移动构造函数没有被调用

Question

我编写了以下代码来演示移动构造。但即使我使用了 std::move()，移动构造函数也没有被调用。谁能看看这里有什么问题。

#include "point.hpp"
using namespace std;
class point; 
d2point add_point(d2point&& p1, d2point&& p2) 
{ 
 cout << "inside add point function" << endl;
 d2point z;
 z.setX(p1.getX() + p2.getX());
 z.setY(p1.getY() + p2.getY());
 return z;
} 
int main()
{

 d2point x(2,3);
 d2point y(2,3);

 d2point z = add_point(move(x), move(y));
 z.print_point();
 x.print_point();

 return 0;
}

以下是来自point.hpp的代码

using namespace std;
class point
{
private:
int x;
public:
point(){cout << "point default constructor gets called" << endl;}
point(int x){ cout << "point param constructor gets called" << endl;
this->x = x;
}   

point(const point& p){   
cout << "point copy constructor gets called" << endl;
this->x = p.x;
}   

point(point&& other):x(std::move(other.x)) {
cout << "point move constructor gets called" << endl;
}   

int getX(){return x;} 
void setX(int x) { this->x = x; }

virtual void print_point() {}
virtual ~point(){cout << "point destructor gets called" << endl;}
};

class d2point: public point
{
private:
int y;

public:
d2point()
{   
cout << "d2point default constructor gets called" << endl;
}   

d2point(int x, int y):point(x) {
cout << "d2point parameterized constructor gets called" << endl;
this->y = y;
}   

d2point(const d2point& rhs): point(rhs) {
cout << "d2point copy constructor gets called" << endl;
this->y = rhs.y;
}

d2point(d2point&& rhs):point(std::move(rhs)) {
cout << "d2point move constructor gets called" << endl;
this->y = std::move(rhs.y);
rhs.y = 0;
}

int getY(){return y;}
void  setY(int y) { this->y = y; }

void print_point(){
cout << "(" << getX()  << "," << y << ")" << endl;
}

~d2point(){ cout << "d2point destructor gets called" << endl;}
};

程序的输出如下：

point param constructor gets called
d2point parameterized constructor gets called
point param constructor gets called
d2point parameterized constructor gets called
inside add point function
point default constructor gets called
d2point default constructor gets called
(4,6)
(2,3)
d2point destructor gets called
point destructor gets called
d2point destructor gets called
point destructor gets called
d2point destructor gets called
point destructor gets called

这里正在调用正在调用右值引用的函数，但没有打印移动构造函数语句，它确实被调用或正在调用其他一些成员函数。请在这里查看问题。

Answer 1

除非您从d2point&& 构造d2point 的新实例，否则不会调用移动构造函数。在您的 add_point 函数中，您将 rvalue references 带到 d2point：这意味着您没有构建任何新实例，而只是引用现有实例：

d2point add_point(d2point&& p1, d2point&& p2) 
//                       ^^            ^^

请记住，std::move 只是对右值引用的强制转换。

d2point z = add_point(move(x), move(y));

在上面的行中，您将x 和y 都转换为d2point&& - 它们被绑定到p1 和p2。还没有施工。最后，将add_point 调用的结果存储到z。由于返回值优化，您极有可能不会看到那里调用了移动构造函数。

---

将add_point 更改为

d2point add_point(d2point p1, d2point p2) 

肯定会调用d2point::d2point(d2point&&)。