如果继承类型受保护，我可以使基类的指针指向派生对象吗？

Question

，如果我有一个基础类（受保护的或私有继承）派生类继承我可以让基类的指针指向派生类对象？ 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<<"\n\nenter 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<<"\n\nenter 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<<"\n\nenter 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(); 
} 

我想补充这两条线在主：

P1 = &c1;
P4 = &s1; 但是这给错误！

Answer 1

你可以做到这一点（投给“基地”型）只能从类（从成员函数和静态成员函数）内，从派生类（仅适用于公共或保护继承的情况下）或者从朋友的功能。若要做到这一点不受限制的能力，您可以：

• 使用公有继承，
• 提供用户指定的转换操作符，
• 提供将返回这样的指针（essentialy与上述相同的功能，但不是“运算符T（）”，而是“T * getBase（）”或者这样的形式） - 成员函数，独立好友函数的静态成员函数。