在下面的代碼中,我無法理解派生類的虛擬方法被調用的方式。 另外,任何人都可以提出一個資源,其中虛擬功能的概念用非常基本的方法進行圖解解釋。混淆派生類虛擬函數被調用的方式
class Base1
{
virtual void fun1() { cout << "Base1::fun1()" << endl; }
virtual void func1() { cout << "Base1::func1()" << endl; }
};
class Base2
{
virtual void fun1() { cout << "Base2::fun1()" << endl; }
virtual void func1() { cout << "Base2::func1()" << endl; }
};
class Base3
{
virtual void fun1() { cout << "Base3::fun1()" << endl; }
virtual void func1() { cout << "Base3::func1()" << endl; }
};
class Derive : public Base1, public Base2, public Base3
{
public:
virtual void Fn()
{
cout << "Derive::Fn" << endl;
}
virtual void Fnc()
{
cout << "Derive::Fnc" << endl;
}
};
typedef void(*Fun)(void);
int main()
{
Derive obj;
Fun pFun = NULL;
// calling 1st virtual function of Base1
pFun = (Fun)*((int*)*(int*)((int*)&obj+0)+0);
pFun();
// calling 2nd virtual function of Base1
pFun = (Fun)*((int*)*(int*)((int*)&obj+0)+1);
pFun();
// calling 1st virtual function of Base2
pFun = (Fun)*((int*)*(int*)((int*)&obj+1)+0);
pFun();
// calling 2nd virtual function of Base2
pFun = (Fun)*((int*)*(int*)((int*)&obj+1)+1);
pFun();
// calling 1st virtual function of Base3
pFun = (Fun)*((int*)*(int*)((int*)&obj+2)+0);
pFun();
// calling 2nd virtual function of Base3
pFun = (Fun)*((int*)*(int*)((int*)&obj+2)+1);
pFun();
// calling 1st virtual function of Drive
pFun = (Fun)*((int*)*(int*)((int*)&obj+0)+2);
pFun();
// calling 2nd virtual function of Drive
pFun = (Fun)*((int*)*(int*)((int*)&obj+0)+3);
pFun();
return 0;
}
一切似乎是未定義行爲,所以沒有太多解釋。 – 2011-12-25 17:55:11