C++ RTTI

C++ runtime type identification。即运行时类型识别

 

//之所以还要比较类型信息 bool equal(const Base &)函数中
//参数Base转化后的类型可能丢失了一些信息
//比如Base参数是Derived的某个派生类对象，
//Derived对象调用equal函数只会比较Derived类的信息
//而忽略了Base参数含有的额外信息，导致误把两者认为是同一个对象

参考 c++ primer 写的一个小例子，实现的对象的equality operator即等价操作符。如果只使用虚函数实现，那么由于派生类中该比较的函数参数与父类参数相同，应该传递的是一个基类的引用，这样如果传递的是派生类对象，使用该成员函数会把派生类对象转成基类对象，会丢失基类中没有的成员。

 

#include<iostream>
using namespace std;

class Base{
	friend bool operator==(const Base &,const Base&);
public:
	// interface members for Base
	Base(int c):color(c){}
protected:
virtual bool equal(const Base &)const;
     int color;
};
class Derived: public Base {
	friend bool operator==(const Base&, const Base&);
public:
	// other interface members for Derived
	Derived(int n,int c):num(n),Base(c){}
private:
	bool equal(const Base&) const;
	// data and other implementation members of Derived
    int num;
};
bool operator==(const Base &lhs, const Base &rhs){
	return typeid(lhs)==typeid(rhs)&&lhs.equal(rhs);
}

bool Derived::equal(const Base &rhs) const
{
	if (const Derived *dp=dynamic_cast<const Derived *>(&rhs))
	{

		return num==dp->num&&color==dp->color;
	}
	else
		return false;
}

bool Base::equal(const Base &rhs) const
{
	// do whatever is required to compare to Base objects
	return color==rhs.color;

}

int main(){

	Base b(1);
	Base *a;
	Derived d(2,1);
	Derived c(2,1);
	a=&c;
	//print double 
	cout<<typeid(1.56).name()<<endl;
	//print class Base *
	cout<<typeid(a).name()<<endl;
	//print class Derived
	cout<<typeid(*a).name()<<endl;
	if(b==d){
		cout<<"object b is equal to d"<<endl;
	}
	else
		cout<<"object b is not equal to d"<<endl;
	//*a and d is equal
	if(*a==d){
		cout<<"the object a points to is equal to d"<<endl;
	}
	int i;
	cin>>i;
}