c++ - 需要帮助使用类获取矩形和圆形的面积

Question

它们需要是类并且获取区域需要是形状的一部分，方法 getArea 需要在形状中，区域需要被保护，并且在它们各自子类的形状、宽度、高度和半径部分(C++)

#include <iostream>
#include <fstream>
#include <string>
#include <cstdlib>
#include <math.h>
using namespace std;

class shape {
    protected:
        double area;
    public:
        double getArea(){return area;};
};

class rectangle:shape {
    private:
        double width;
        double height;

    public:
        rectangle(){
            width=3;
            height=4;
        }
        double getHeight(){return height;};
        double getWidth(){return width;};

        void setHeight(double h){height=h;};
        void setWidth(double w){width=w;};
        void setArea(double width, double height){area=height*width;};
};

class circle:shape {
    private:
        double radius;

    public:
        circle(){
            radius=1;
        }

        double getRadius(){return radius;};
        void setRadius(double r){radius=r;};
        void setArea(double width, double height){area=M_PI*(pow(radius,2));};
};

int main () {
    rectangle miRectangulo;
    circle miCirculo;
    cout<<"Area of the rectangle is "<<miRectangulo.getArea()<<endl;
    cout<<"Area of the circle is "<<miCirculo.getArea();

    return 0;
}

他们确实需要遵循这些特定条件，现在我收到“double::shape getArea() is inaccesible”错误

Answer 1

默认情况下，继承是私有(private)的，这意味着 getArea 将不可访问。改为作为公共(public)基础继承。

class circle: public shape
{
/* snip */
};