如果继承类型受到保护,我可以制作基类的指针以指向派生对象吗?

can I make a pointer of base class to point to a derived object if inheritance type is protected?

本文关键字:指针 派生 对象 基类 类型 继承 保护 我可以 如果      更新时间:2023-10-16

如果我有一个从基类继承的派生类(受保护的或私有的继承),我可以制作基类的指针指向派生对象吗?这是我的 C++ 代码:

#include<iostream.h>
#include<conio.h>
class Geoshape
{
protected:
    int dim1;
    int dim2;
public:
    Geoshape()
        {dim1=0;dim2=0;}
    Geoshape(int x, int y)
        {dim1=x ; dim2=y;}
    void setDim1(int x)
        {dim1=x;}
    int getDim1()
        {return dim1;}
    void setDim2(int y)
        {dim2=y;}
    int getDim2()
        {return dim2;}
    int calculateArea()
        {return dim1*dim2;}
};
class Circle:protected Geoshape
{
public:
    Circle()
        {}
    Circle(int r):Geoshape(r,r)
        {dim1=r;dim2=r;}
    void setR(int r)
        {dim1=dim2=r;}
    int getR()
        {return dim1;}
    float calculateArea()
        {return 22.0/7*dim1*dim2;}
};
class Triangle:public Geoshape
{
public:
    Triangle()
        {}
    Triangle(int x, int y):Geoshape(x,y)
        {}
    void setH(int h)
        {dim2=h;}
    int getH()
        {return dim2;}
    void setB(int b)
        {dim1=b;}
    int getB()
        {return dim1;}
    float calculateArea()
        {return .5*dim1*dim2;}
};
class Rectangle:public Geoshape
{
public:
    Rectangle()
        {}
    Rectangle(int x, int y):Geoshape(x,y)
        {}
    void setL(int l)
        {dim1=l;}
    int getL()
        {return dim1;}
    void setH(int h)
        {dim2=h;}
    int getH()
        {return dim2;}
};
class Square:protected Rectangle
{
public:
    Square()
        {}
    Square(int l):Rectangle(l,l)
        {dim1=l;dim2=l;}
    void setL(int l)
        {dim1=dim2=l;}
    int getL()
        {return dim1;}
    float calculateArea()
        {return dim1*dim1;}
};
void main()
{
clrscr();
cout<<"enter circle raduis: ";
int raduis;
cin>>raduis;
Circle c1(raduis);
cout<< "this is area of Circle: "<<c1.calculateArea();
getch();
cout<<"nnenter base of triangle: ";
int base;
cin>>base;
cout<<"enter height of triangle: ";
int height;
cin>>height;
Triangle t1(base,height);
cout<< "this is area of Triangle: "<<t1.calculateArea();
getch();
cout<<"nnenter length of rectangle: ";
int length;
cin>>length;
cout<<"enter height of rectangle: ";
int height1;
cin>>height1;
Rectangle r1(length,height1);
cout<< "this is area of Rectangle: "<<r1.calculateArea();
getch();
cout<<"nnenter length of square: ";
int len;
cin>>len;
Square s1(len);
cout<< "this is area of Square: "<<s1.calculateArea();
Geoshape *p1;Geoshape *p2;Geoshape *p3;Geoshape *p4;
p2=&t1;
p3=&r1;
getch();
}

我想在主要中添加这两行:

p1=&c1;
p4=&s1;但这会产生错误!

您只能从类内部(从成员函数和静态成员函数)、派生类(仅在公共或受保护继承的情况下)或从友元函数执行此操作(强制转换为"基"类型)。要不受限制地做到这一点,您可以:

  • 使用公共继承,
  • 提供用户指定的转换运算符,
  • 提供一个函数,该函数将返回这样的指针(本质上与上面相同,但不是以"运算符 T()"的形式,而是"T* getBase()"或类似的 sth 形式) - 成员函数,独立友元函数的静态成员函数。