避免使用“返回"复制对象陈述

问题描述

我在 C++ 中有一个非常基本的问题.返回对象时如何避免复制?

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

这是一个例子:

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;
}

据我了解 C++ 的工作原理，此函数将创建 2 个向量:本地向量 (x) 和将返回的 x 副本.有没有办法避免复制?(而且我不想返回指向对象的指针，而是返回对象本身)

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)

使用移动语义"(在评论中说明)的函数的语法是什么?

What would be the syntax of that function using "move semantics" (which was stated in the comments)?

推荐答案

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

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

在 C++11 中有移动构造函数和赋值，它们也很便宜.您可以在此处阅读教程:http://thbecker.net/articles/rvalue_references/section_01.html

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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