我可以在Rust中按值有效地返回对象吗? [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>>) { /* ... */ }

我希望拥有:

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

我听说C ++通常会实现返回值优化(RVO)，如果您幸运的话，并且具有良好的编译器.我们可以在此处禁用复制，并通过传递给函数的隐藏指针将其返回吗? 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，您一定要写

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具有RVO，并且此在指南中刊登广告.您可以使用以下代码自己看到它:

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();
}

如果您以发布模式，输出程序集，您将看到以下内容:

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

内联了

Vec::new()，但是您可以看到一个想法-新鲜的Vec实例的地址传递到%rdi中的函数中，并且该函数将Vec字段直接存储到此内存中，避免了不必要的操作通过堆栈复制.这就是它的称呼方式:

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

您可以看到最终Vec实例将直接放入堆栈存储器中.

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

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