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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， < / 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#转移所有权" > 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 class = "playground" > < code class = "language-rust edition2021" > fn main() {
let x = 1;
let y = x;
// 不会报错
println!("我(x)的值仅仅是被复制了，我还拥有值的所有权，不信你看：{:?}",x);
let s = "aaa".to_string();
let s1 = s;
// 会报错
println!("我(s)的值被转移给了s1，
}< / code > < / 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 class = "playground" > < code class = "language-rust edition2021" > struct LargeArray {
a: [i128; 10000],
}
impl LargeArray {
#[inline(always)]
fn transfer(mut self) -> Self {
println!("{:?}", & 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!("{:?}", & mut f.a[1] as *mut i128);
let mut f2 = f.transfer();
println!("{:?}", & mut f2.a[1] as *mut i128);
}< / code > < / 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， ， ， ！ ？ ！ ！ ， ？ < / p >
< p > 在福建有一个武夷山5A景区， ， ， ， ？ < / 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， ？ ， ： < / 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？ ， ， ， ？ < / 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 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 >
< 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#创建数组" > 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， ， ， ： < 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 class = "playground" > < code class = "language-rust edition2021" > < span class = "boring" > #![allow(unused)]
< / span > < span class = "boring" > fn main() {
< / span > println!("{:?}", & mut f.a[1] as *mut i128);
< span class = "boring" > }< / span > < / code > < / 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" > & mut self< / a > < / h4 >
< p > 其实办法也很多，首当其冲的就是使用< code > & 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 class = "playground" > < code class = "language-rust edition2021" > struct LargeArray {
a: Box< [i128; 10000]> ,
}
impl LargeArray {
#[inline(always)]
fn transfer(mut self) -> Self {
println!("{:?}", & 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!("{:?}", & mut f.a[1] as *mut i128);
let mut f2 = f.transfer();
println!("{:?}", & mut f2.a[1] as *mut i128);
}< / code > < / 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， < / p >
< h4 id = "神鬼莫测的编译器优化" > < a class = "header" href = "#神鬼莫测的编译器优化" > 神鬼莫测的编译器优化< / 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 >
< / main >
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