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添加 Rust 语言圣经（镜像）
2 years ago
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<h1 id="rust所有权转移时发生了奇怪的深拷贝"><a class="header" href="#rust所有权转移时发生了奇怪的深拷贝">Rust所有权转移时发生了奇怪的深拷贝</a></h1>
<p>深拷贝可以说是Rust性能优化的禁忌之词，但是在最不该发生深拷贝的地方却发生了, 本文带领大家来深入分析下原因。</p>
<p>在所有权章节中，我们详细介绍过<a href="https://course.rs/basic/ownership/ownership.html#%E8%BD%AC%E7%A7%BB%E6%89%80%E6%9C%89%E6%9D%83">所有权转移(move)</a>, 里面提到过一个重点：当类型实现<code>Copy</code>特征时，不会转移所有权，而是直接对值进行拷贝：</p>
<pre><pre class="playground"><code class="language-rust edition2021">fn main() {
let x = 1;
let y = x;
// 不会报错
println!(&quot;我(x)的值仅仅是被复制了，我还拥有值的所有权，不信你看：{:?}&quot;,x);
let s = &quot;aaa&quot;.to_string();
let s1 = s;
// 会报错
println!(&quot;我(s)的值被转移给了s1，我已经失去所有权了: {}&quot;,s);
}
</code></pre></pre>
<p>这里的<code>x</code>是数值类型，因此实现了<code>Copy</code>特征，当赋值给<code>y</code>时，仅仅是复制了值，并没有把所有权转移给<code>y</code>，但是<code>s</code>恰好相反，它没有实现<code>Copy</code>特征，当赋值后，所有权被转移给<code>s1</code>，最终导致了最后一行代码的报错.</p>
<p>根据之前的所有权学习章节，所有权转移时的仅仅是复制一个引用，并不会复制底层的数据，例如上面代码中，<code>s</code>的所有权转移给<code>s1</code>时，仅仅是复制了一个引用，该引用继续指向之前的字符串底层数据，因此<strong>所有权转移的性能是非常高的</strong>。</p>
<p>但是如果一切都这么完美，也不会出现这篇文章了，实际上是怎么样？先来看一段代码.</p>
<h2 id="move时发生了数据的深拷贝"><a class="header" href="#move时发生了数据的深拷贝">move时发生了数据的深拷贝</a></h2>
<pre><pre class="playground"><code class="language-rust edition2021">struct LargeArray {
a: [i128; 10000],
}
impl LargeArray {
#[inline(always)]
fn transfer(mut self) -&gt; Self {
println!(&quot;{:?}&quot;, &amp;mut self.a[1] as *mut i128);
// 改变数组中的值
self.a[1] += 23;
self.a[4] += 24;
// 返回所有权
self
}
}
fn main() {
let mut f = LargeArray { a: [10i128; 10000] };
println!(&quot;{:?}&quot;, &amp;mut f.a[1] as *mut i128);
let mut f2 = f.transfer();
println!(&quot;{:?}&quot;, &amp;mut f2.a[1] as *mut i128);
}
</code></pre></pre>
<p>上面的例子很简单，创建了一个结构体<code>f</code>(内部有一个大数组)，接着将它的所有权转移给<code>transfer</code>方法，最后再通过<code>Self</code>返回，转移给<code>f2</code>，在此过程中，观察结构体中的数组第二个元素的内存地址如何变化。</p>
<p>这里还有几个注意点：</p>
<ul>
<li><code>LargeArray</code>没有实现<code>Copy</code>特征，因此在所有权转移时， <strong>本应该</strong>只是复制一下引用，底层的数组并不会被复制</li>
<li><code>transfer</code>方法的参数<code>self</code>表示接收所有权，而不是借用，返回类型<code>Self</code>也表示返回所有权，而不是返回借用, 具体内容在<a href="https://course.rs/basic/method.html">方法</a>章节有介绍</li>
</ul>
<p>从上可知，我们并不应该去复制底层的数组，那么底层数组的地址也不应该变化，换而言之三次内存地址输出应该是同一个地址。但是真的如此吗？世事难料：</p>
<pre><code class="language-console">0x16f9d6870
0x16fa4bbc0
0x16fa24ac0
</code></pre>
<p>果然，结果让人大跌眼镜，竟然三次地址都不一样，意味着每次转移所有权都发生了底层数组的深拷贝！什么情况？！！如果这样，我们以后还能信任Rust吗？完全不符合官方的宣传。</p>
