c++ - 为什么在设置对象等于另一个函数的返回值时调用移动构造函数/赋值

Question

#include <iostream>

class Box{
public:
int x;
Box(){
    x=0;
    std::cout << "Constructor" << std::endl;
}
Box(const Box& other){
    x = other.x;
    std::cout << "Copy constructor" << std::endl;
}
Box(Box&& other){
    x = other.x;
    other.x = 0;
    std::cout << "Move constructor" << std::endl;
}

Box& operator=(Box&& other) {
    x = other.x;
    other.x = 0;
    std::cout << "Move assignment" << std::endl;
    return *this;
}

Box& operator=(const Box &other){
    x = other.x;
    std::cout << "Copy assignment" << std::endl;
}

~Box(){
    std::cout << "Destructor" << std::endl;
    x=0;
}
};
Box send(Box b){
std::cout << "Sending" << std::endl;
return b;
}
int main(){
Box b1, b2;
b1 = send(b2);
return 0;
}   

在输出中:

Constructor
Constructor
Copy Constructor
Sending
Move Constructor
Move Assignment
Destructor
Destructor
Destructor
Destructor

我不太清楚为什么在执行 b1 = send(b2) 时使用移动构造函数然后赋值。

Answer 1

Why is the move constructor/assignment called when setting an object equal to the return value of another function

您的send 函数返回b，它是一个参数，而不是局部变量:

Box send(Box b) {
    std::cout << "Sending" << std::endl;
    return b; // b is a parameter
}

因此没有发生 RVO 优化，因为 returning a parameter disables RVO .同时，还有另一个可用的优化选项，隐式 std::move:

Box send(Box b) {
    std::cout << "Sending" << std::endl;
    return std::move(b); // implicit std::move is applied.
}

所以它是使用你的移动构造函数构建的:

Box(Box&& other);

如果您改为创建并返回局部变量 c，您将看不到移动构造:

Box send(Box b) {
    Box c;
    std::cout << "Sending" << std::endl;
    return c; // Returns c, not b -> No move construction thanks to RVO
}

因为 RVO 随后跳入。