我可以在 Rust 中按值有效地返回一个对象吗? [英] Can I efficiently return an object by value in Rust?

问题描述

我想用一个函数初始化一个大对象.目前我有:

I would like to initialize a large object with a function. Currently I have:

fn initialize(mydata: &mut Vec<Vec<MyStruct>>) { /* ... */ }

我更喜欢:

fn initialize() -> Vec<Vec<MyStruct>> { /* ... */ }

我听说 C++ 经常实现返回值优化 (RVO)，如果你很幸运并且有一个好的编译器.我们可以在这里禁用复制并让它由传递给函数的隐藏指针返回吗?RVO 是语言的一部分还是可选的优化?

I've heard that C++ often implements return value optimization (RVO), if you are lucky and have a good compiler. Can we disable copying here and have it returned by a hidden pointer that is passed into the function? Is RVO part of the language or an optional optimization?

推荐答案

是的，无论如何，你应该写

Yes, by all means, you should write

fn initialize() -> Vec<Vec<MyStruct>> { ... }

(顺便说一句，Vec 并没有那么大——它只有 3 个指针大小的整数)

(By the way, a Vec is not that large - it's only 3 pointer-sized integers)

Rust 有 RVO，这个 在指南中做了广告.您可以使用以下代码自行查看:

Rust has RVO, and this is advertised in guides. You can see it yourself with this code:

#[inline(never)]
fn initialize() -> Vec<i32> {
    Vec::new()
}

fn main() {
    let v = initialize();
}

如果你在发布模式下编译这个程序 在操场上，输出程序集，除此之外你会看到:

If you compile this program in release mode on the playground, outputting assembly, among everything else you will see this:

playground::initialize:
    movq    $4, (%rdi)
    xorps   %xmm0, %xmm0
    movups  %xmm0, 8(%rdi)
    retq

Vec::new() 被内联了，但你可以看到这个想法 - 新的 Vec 实例的地址被传递到 % 的函数中rdi，并且该函数将Vec 字段直接存储到此内存中，避免通过堆栈进行不必要的复制.它是这样叫的:

Vec::new() was inlined, but you can see the idea - the address for the fresh Vec instance is passed into the function in %rdi, and the function stores Vec fields directly into this memory, avoiding unnecessary copying through the stack. This is how it is called:

playground::main:
    subq    $24, %rsp
    movq    %rsp, %rdi
    callq   playground::initialize

可以看到最终Vec实例会被直接放入栈内存中.

You can see that eventually the Vec instance will be put directly into the stack memory.

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