基类到子类的强制转换问题，管理员（刚才那个帖子忘了给分，可以删掉，不好意思）

[fibers]

假如有如下关系：
class a{};
class b:public a
{
private: int x;
public: b(); int method1()const{return x;}
};
class c:public b
{
private: char ar[10];
public:c(); const char * method2(){return ar;}
};

b obj1;
b * pb = & obj1;
c * pc;
pc = dynamic_cast <c*> (pb);
pc-> method2(); // pc可以调用method2方法么？？？？？

这种基类到子类的向下转换是怎么执行的？？比如说子类对象里有基类对象没有的char 型私有成员以及method2（）方法。。在向下转换的时候。会自动生成这些成员？
还是说忽略这些成员?

[ckt]

你的转换很有问题，
1.类b至少要有一个virtual，才能实现多态。
2.dynamic_cast是用于将基类的指针（引用）转为子类的对象的指针（引用）

[SimonFu]

dynamic_cast是用于将子类的指针（引用）转为基类的对象的指针（引用）吧,这样才是安全的

[ryfdizuo]

dynamic_cast
dynamic_cast can be used only with pointers and references to objects. Its purpose is to ensure that the result of the type conversion is a valid complete object of the requested class.

Therefore, dynamic_cast is always successful when we cast a class to one of its base classes:

class CBase { };
class CDerived: public CBase { };

CBase b; CBase* pb;
CDerived d; CDerived* pd;

pb = dynamic_cast<CBase*>(&d);     // ok: derived-to-base
pd = dynamic_cast<CDerived*>(&b);  // wrong: base-to-derived
 


The second conversion in this piece of code would produce a compilation error since base-to-derived conversions are not allowed with dynamic_cast unless the base class is polymorphic.

When a class is polymorphic, dynamic_cast performs a special checking during runtime to ensure that the expression yields a valid complete object of the requested class:

// dynamic_cast
#include <iostream>
#include <exception>
using namespace std;

class CBase { virtual void dummy() {} };
class CDerived: public CBase { int a; };

int main () {
  try {
    CBase * pba = new CDerived;
    CBase * pbb = new CBase;
    CDerived * pd;

    pd = dynamic_cast<CDerived*>(pba);
    if (pd==0) cout << "Null pointer on first type-cast" << endl;

    pd = dynamic_cast<CDerived*>(pbb);
    if (pd==0) cout << "Null pointer on second type-cast" << endl;

  } catch (exception& e) {cout << "Exception: " << e.what();}
  return 0;
}

[SimonFu]

如果你强制转换了,存储空间是将基类当子类型读取,编译器认为数据存在

[goodmrning]

接分！！！！
帮楼主顶下

[ttkk_2007]

b       obj1;   
b       *       pb       =       &       obj1;   
c       *       pc;   
pc       =       dynamic_cast   <c*>   (pb);   
pc->   method2();       //       pc可以调用method2方法么？？？？？   

这种根本不能向下转换，pc肯定为NULL，b要有virtual函数

b *pb = new c;
c *pc = dynamic_cast<c*>(pb);