一、dynamic_cast运算符
dynamic_cast运算符是最常用的RTTI的组件，它不能回答“指针指向的是那类对象”这样的问题，
但能够回答“是否可以安全地将对象的地址赋给特定类型的指针”这样的问题。

class A {...};
class B :public A {...};
class C :public B {...};

B *p = dynamoc_cast<B *>(q);
//上述语句描述了指针q的类型是否可以安全的被转换为B*？如果可以，运算符将返回对象q的地址，否则返回一个空指针。

二、demon

#include<iostream>
#include<cstdlib>
#include<ctime>

using std::cout;
class A
{
private:
	int hold;
public:
	A(int h = 0) : hold(h){}
	virtual void Speak() const { cout << "我是A类！\n"; }
	virtual int Value() const { return hold; }
};

class B : public A
{
public:
	B(int h = 0 ) : A(h) {}
	void Speak() const { cout << "我是B类\n" ;}
	virtual void Say() const
	{
		cout << "I hold the B value of " << Value() << "!\n";
	}
};

class C : public B
{
private:
	char ch;
public:
	C(int h = 0,char c ='A') : B(h),ch(c){}
	void Speak() const { cout << "我是C类\n"; }
	void Say() const
	{
		cout << "i hold the character " << ch << "and the integer " <<Value() << "!\n";
	}
};

A *GetOne();

int main()
{
	std::srand(std::time(0));
	A * pg;
	B * ps;
	for(int i = 0;i<5;i++)
	{
		pg  = GetOne();
		pg->Speak();
		if(ps = dynamic_cast<B *>(pg))
		{
			ps->Say();
			cout << "可以转换B类======" <<'\n';
		}else
			cout << "不可以转换" <<"\n";
		cout <<"================I is "<< i << "\n\n";
	}
	return 0;
}

A *GetOne()
{
	A *p;
	switch(std::rand()%3)
	{
		case 0:
			p = new A(std::rand()%100);
			cout << "p是A类\n";
			break;
		case 1:
			p = new B(std::rand()%100);
			cout << "p是B类\n";
			break;
		case 2:
			p = new C(std::rand()%100,'A'+std::rand()%26);
			cout << "p是C类\n";
			break;
	}
	return p;
}

运行结果：