ATL布幔下的秘密之虚函数背后的东西

现在，你会发现在基类中含有多个虚函数的情况下，派生类并不能完全重写它们。

　　程序26.

#include ＜iostream＞ using namespace std; class Base { 　public: 　Base() { 　　cout ＜＜ "In Base" ＜＜ endl; 　　cout ＜＜ "Virtual Pointer = " ＜＜ (int*)this ＜＜ endl; 　　cout ＜＜ "Address of Vtable = " ＜＜ (int*)*(int*)this ＜＜ endl; 　　cout ＜＜ "Value at Vtable 1st entry = " ＜＜ (int*)*((int*)*(int*)this+0) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 2nd entry = " ＜＜ (int*)*((int*)*(int*)this+1) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 3rd entry = " ＜＜ (int*)*((int*)*(int*)this+2) ＜＜ endl; 　　cout ＜＜ endl; 　} 　virtual void f1() { cout ＜＜ "Base::f1" ＜＜ endl; } 　virtual void f2() { cout ＜＜ "Base::f2" ＜＜ endl; } }; class Drive : public Base { 　public: 　Drive() { 　　cout ＜＜ "In Drive" ＜＜ endl; 　　cout ＜＜ "Virtual Pointer = " ＜＜ (int*)this ＜＜ endl; 　　cout ＜＜ "Address of Vtable = " ＜＜ (int*)*(int*)this ＜＜ endl; 　　cout ＜＜ "Value at Vtable 1st entry = " ＜＜ (int*)*((int*)*(int*)this+0) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 2nd entry = " ＜＜ (int*)*((int*)*(int*)this+1) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 3rd entry = " ＜＜ (int*)*((int*)*(int*)this+2) ＜＜ endl; 　　cout ＜＜ endl; 　} 　virtual void f1() { cout ＜＜ "Drive::f1" ＜＜ endl; } }; int main() { 　Drive d; 　return 0; }

　　程序的输出为：

In Base Virtual Pointer = 0012FF7C Address of Vtable = 0046C0E0 Value at Vtable 1st entry = 004010F0 Value at Vtable 2nd entry = 00401145 Value at Vtable 3rd entry = 00000000 In Drive Virtual Pointer = 0012FF7C Address of Vtable = 0046C0C8 Value at Vtable 1st entry = 0040121C Value at Vtable 2nd entry = 00401145 Value at Vtable 3rd entry = 00000000

　　这个程序的输出表明基类的虚函数在派生类中并未被重写，然后，派生类的构造函数没有对虚函数的入口做任何的事情。
　　
　　那么现在，让我邀请纯虚函数来加入这一游戏，再来看看它的行为吧。请看以下的程序：

　　程序27.

#include ＜iostream＞ using namespace std; class Base { 　public: 　Base() { 　　cout ＜＜ "In Base" ＜＜ endl; 　　cout ＜＜ "Virtual Pointer = " ＜＜ (int*)this ＜＜ endl; 　　cout ＜＜ "Address of Vtable = " ＜＜ (int*)*(int*)this ＜＜ endl; 　　cout ＜＜ "Value at Vtable 1st entry = " ＜＜ (int*)*((int*)*(int*)this+0) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 2nd entry = " ＜＜ (int*)*((int*)*(int*)this+1) ