
大家知道虚函数是通过一张虚函数表来实现的。在这个表中,主要是一个类的虚函数的地址表,这张表解决了继承、覆盖的问题,其内容真是反应实际的函数。这样,在有虚函数的类的实例中,这个表分配在了这个实例的内存中,所以,当用父类的指针来操作一个子类的时候,这张虚函数表就显得尤为重要了。它就像一个地图一样,指明了实际所应该调用的函数。
C++的标准规则中说到,编译器必须保证虚函数表的指针存在于对象实例中最前面的位置(这样是为了保证正确取到虚函数的偏移量)。这意味着通过对象实例的地址得到这张虚函数表,然后可以遍历其中的函数指针,并调用相应的函数。
#include <iostream>
using namespace std;
class Base
{
public:
virtual void fun1(){cout<<"Base::fun1\n";}
virtual void fun2(){cout<<"Base::fun2\n";}
virtual void fun3(){cout<<"Base::fun3\n";}
private:
int num1;
int num2;
}; typedef void (*Fun)(void); int main()
{
Base b;
Fun pFun;
//通过指针分别调用了对象b的3个虚函数。
pFun = (Fun)* ( (int*)*(int*)(&b)+ );
pFun();
pFun = (Fun)* ( (int*)*(int*)(&b)+ );
pFun();
pFun = (Fun)* ( (int*)*(int*)(&b)+ );
pFun();
return ;
}
/*
程序执行结果如下:
Base::fun1
Base::fun2
Base::fun3
Press <RETURN> to close this window...
*/
程序中的Base对象b内存结构如下:
一个类会有多少张虚函数表呢?
对于一个单继承的类,如果它有虚函数,则只有一张虚函数表。对于多重继承的类,它可能有多张虚函数的表。
#include <iostream>
using namespace std;
class Base1
{
public:
Base1(int num):num_1(num){}
virtual void fun1(){cout<<"Base1::fun1 "<<num_1<<endl;}
virtual void fun2(){cout<<"Base1::fun2 "<<num_1<<endl;}
virtual void fun3(){cout<<"Base1::fun3 "<<num_1<<endl;}
private:
int num_1;
};
class Base2
{
public:
Base2(int num):num_2(num){}
virtual void fun1(){cout<<"Base2::fun1 "<<num_2<<endl;}
virtual void fun2(){cout<<"Base2::fun2 "<<num_2<<endl;}
virtual void fun3(){cout<<"Base2::fun3 "<<num_2<<endl;}
private:
int num_2;
};
class Base3
{
public:
Base3(int num):num_3(num){}
virtual void fun1(){cout<<"Base3::fun1 "<<num_3<<endl;}
virtual void fun2(){cout<<"Base3::fun2 "<<num_3<<endl;}
virtual void fun3(){cout<<"Base3::fun3 "<<num_3<<endl;}
private:
int num_3;
};
class Derived1:public Base1
{
public:
Derived1(int num):Base1(num){}
virtual void fDer1_1(){cout<<"Derived1::fDer1_1\n";}//无覆盖
virtual void fDer1_2(){cout<<"Derived1::fDer1_2\n";}
};
class Derived2:public Base1
{
public:
Derived2(int num):Base1(num){}
virtual void fun2(){cout<<"Derived2::fun2 "<<endl;}//只覆盖了Base1::fun2
virtual void fDer2_1(){cout<<"Derived2::fDer2_1\n";}
virtual void fDer2_2(){cout<<"Derived2::fDer2_2\n";}
};
class Derived3:public Base1,public Base2,public Base3//多重继承,无覆盖
{
public:
Derived3(int num_1,int num_2,int num_3):Base1(num_1),Base2(num_2),Base3(num_3){}
virtual void fDer3_1(){cout<<"Derived3::fDer3_1\n";}
virtual void fDer3_2(){cout<<"Derived3::fDer3_2\n";} };
class Derived4:public Base1,public Base2,public Base3//多重继承,有覆盖
{
public:
Derived4(int num_1,int num_2,int num_3):Base1(num_1),Base2(num_2),Base3(num_3){}
virtual void fun1(){cout<<"Derived4::fun1\n";}//覆盖了所有基类的fun1函数
virtual void fDer4_1(){cout<<"Derived4::fDer4_1\n";} };
int main()
{
Base1 *pBase1 = NULL;
Base2 *pBase2 = NULL;
Base3 *pBase3 = NULL; cout<<"----- Generally inherited from Base1, no cover------\n";
Derived1 d1();
pBase1 = &d1;
pBase1->fun1(); cout<<"----- Generally inherited from Base1, covering fun2---\n";
Derived2 d2();
pBase1 = &d2;
pBase1->fun2(); cout<<"----- Multiple inheritance, no cover-----------------\n";
Derived3 d3(,,);
pBase1 = &d3;
pBase2 = &d3;
pBase3 = &d3;
pBase1->fun1();
pBase2->fun1();
pBase3->fun1(); cout<<"----- Multiple inheritance, covering fun1-------------\n";
Derived4 d4(,,);
pBase1 = &d4;
pBase2 = &d4;
pBase3 = &d4;
pBase1->fun1();
pBase2->fun1();
pBase3->fun1();
return ;
} /*
* 程序运行结果如下:
----- Generally inherited from Base1, no cover------
Base1::fun1 1
----- Generally inherited from Base1, covering fun2---
Derived2::fun2
----- Multiple inheritance, no cover-----------------
Base1::fun1 1
Base2::fun1 2
Base3::fun1 3
----- Multiple inheritance, covering fun1-------------
Derived4::fun1
Derived4::fun1
Derived4::fun1
Press <RETURN> to close this window...
*/
一般继承(无虚函数覆盖)
Derived1类继承自Base1类,没有任何覆盖基类的函数,因此Dervied1的两个虚拟函数被依次添加到了虚函数表的尾部。Derived1的虚函数表如下图:
一般继承(有虚函数的覆盖)
Derived2继承自Base1类,并对基类中的fun2()进行了覆盖。所以虚函数表中的Derived2::fun2代替了Base::fun2,用时派生类中新的虚函数添加到虚函数的表尾。Derived2的虚函数表如下图:
多重继承(无虚函数覆盖)
Derived3继承自Base1,Base2,Base3,其虚函数表如下:
Derived3的每个父类都有自己的虚表,所以Derived3也就有了3个虚表。这里父类虚表的顺序与声明继承父类的顺序一致。这样做就是为了解决不同的父类类型的指针指向同一个子类实例,而能够调用到实际的函数。例如:
Base2 *pBase2 = new Derived3();
pBase->fun2();
把Base2类型的指针指向Derived3实例,那么调用将是对应Base2虚表里的那些函数.
多重继承(有虚函数覆盖)
Derived4类继承自Base1,Base2,Base3并对3个基类的fun1函数进行了覆盖。其虚函数表如下:
可以看见基类中的fun1都被替换成了Derived4::fun1,这样,我们就可以把任意一个静态类型的父类指向子类,并调用子类的f()了。
Base1 *pBase1 = new Derived4();
pBase1->fun1();