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？

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。

解决方案

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）

如果你自己的编译器文档没有说它是否支持，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。

还有一个非常广泛的讨论这里 p>

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

Example code:

std::vector<A> MyFunc();

...

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

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?

解决方案

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.

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.

There's also an excellent broader discussion here

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