比较和输出运算符重载的 C++ 继承

Question

我有一些带有几何图形的 C++ 类，我需要重载比较运算符（例如 >、<、= 等）和输出运算符 <<当我需要打印或比较这些类的任何实例时，我会参考它们区域的值（我使用getArea() 得到）。

我试图将这些运算符重载放在父类 GeometricShape 中，但我遇到了各种编译器错误，这对我来说没有多大意义。

如果我将运算符重载放在每个子类中，它可以正常工作，但它非常丑陋且不理想（大量代码重复）。

如何删除此代码重复并将运算符重载移动到父类？

using namespace std;

#include <iostream>
#include <stdlib.h>
#include <math.h>

class GeometricShape
{
    public:

    double getArea();
};

class Poligon : public GeometricShape
{
    protected:

    double base;
    double height;

    public:

    Poligon(double base, double height) : base(base), height(height) {}
    Poligon(float base, float height) : Poligon(double(base), double(height)) {}
    Poligon(int base, int height) : Poligon(double(base), double(height)) {}

    double getBase()
    {
        return this->base;
    }

    double getHeight()
    {
        return this->height;
    }

    void setBase(double b)
    {
        this->base = b;
    }

    void setHeight(double a)
    {
        this->height = a;
    }
};


class Rectangle : public Poligon
{
    public:

    Rectangle(double base, double height) : Poligon(base, height) {}
    Rectangle(float base, float height) : Poligon(base, height) {}
    Rectangle(int base, int height) : Poligon(base, height) {}

    double getArea()
    {
        return base * height;
    }

    // HOW TO MOVE THIS CODE INTO GeometricShape and reuse it???
    friend ostream& operator<<(ostream& out, Rectangle &obj)
    {
        out << obj.getArea();
        return out;
    }
    friend bool operator==(Rectangle &obj1, Rectangle &obj2)
    {
        return obj1.getArea() == obj2.getArea();
    }
    friend bool operator>(Rectangle &obj1, Rectangle &obj2)
    {
        return obj1.getArea() > obj2.getArea();
    }
    friend bool operator<(Rectangle &obj1, Rectangle &obj2)
    {
        return obj1.getArea() < obj2.getArea();
    }
    friend bool operator <= (Rectangle &obj1, Rectangle &obj2)
    {
        return obj1.getArea() <= obj2.getArea();
    }
    friend bool operator >= (Rectangle &obj1, Rectangle &obj2)
    {
        return obj1.getArea() >= obj2.getArea();
    }
};

class Triangle : public Poligon
{
    public:

    Triangle(double base, double height) : Poligon(base, height) {}
    Triangle(float base, float height) : Poligon(base, height) {}
    Triangle(int base, int height) : Poligon(base, height) {}

    double getArea()
    {
        return (base * height)/2;
    }
    
    // HOW TO MOVE THIS CODE INTO GeometricShape and reuse it???
    friend ostream& operator<<(ostream& out, Triangle &obj)
    {
        out << obj.getArea();
        return out;
    }
    friend bool operator==(Triangle &obj1, Triangle &obj2)
    {
        return obj1.getArea() == obj2.getArea();
    }
    friend bool operator>(Triangle &obj1, Triangle &obj2)
    {
        return obj1.getArea() > obj2.getArea();
    }
    friend bool operator<(Triangle &obj1, Triangle &obj2)
    {
        return obj1.getArea() < obj2.getArea();
    }
    friend bool operator <= (Triangle &obj1, Triangle &obj2)
    {
        return obj1.getArea() <= obj2.getArea();
    }
    friend bool operator >= (Triangle &obj1, Triangle &obj2)
    {
        return obj1.getArea() >= obj2.getArea();
    }
};

class Circle : public GeometricShape
{
    double radius;

    public:

    Circle(double radius) : radius(radius) {}

    double getArea()
    {
        return (radius * radius) * M_PI;
    }

    double getRadius()
    {
        return this->radius;
    }

    void setRadius(double r)
    {
        this->radius = r;
    }

    // HOW TO MOVE THIS CODE INTO GeometricShape and reuse it???
    friend ostream& operator<<(ostream& out, Circle &obj)
    {
        out << obj.getArea();
        return out;
    }
    friend bool operator==(Circle &obj1, Circle &obj2)
    {
        return obj1.getArea() == obj2.getArea();
    }
    friend bool operator>(Circle &obj1, Circle &obj2)
    {
        return obj1.getArea() > obj2.getArea();
    }
    friend bool operator<(Circle &obj1, Circle &obj2)
    {
        return obj1.getArea() < obj2.getArea();
    }
    friend bool operator <= (Circle &obj1, Circle &obj2)
    {
        return obj1.getArea() <= obj2.getArea();
    }
    friend bool operator >= (Circle &obj1, Circle &obj2)
    {
        return obj1.getArea() >= obj2.getArea();
    }
};

Answer 1

这段代码有很多问题需要修复。

外部独立的operator< 应该不会出现编译错误：

bool operator<(GeometricShape& s1, GeometricShape& s2) {
    return s1.getArea() < s2.getArea();
}

1. 您应该将 operator
2. getArea 应在 GeometricShape 中标记为虚拟。所有子类也应该用override 标记方法。否则，将调用 GeometricShape 中的 getArea。
3. 在几何形状中，您应该将getArea() 函数设为纯虚函数。
4. 为确保派生类的对象调用了正确的析构函数，应将 GeometricShape 中的析构函数标记为虚拟。

这将使您的方法签名看起来像：

对于几何形状：

virtual double getArea() const = 0;

对于派生类：

double getArea() const override;

编辑：cmets 中有很多关于如何改进这段代码的建议。我已经尽力在这里总结了。