为什么不调用移动构造函数？ [英] Why is the move constructor not called?

问题描述

根据我的理解，移动构造函数将在创建临时对象时被调用。这里的 getA（）函数返回一个临时对象，但我的程序不打印来自move构造函数的消息：

As per my understanding, the move constructor will be called when there is a temporary object created. Here the getA() function is returning a temporary object but my program is not printing the message from the move constructor:

#include <iostream>

using namespace std;

class A
{
    public:
    A()
    {
        cout<<"Hi from default\n";
    }

    A(A && obj)
    {
        cout<<"Hi from move\n";
    } 
};

A getA()
{
    A obj;
    cout<<"from getA\n";
    return obj;
}

int main()
{
    A b(getA());

   return 0;
}

推荐答案

输出实例 obj ，并将对象直接发送回调用者，而不进行概念值复制。

The compiler is allowed to optimise out the instance obj and send the object directly back to the caller without a conceptual value copy being taken.

称为命名返回值优化（NRVO）。这是一个比传统的返回值优化（RVO）更积极的优化，编译器可以调用它来避免匿名临时值的副本。

This is called named return value optimisation (NRVO). It's a more aggressive optimisation than classical return value optimisation (RVO) that a compiler can invoke to obviate the value copy of an anonymous temporary.

为了避免疑问，编译器可以这样做，即使有这样做的副作用（在您的情况下缺乏控制台输出）。

For the avoidance of doubt the compiler can do this even if there is a side-effect in doing so (in your case the lack of console output).

这篇关于为什么不调用移动构造函数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！