继承、封装和多态是面向程序设计(OOP)的三大特点,而它们三者之中最具实际操作性的当属继承。通过继承可以实现简单功能的组合和定制,而多重继承更将这种能力发挥到更高的境界。不过事事都有弊端,如果使用多重继承不当很容易造成菱形继承问题(diamond problem)。Bjarne Stroustrup用下面这个例子描述菱形继承问题:
- class storable //this is the our base class inherited by transmitter and receiver classes
- {
- public:
- virtual void read();
- virtual void write();
- private:
- ....
- }
- class transmitter: public storable
- {
- public:
- void write();
- ...
- }
- class receiver: public storable
- {
- public:
- void read();
- ...
- }
- class radio: public transmitter, public receiver
- {
- public:
- void read();
- ....
- }
transmitter和receiver都从storable派生而来,而radio则从transmitter和receiver派生而来,是最终派生类。这样的继承关系很特殊,因为从storable到radio存在两条不同的路径,即radio将从两条不同的继承路径获得storable的成员。如果不采取措施应对菱形继承问题的话,编译程序时将收到许多二义性的错误提示。如在上述示例中,如果radio类的对象调用write()方法,编译器将无法确定应当调用transmitter中的write()方法或是receiver中的write()方法。解决的方法很简单:只需在transmiiter和receiver类的继承列表前添加virtual关键字即可,如下列代码所示:
- class storable
- {
- public:
- int istorable;
- virtual void read()
- {
- cout<<"read() in storable called.\n";
- }
- virtual void write()
- {
- cout<<"write() int storable called.\n";
- }
- };
- // class transmitter
- class transmitter : virtual public storable
- {
- public:
- int itransmitter;
- void read()
- {
- cout<<"read() in transmitter called.\n";
- }
- };
- // class receiver
- class receiver : virtual public storable
- {
- public:
- int ireiceiver;
- void write()
- {
- cout<<"write() in receiver called.\n";
- }
- };
- // class radio
- class radio : public transmitter, public receiver
- {
- public:
- int iradio;
- };
为了方便描述使用虚拟继承后radio类对象的内存布局,我在各类中都增加了一个整型的成员变量。
现在编写主函数来打印radio类对象的内存布局,代码如下所示:
- typedef void (*FUN)();
- // main function
- int main()
- {
- radio r;
- r.istorable = 1;
- r.itransmitter = 2;
- r.ireiceiver = 3;
- r.iradio = 4;
- cout<<"address of r:"<<(INT_PTR*)&r<<endl;
- cout<<"vbtab of transmitter:\n";
- cout<<" [1] "<<((INT_PTR*)(*(INT_PTR*)&r))[0]<<endl;
- cout<<" [2] "<<((INT_PTR*)(*(INT_PTR*)&r))[1]<<endl;
- cout<<"itransmitter = "<<*((INT_PTR*)&r + 1)<<endl;
- cout<<"vbtab of receiver:\n";
- cout<<" [1] "<<((INT_PTR*)*((INT_PTR*)&r + 2))[0]<<endl;
- cout<<" [2] "<<((INT_PTR*)*((INT_PTR*)&r + 2))[1]<<endl;
- cout<<"ireiceiver = "<<*((INT_PTR*)&r + 3)<<endl;
- cout<<"iradio = "<<*((INT_PTR*)&r + 4)<<endl;
- cout<<"address of storable part:"<<(INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1])<<endl;
- cout<<"_vptr of storable:"<<(INT_PTR*)*(INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1])<<endl;
- cout<<" [1] "<<(INT_PTR*)*(INT_PTR*)*(INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1])<<" ";
- ((FUN)(*(INT_PTR*)*(INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1])))();
- cout<<" [2] "<<(INT_PTR*)*((INT_PTR*)*(INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1]) + 1)<<" ";
- ((FUN)*((INT_PTR*)*(INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1]) + 1))();
- cout<<"istorable = "<<*((INT_PTR*)((char*)&r + ((INT_PTR*)(*(INT_PTR*)&r))[1]) + 1)<<endl;
- system("pause");
- return 0;
- }
运行结果如下所示:
- address of r:0021F73C
- vbtab of transmitter:
- [1] 0
- [2] 20
- itransmitter = 2
- vbtab of receiver:
- [1] 0
- [2] 12
- ireiceiver = 3
- iradio = 4
- address of storable part:0021F750
- _vptr of storable:002F790C
- [1] 002F12A8 read() in transmitter called.
- [2] 002F100F write() in receiver called.
- istorable = 1
- 请按任意键继续. . .
从上面的结果不难看出,VC++编译器引入虚基类表(virtual base table)来解决菱形继承问题。这样做避免了菱形继承造成的二义性以及内存浪费问题,但增加了内存布局的复杂性,此外引入新的虚基类表又不可避免使用指针进行数据的存取,这将降低程序的执行效率。