<p>在福建有一个武夷山5A景区，不仅美食特别好吃，而且风景非常优美，其中最著名的就是历时1个多小时的九曲十八弯漂流，而我们的结论是否也能像漂游一样来个大转折？大家拭目以待。</p>
<h2 id="罪魁祸首println"><a class="header" href="#罪魁祸首println">罪魁祸首println?</a></h2>
<p>首先，通过谷歌搜索，我发现了一些蛛丝马迹，有文章提到如果通过<code>println</code>输出内存地址，可能会导致编译器优化失效，也就是从本该有的所有权转移变成了深拷贝，不妨来试试。</p>
<p>但是问题又来了，如果不用<code>println</code>或者类似的方法，我们怎么观察内存地址？好像陷入了绝路。。。只能从Rust之外去想办法了，此时大学学过的汇编发挥了作用：</p>
<pre><code class="language-asm">.LCPI0_0:
.quad 10
.quad 0
example::xxx:
mov eax, 160000
call __rust_probestack
sub rsp, rax
mov rax, rsp
lea rcx, [rsp + 160000]
vbroadcasti128 ymm0, xmmword ptr [rip + .LCPI0_0]
.LBB0_1:
vmovdqu ymmword ptr [rax], ymm0
vmovdqu ymmword ptr [rax + 32], ymm0
vmovdqu ymmword ptr [rax + 64], ymm0
vmovdqu ymmword ptr [rax + 96], ymm0
vmovdqu ymmword ptr [rax + 128], ymm0
vmovdqu ymmword ptr [rax + 160], ymm0
vmovdqu ymmword ptr [rax + 192], ymm0
vmovdqu ymmword ptr [rax + 224], ymm0
vmovdqu ymmword ptr [rax + 256], ymm0
vmovdqu ymmword ptr [rax + 288], ymm0
vmovdqu ymmword ptr [rax + 320], ymm0
vmovdqu ymmword ptr [rax + 352], ymm0
vmovdqu ymmword ptr [rax + 384], ymm0
vmovdqu ymmword ptr [rax + 416], ymm0
vmovdqu ymmword ptr [rax + 448], ymm0
vmovdqu ymmword ptr [rax + 480], ymm0
vmovdqu ymmword ptr [rax + 512], ymm0
vmovdqu ymmword ptr [rax + 544], ymm0
vmovdqu ymmword ptr [rax + 576], ymm0
vmovdqu ymmword ptr [rax + 608], ymm0
add rax, 640
cmp rax, rcx
jne .LBB0_1
mov rax, qword ptr [rsp + 16]
mov rdx, qword ptr [rsp + 24]
add rax, 69
adc rdx, 0
add rsp, 160000
vzeroupper
ret
</code></pre>
<p>去掉所有<code>println</code>后的汇编生成如上所示(大家可以在godbolt上自己尝试)，以我蹩脚的汇编水平来看，貌似没有任何数组拷贝的发生，也就是说：
如同量子的不可观测性，我们的<code>move</code>也这么傲娇？我们用println观测，它就傲娇去复制，不观测时，就老老实实转移所有权？WTF!</p>
<p>事情感觉进入了僵局，下一步该如何办？</p>
<h2 id="栈和堆的不同move行为"><a class="header" href="#栈和堆的不同move行为">栈和堆的不同move行为</a></h2>
<p>我突然灵光一现，想到一个问题，之前的所有权转移其实可以分为两类：<strong>栈上数据的复制和堆上数据的转移</strong>，这也是非常符合直觉的，例如<code>i32</code>这种类型实现了<code>Copy</code>特征，可以存储在栈上，因此它就是复制行为，而<code>String</code>类型是引用存储在栈上，底层数据存储在堆上，因此转移所有权时只需要复制一下引用即可。</p>
<p>那问题来了，我们的<code>LargeArray</code>存在哪里？这也许就是一个破局点！</p>
<pre><pre class="playground"><code class="language-rust edition2021">
<span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>struct LargeArray {
a: [i128; 10000],
}
<span class="boring">}
</span></code></pre></pre>
<p>结构体是一个复合类型，它内部字段的数据存在哪里，就大致决定了它存在哪里。而该结构体里面的<code>a</code>字段是一个数组，而不是动态数组<code>Vec</code>，从<a href="https://course.rs/basic/compound-type/array.html#%E5%88%9B%E5%BB%BA%E6%95%B0%E7%BB%84">数组</a>章节可知：数组是存储在栈上的数据结构！</p>
