函数的覆盖和函数的隐藏

多看看书..写写代码....
#include"iostream.h"
class father
{
public:
    void demo1()
    {
        cout<<"this is father"<<endl;
    }
    virtual void demo2()
    {
        cout<<"this is virtual father"<<endl;
    }
};
class son:public father
{
public:
    void demo1()
    {
        cout<<"this is son"<<endl;
    }
     void demo2()
    {
        cout<<"this is virtual son"<<endl;
    }
};
int main()
{
    father a;
    son b;
    a.demo1();
    b.demo1();
    father &c=b;
    c.demo1();
    c.demo2();
    return 1;
}

不是....自己看看书..

http://hi.baidu.com/pcrazyc/blog/item/31d9113dcc3c80ea3d6d97b0.html

自己看

If u use key word "virtual" to specify a function in base class, u may override(same name, same return & same arguments) it in derived class  if u want use the polymorphism in cpp.

If u define a function in derived class and use just a same name(return or argument may differ) as in base class, whatever u sepcify "virtual" to the function in base class or not, a derived class object can only call it's own version function, then the same name function in base class is hidden.