dynamic_cast

//转载 http://blog.youkuaiyun.com/boluo1982107/article/details/3239793 的一个例子  修改运行了下

#include <iostream>
using namespace std;

class base1
{
public:
     void same_name_fun()               ----------------------1，//virtual void same_name_fun() 
    {
        cout<<"in base1 same_name_fun"<<endl;
    }
    virtual ~base1(){}
};
class base2
{
public:
    void ba2_fun()
    {
        cout<<"base2_fun"<<endl;
    }
};
class derived : public base1,base2
{
public:

    void same_name_fun()         ------------------------2，//virtual void same_name_fun()       

    {
        cout<<"in derived same_name_fun"<<endl;
    }
    void de_fun()
    {
        cout<<"de_fun"<<endl;
    }
};
int main()
{
  derived  * obj;     // note: **1没有 new
        obj->same_name_fun();         //      ** 均加上virtual 关键字  ** Segmentation fault  去掉virtual 关键字 ok

       (dynamic_cast<base1 *>(obj))->same_name_fun();

        return 0;

}

结果：

in derived same_name_fun
in base1 same_name_fun

1,2, 均加上virtual 关键字  **2 Segmentation fault  去掉virtual 关键字 ok   .

#include <iostream>
using namespace std;

class base1
{
public:
    void same_name_fun()
    {
        cout<<"in base1 same_name_fun"<<endl;
    }
    virtual ~base1(){}
};
class base2
{
public:
    void ba2_fun()
    {
        cout<<"base2_fun"<<endl;
    }
};
class derived : public base1,base2
{
public:
    void same_name_fun()
    {
        cout<<"in derived same_name_fun"<<endl;
    }
    void de_fun()
    {
        cout<<"de_fun"<<endl;
    }
};
int main()
{
    derived obj;
    obj.same_name_fun();
    (static_cast<base1>(obj)).same_name_fun();   **  //(dynamic_cast<base1>(obj)).same_name_fun(); error: cannot dynamic_cast ‘obj’ (of type ‘class derived’) to type ‘class base1’ (target is not pointer or reference)
    return 0;
}

运行结果：

in derived same_name_fun
in base1 same_name_fun

** 如果换成dynamic_cast会有error

正常的程序
#include <iostream>
using namespace std;

class base1
{
public:
    void same_name_fun()                                   //---------------------------1 加上virtual关键字  virtual  void same_name_fun()
    {
        cout<<"in base1 same_name_fun"<<endl;
    }
    virtual ~base1(){}
};
class base2
{
public:
    void ba2_fun()
    {
        cout<<"base2_fun"<<endl;
    }
};
class derived : public base1,base2
{
public:
     void same_name_fun()                         // ---------------------------2 加virtual关键字  virtual  void same_name_fun()
    {
        cout<<"in derived same_name_fun"<<endl;
    }
    void de_fun()
    {
        cout<<"de_fun"<<endl;
    }
};
int main()
{
  derived  * obj = new derived();  //note **1 有new关键字
        obj->same_name_fun();
        (dynamic_cast<base1 *>(obj))->same_name_fun();
        return 0;
}

运行结果

in derived same_name_fun
in base1 same_name_fun

1 2都加virtual关键字运行结果

in derived same_name_fun
in derived same_name_fun