#include <iostream>
using namespace std;

class a
{
public:
a();
void display();
};

a::a()
{
cout<<"a.............."<<endl;
}

void a::display()
{
cout<<"a::display"<<endl;
}

class b : public a
{
public:
b();
void display();
};

b::b()
{
cout<<"b.............."<<endl;
}
void b::display()
{
cout<<"b::display"<<endl;
}

class c : public a
{
public:
c();
void display();
};

c::c()
{
cout<<"c.............."<<endl;
}
void c::display()
{
cout<<"c::display"<<endl;
}
int main(void)
{
b* objb = new b;
c* objc = (c*)objb;
objc->display();
return 0;
}

执行结果:
a..............
b..............
c::display

我想知道在转换的时候它是如何转换的,原理是什么,在内存中是如何来分配的?

如果将程序改成这样,将a中的display改在虚函数,则这样:
#include <iostream>
using namespace std;

class a
{
public:
a();
virtual void display();
};

a::a()
{
cout<<"a.............."<<endl;
}

void a::display()
{
cout<<"a::display"<<endl;
}

class b : public a
{
public:
b();
void display();
};

b::b()
{
cout<<"b.............."<<endl;
}
void b::display()
{
cout<<"b::display"<<endl;
}

class c : public a
{
public:
c();
void display();
};

c::c()
{
cout<<"c.............."<<endl;
}
void c::display()
{
cout<<"c::display"<<endl;
}
int main(void)
{
b* objb = new b;
c* objc = (c*)objb;
objc->display();
return 0;
}

执行结果:
a..............
b..............
b::display

我也这个时候肯定与晚绑定,多态有关,但还是不太明白,请哪位高手指点一下.谢谢!

首先谢谢上面给我答复的, thank you very much!

我想这不是偶然,请看下面情况:

#include <iostream>
using namespace std;

class a
{
public:
void display();
};
void a::display()
{
cout<<"a::display"<<endl;
}

class b : public a
{
public:
void display();
};
void b::display()
{
cout<<"b::display"<<endl;
}

class c : public a
{
public:
void display();
void show();
private:
int i;
};
void c::display()
{
i = 10;
cout<<"c::display"<<endl;

}
void c::show()
{
cout<<"c::show()"<<endl;
cout<<i<<endl;
}

int main(void)
{

b* objb = new b;
c* objc = (c*)objb;

objc->display();
objc->show();

return 0;
}

我在 c 类中增加了一个函数与一个私有的成员变量,而在进行 c* objc = (c*)objb 后,这个私有变量也就变成可用的了.

我觉得这个问题值得深究哟!呵呵