返回一个没有副本的c ++ std :: vector？ [英] Returning a c++ std::vector without a copy?

问题描述

是否可以从函数返回标准容器而不进行复制？

Is it possible to return a standard container from a function without making a copy?

示例代码：

std::vector<A> MyFunc();

...

std::vector<A> b = MyFunc();

据我所知，这将返回值复制到一个新的向量b。

As far as I understand, this copies the return value into a new vector b. Does making the function return references or something like that allow avoiding the copy?

推荐答案

如果你的编译器支持NRVO，那么没有副本前提是在返回对象的函数中满足某些条件。幸运的是，这最终添加到 Visual C ++ 2005（v8。 0）如果容器很大，这可能对perf造成很大的影响。

If your compiler supports the NRVO then no copy will be made, provided certain conditions are met in the function returning the object. Thankfully, this was finally added in Visual C++ 2005 (v8.0) This can have a major +ve impact on perf if the container is large, obviously.

如果你自己的编译器文档不说是否它支持，你应该能够编译C ++代码到汇编器（在优化/释放模式），并检查使用简单的示例函数做了什么。

If your own compiler docs do not say whether or not it's supported, you should be able to compile the C++ code to assembler (in optimized/release mode) and check what's done using a simple sample function.

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