C++虚函数表解析【二】

接c++虚函数表解析【一】

如果一个类没有继承关系，且该类中定义了虚函数，那么这个类的实例只有一张虚函数表，如【一】所说明；对于一个单继承的类，即派生类的基类只有一个，如果它有虚函数，也只有一张虚函数表。但是，对于多重继承的类，它可能会有多张虚函数表。下面一一具体说明：

还是以代码为例，看下面代码：

class Base1{
public:
Base1(int num):num_1(num){}
virtual void foo1(){cout<<"Base::foo1"<<num_1<<endl;}
virtual void foo2(){cout<<"Base::foo2"<<num_1<<endl;}
virtual void foo3(){cout<<"Base::foo2"<<num_1<<endl;}
private:
int num_1;
};
class Base2{
public:
Base1(int num):num_2(num){}
virtual void foo1(){cout<<"Base::foo1"<<num_2<<endl;}
virtual void foo2(){cout<<"Base::foo2"<<num_2<<endl;}
virtual void foo3(){cout<<"Base::foo2"<<num_2<<endl;}
private:
int num_2;
};
class Base3{
public:
Base1(int num):num_3(num){}
virtual void foo1(){cout<<"Base::foo1"<<num_3<<endl;}
virtual void foo2(){cout<<"Base::foo2"<<num_3<<endl;}
virtual void foo3(){cout<<"Base::foo2"<<num_3<<endl;}
private:
int num_3;
};
class Derived1:public Base1{
public:
Derived(int num):Base1(num){}
virtual void faa1(){cout<<"Derived1::faa1"<<endl;}//未覆盖Base1中定义的虚函数
virtual void faa2(){cout<<"Derived1::faa2"<<endl;}

};
class Derived2:public Base1{
public:
Derived2(int num):Base1(num){}
virtual void foo2(){cout<<"Base::foo2"<<endl;} //只覆盖了Base1::foo2
virtual void fbb2(){cout<<"Derived2::fbb2"<<endl;}
virtual void fbb3(){cout<<"Derived2::fbb3"<<endl;}

};
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 fcc1(){cout<<"Derived3::fcc1"<<endl;}
virtual void fcc2(){cout<<"Derived3::fcc2"<<endl;}

};
class Derived4:public Base1,public Base2,public Base3{  //多重继承，有覆盖
public:
Derived3(int num_1,int num_2,int num_3):Base1(num_1),Base2(num_2),Base3(num_3){}
virtual void foo1(){cout<<"Derived3::foo1"<<endl;}//覆盖了Base1::fool,Base2::fool,Base::fool
virtual void fdd(){cout<<"Derived4::fdd"<<endl;}

};

上面代码说明了四种继承情况下的虚函数表：

1）一般继承（无虚函数覆盖）：Derived1类继承自Base1类，Derived1的虚函数表如图1所示。

Derived1类内没有任何覆盖基类Base1的函数，因此虚函数依次被添加到虚函数表中。



2）一般继承（有虚函数覆盖）:Deirved2类继承自Base1类，并对Base1类中的虚函数foo2()进行了覆盖。Derived2的虚函数表如图二所示。



Derived2覆盖了基类Base1的foo1()，因此其虚函数表中Derived2::foo2()替换了Base::foo2()。

未完待续。