返回没有副本的c ++ std::vector?
是否可以在不复制的情况下从函数返回标准容器?
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();
据我了解，这会将返回值复制到一个新向量 b 中.使函数返回引用或类似的东西可以避免复制吗?
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
如果您的编译器支持 NRVO，那么只要返回对象的函数满足某些条件，就不会进行复制.值得庆幸的是，这最终被添加到 Visual C++ 2005 (v8.0) 显然，如果容器很大，这会对性能产生重大的 +ve 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.
如果您自己的编译器文档没有说明它是否受支持，您应该能够将 C++ 代码编译为汇编器(在优化/发布模式下)并使用简单的示例函数检查所做的工作.
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.
还有一个很棒的更广泛的讨论 这里
There's also an excellent broader discussion here
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