【編程語言】如何解決菱形繼承問題

繼承、封裝和多態是面向程序設計(OOP)的三大特點，而它們三者之中最具實際操作性的當屬繼承。通過繼承可以實現簡單功能的組合和定製，而多重繼承更將這種能力發揮到更高的境界。不過事事都有弊端，如果使用多重繼承不當很容易造成菱形繼承問題(diamond problem)。Bjarne Stroustrup用下面這個例子描述菱形繼承問題：

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都從storable派生而來，而radio則從transmitter和receiver派生而來，是最終派生類。這樣的繼承關係很特殊，因爲從storable到radio存在兩條不同的路徑，即radio將從兩條不同的繼承路徑獲得storable的成員。如果不採取措施應對菱形繼承問題的話，編譯程序時將收到許多二義性的錯誤提示。如在上述示例中，如果radio類的對象調用write()方法，編譯器將無法確定應當調用transmitter中的write()方法或是receiver中的write()方法。解決的方法很簡單：只需在transmiiter和receiver類的繼承列表前添加virtual關鍵字即可，如下列代碼所示：

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;
};

爲了方便描述使用虛擬繼承後radio類對象的內存佈局，我在各類中都增加了一個整型的成員變量。

現在編寫主函數來打印radio類對象的內存佈局，代碼如下所示：

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;
}

運行結果如下所示：

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
請按任意鍵繼續. . .

從上面的結果不難看出，VC++編譯器引入虛基類表(virtual base table)來解決菱形繼承問題。這樣做避免了菱形繼承造成的二義性以及內存浪費問題，但增加了內存佈局的複雜性，此外引入新的虛基類表又不可避免使用指針進行數據的存取，這將降低程序的執行效率。