的,把<code>a</code>的递增部分复用下。</p>
<p>代码说实话。。。更丑了,但是更强了吗?</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><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></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><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></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><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></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="闭包中的例子"><a class="header" href="#闭包中的例子">闭包中的例子</a></h2>
<p>再来看一个使用了闭包的例子:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>use tokio::runtime::Runtime;
struct Server {
number_of_connections : u64
}
impl Server {
pub fn new() -&gt; Self {
Server { number_of_connections : 0}
}
pub fn increase_connections_count(&amp;mut self) {
self.number_of_connections += 1;
}
}
struct ServerRuntime {
runtime: Runtime,
server: Server
}
impl ServerRuntime {
pub fn new(runtime: Runtime, server: Server) -&gt; Self {
ServerRuntime { runtime, server }
}
pub fn increase_connections_count(&amp;mut self) {
self.runtime.block_on(async {
self.server.increase_connections_count()
})
}
}
<span class="boring">}</span></code></pre></pre>
<p>代码中使用了<code>tokio</code>,在<code>increase_connections_count</code>函数中启动了一个异步任务,并且等待它的完成。这个函数中分别引用了<code>self</code>中的不同字段:<code>runtime</code><code>server</code>,但是可能因为闭包的原因,编译器没有像本文最开始的例子中那样聪明,并不能识别这两个引用仅仅引用了同一个结构体的不同部分,因此报错了:</p>
<pre><code class="language-console">error[E0501]: cannot borrow `self.runtime` as mutable because previous closure requires unique access
--&gt; the_little_things\src\main.rs:28:9
|
28 | self.runtime.block_on(async {
| __________^____________--------_______-
| | | |
| | _________| first borrow later used by call
| ||
29 | || self.server.increase_connections_count()
| || ---- first borrow occurs due to use of `self` in generator
30 | || })
| ||_________-^ second borrow occurs here
| |__________|
| generator construction occurs here
</code></pre>
<h4 id="解决办法-1"><a class="header" href="#解决办法-1">解决办法</a></h4>
<p>解决办法很粗暴,既然编译器不能理解闭包中的引用是不同的,那么我们就主动告诉它:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub fn increase_connections_count(&amp;mut self) {
let runtime = &amp;mut self.runtime;
let server = &amp;mut self.server;
runtime.block_on(async {
server.increase_connections_count()
})
}
<span class="boring">}</span></code></pre></pre>
<p>上面通过变量声明的方式,在闭包外声明了两个变量分别引用结构体<code>self</code>的不同字段,这样一来,编译器就不会那么笨,编译顺利通过。</p>
<p>你也可以这么写:</p>
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>pub fn increase_connections_count(&amp;mut self) {
let ServerRuntime { runtime, server } = self;
runtime.block_on(async {
server.increase_connections_count()
})
}
<span class="boring">}</span></code></pre></pre>
<p>当然,如果难以解决,还有一个笨办法,那就是将<code>server</code><code>runtime</code>分离开来,不要放在一个结构体中。</p>
<h2 id="总结"><a class="header" href="#总结">总结</a></h2>
<p>心中有剑,手中无剑,是武学至高境界。</p>
<p>本文列出的那条编译规则,在未来就将是大家心中的那把剑,当这些心剑招式足够多时,量变产生质变,终将天下无敌。</p>
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