运算符重载

Question

我对有关运算符重载的话题感到困惑。见以下代码：

#include <iostream>;

class Vectors {
public:
    int x, y;
    Vectors() {};
    Vectors(int a,int b) {
        x = a, y = b;
    }
    Vectors operator+(Vectors aso) {
        Vectors brandNew;
        std::cout << "ASO X " << aso.x << std::endl;
        std::cout << "ASO Y " << aso.y << std::endl;
        brandNew.x = brandNew.x + aso.x;
        brandNew.y = brandNew.y + aso.y;
        return (brandNew);
    };
};
int main() {
    Vectors v1(2,3);
    Vectors v2(4,5);
    Vectors v3;
    v3 = v1 + v2;

    std::cout << "Vector V3 X : " << v3.x << std::endl;
    std::cout << "VECTOR V3 Y : " << v3.y << std::endl;
}

当我打印 aso.x 时，y 给了我 4 和 5。我想做的是同时添加 v1 和 v2；表示 v1 和 v2 的 x，以及 v1 和 v2 的 y。然后，将其传递给 Vectors 对象和 return 该对象。

根据我的上述情况，我该如何实现？

Answer 1

您的代码创建了新的 Vectors 对象 brandNew，然后将其初始化为 0 的值添加到您传入的 Vectors 对象中的值，而不是当前对象中的值。这就是为什么将 v1 添加到 v2 结果与 v2.

中的值相同的原因

替换

brandNew.x = brandNew.x + aso.x;
brandNew.y = brandNew.y + aso.y;

和

brandNew.x = x + aso.x;
brandNew.y = y + aso.y;

Answer 2

你的意思可能是

Vectors operator+(const Vectors& aso) {
        Vectors brandNew;
        std::cout << "ASO X " << aso.x << std::endl;
        std::cout << "ASO Y " << aso.y << std::endl;
        brandNew.x = x + aso.x;
        brandNew.y = y + aso.y;
        return (brandNew);
    };

或

Vectors operator+(const Vectors& aso) {
    Vectors brandNew(x + aso.x, y + aso.y);
    std::cout << "ASO X " << aso.x << std::endl;
    std::cout << "ASO Y " << aso.y << std::endl;
    return (brandNew);
};

根据您的评论

But about if there were three.

像这样链接运算符

int main() {
    Vectors v1(2,3);
    Vectors v2(4,5);
    Vectors v3(7,11);
    Vectors v4;
    v4 = v1 + v2 + v3;

    std::cout << "Vector V4 X : " << v4.x << std::endl;
    std::cout << "Vector V4 X : " << v4.y << std::endl;
}

见Live Demo

Answer 3

您尚未初始化 brandNew 的 x 和 y 成员。您会得到随机结果。

Vectors operator+(Vectors aso) {
    Vectors brandNew;
    std::cout << "ASO X " << aso.x << std::endl;
    std::cout << "ASO Y " << aso.y << std::endl;
    brandNew.x = x + aso.x;
    brandNew.y = y + aso.y;
    return (brandNew);
};

Answer 4

你也可以使用外部运算符+ :

class Vectors {
public:
    int x, y;
    Vectors() {};
    Vectors(int a,int b) {
        x = a, y = b;
    }

    friend Vectors operator+(const Vectors& v1, const Vectors& v2);
};

Vectors operator+(const Vectors& v1, const Vectors& v2) {
        Vectors brandNew;
        brandNew.x = v1.x + v2.x;
        brandNew.y = v1.y + v2.y;
        return (brandNew);
    };

int main() {
    Vectors v1(2,3);
    Vectors v2(4,5);
    Vectors v3;
    v3 = v1 + v2;

    std::cout << "Vector V3 X : " << v3.x << std::endl;
    std::cout << "VECTOR V3 Y : " << v3.y << std::endl;
}

Answer 5

+运算符的正确定义如下：

  Vectors &operator+(const Vectors& aso)
  {
    std::cout << "ASO X " << aso.x << std::endl;
    std::cout << "ASO Y " << aso.y << std::endl;
    x = x + aso.x;
    y = y + aso.y;
    return (*this);
  }

以上代码不需要临时变量，也不需要像通过引用传递的那样对参数进行不必要的复制。

运算符重载

Operator Overloading

c++

operator-overloading

output