C ++：避免​​复制与“返回”语句 [英] C++ : Avoiding copy with the "return" statement

问题描述

我在C ++中有一个非常基本的问题。
如何在返回对象时避免复制？

下面是一个示例：

  std :: vector< unsign int> test（const unsigned int n）
 {
 std :: vector< unsigned int> X; 
 for（unsigned int i = 0; i  x.push_back（i）; 
} 
 return x; 
} 
  

根据我了解C ++的工作原理，此函数将创建2个向量：local one（x），以及将返回的x的副本。有没有办法避免复制？ （我不想返回指向对象的指针，而是对象本身）

非常感谢。

编辑：根据第一个答案的额外问题：使用move semantics的那个函数的语法是什么？

解决方案

此程序可以利用命名返回值优化（NRVO）。请参阅： http://en.wikipedia.org/wiki/Copy_elision

在C ++ 11中有move constructors和assignment也很便宜。您可以在这里阅读教程： http://thbecker.net/articles/rvalue_references/section_01.html p>

I have a very basic question in C++. How to avoid copy when returning an object ?

Here is an example :

std::vector<unsigned int> test(const unsigned int n)
{
    std::vector<unsigned int> x;
    for (unsigned int i = 0; i < n; ++i) {
        x.push_back(i);
    }
    return x;
}

As I understand how C++ works, this function will create 2 vectors : the local one (x), and the copy of x which will be returned. Is there a way to avoid the copy ? (and I don't want to return a pointer to an object, but the object itself)

Thank you very much.

EDIT : extra question according to the first answers : what would be the syntax of that function using "move semantics" ?

解决方案

This program can take advantage of named return value optimization (NRVO). See here: http://en.wikipedia.org/wiki/Copy_elision

In C++11 there are move constructors and assignment which are also cheap. You can read a tutorial here: http://thbecker.net/articles/rvalue_references/section_01.html

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