<p>再想想，栈上的数据在<code>move</code>的时候，是要从一个栈复制到另外一个栈的，那是不是内存地址就变了？！因此，就能完美解释，为什么使用<code>println</code>时，数组的地址会变化了，是因为栈上的数组发生了复制。</p>
<p>但是问题还有，为什么不使用<code>println</code>，数组地址就不变？要解释清楚这个问题，先从编译器优化讲起。</p>
<h2 id="编译器对move的优化"><a class="header" href="#编译器对move的优化">编译器对move的优化</a></h2>
<p>从根本上来说，<code>move</code>就意味着拷贝复制，只不过看是浅拷贝还是深拷贝，对于堆上的数据来说，浅拷贝只复制引用，而栈上的数据则是整个复制。</p>
<p>但是在实际场景中，由于编译器的复杂实现，它能优化的场景远比我们想象中更多，例如对于<code>move</code>过程中的复制，编译器有可能帮你优化掉，在没有<code>println</code>的代码中，该<code>move</code>过程就被Rust编译器优化了。</p>
<p>但是这种编译器优化非常复杂，而且随着Rust的版本更新在不停变化，因此几乎没有人能说清楚这里面的门门道道，但是有一点可以知道：<strong><code>move</code>确实存在被优化的可能性，最终避免了复制的发生</strong>.</p>
<p>那么<code>println</code>没有被优化的原因也呼之欲出了: 它阻止了编译器对<code>move</code>的优化。</p>
<h2 id="println阻止了优化"><a class="header" href="#println阻止了优化">println阻止了优化</a></h2>
<p>编译器优化的一个基本准则就是：中间过程没有其它代码在使用，则可以尝试消除这些中间过程。</p>
<p>回头来看看<code>println</code>:</p>
<pre><pre class="playground"><code class="language-rust edition2021">
<span class="boring">#![allow(unused)]
</span><span class="boring">fn main() {
</span>println!(&quot;{:?}&quot;, &amp;mut f.a[1] as *mut i128);
<span class="boring">}
</span></code></pre></pre>
<p>它需要打印数组在各个点的内存地址，假如编译器优化了复制，那这些中间状态的内存地址是不是就丢失了？对于这种可能会导致状态丢失的情况，编译器是不可能进行优化的，因此<code>move</code>时的栈上数组复制就顺理成章的发生了, 还是2次。</p>
<h2 id="最佳实践"><a class="header" href="#最佳实践">最佳实践</a></h2>
<p>那么，在实践中遇到这种情况怎么办？</p>
<h4 id="mut-self"><a class="header" href="#mut-self">&amp;mut self</a></h4>
<p>其实办法也很多，首当其冲的就是使用<code>&amp;mut self</code>进行可变借用，而不是转移进来所有权，再转移出去。</p>
<h4 id="box分配到堆上"><a class="header" href="#box分配到堆上">Box分配到堆上</a></h4>
<p>如果你确实需要依赖所有权的转移来实现某个功能(例如链式方法调用:<code>x.a().b()...</code>)，那么就需要使用<code>Box</code>把该数组分配在堆上，而不是栈上：</p>
<pre><pre class="playground"><code class="language-rust edition2021">struct LargeArray {
a: Box&lt;[i128; 10000]&gt;,
}
impl LargeArray {
#[inline(always)]
fn transfer(mut self) -&gt; Self {
println!(&quot;{:?}&quot;, &amp;mut self.a[1] as *mut i128);
//do some stuff to alter it
self.a[1] += 23;
self.a[4] += 24;
//return the same object
self
}
}
fn main() {
let mut f = LargeArray { a: Box::new([10i128; 10000] )};
println!(&quot;{:?}&quot;, &amp;mut f.a[1] as *mut i128);
let mut f2 = f.transfer();
println!(&quot;{:?}&quot;, &amp;mut f2.a[1] as *mut i128);
}
</code></pre></pre>
<p>输出如下:</p>
<pre><code>0x138008010
0x138008010
0x138008010
</code></pre>
<p>完美符合了我们对堆上数据的预期，hooray!</p>
<h4 id="神龟莫测的编译器优化"><a class="header" href="#神龟莫测的编译器优化">神龟莫测的编译器优化</a></h4>
<p>当然，你也可以选择相信编译器的优化，虽然很难识它真面目，同时它的行为也神鬼莫测，但是总归是在之前的例子中证明了，它确实可以，不是嘛？ = , =</p>
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