Question

与this指针有关，我希望有人认为以下两个代码段是否实现相同的功能。

Position Position::operator = (Position pos)
{
     this->x = pos.x;
     this->y = pos.y;
     return *this;
}

Position & Position::operator = (Position pos)
{
     this->x = pos.x;
     this->y = pos.y;
}

我知道第一个代码段更常用。但是，我想确认第二个代码段是否具有与我使用this传递对&对象的引用相同的功能。

Answer 1

在选择所需选项之前，请注意@RemyLebeau提到的这篇文章：What are the basic rules and idioms for operator overloading?

两者之间差异的示例：

class Position1 {
public:
    int x, y;

    Position1 operator=(Position1 pos)
    {
        this->x = pos.x;
        this->y = pos.y;
        return *this;
    }
};

class Position2 {
public:
    int x, y;

    Position2& operator=(Position2 pos)
    {
        this->x = pos.x;
        this->y = pos.y;
        return *this;
    }
};

int main() {
    Position1 p11, p12;
    Position2 p21, p22;

    //Position1 *p13 = &(p12 = p11); // Compilation error: Taking address of temporary object..
    p21.y = 20;
    p21.x = 10;
    Position2 *p23 = &(p22 = p21); // Working - the `temporary object` is a reference to non temporary object.
    p23->x = 99;
    p23->y = 84;
    cout << p21.x << " " << p21.y; // didn't change
    cout << p22.x << " " << p22.y; // changed
    cout << p23->x << " " << p23->y; // changed

    //===================================================

    //(p12 = p11).y = 3; // Compiling error: Expression is not assignable
    (p22 = p21).y = 3; // works
    cout << p21.x << " " << p21.y << endl; // didn't change
    cout << p22.x << " " << p22.y << endl; // changed

    return 0;
}

此指针-按值返回vs按引用返回

1 个答案: