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（），也没有调用move构造函数。有人可以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point.hpp

中的代码

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;}
};

该计划的输出如下：

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

这里调用了带右值引用的函数，但是没有打印移动构造函数语句，它实际上被调用或者正在调用其他一些成员函数。请在这里看到问题。

Answer 1

除非您从d2point构建d2point&&的新实例，否则不会调用移动构造函数。在add_point功能中，您将 rvalue 引用 带到d2point：这意味着您没有构建任何新的例如，但仅仅是指现有的：

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

请记住，std::move只是对右值参考的强制转换。

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

在上面一行中，您将x和y投射到d2point&& - 它们将被绑定到p1和p2。还没有施工。最后，您将add_point来电的结果存储到z。由于返回值优化，您很可能不会在那里看到调用的移动构造函数。

将add_point更改为

d2point add_point(d2point p1, d2point p2)

肯定会调用d2point::d2point(d2point&&)。

移动构造函数未被调用

1 个答案: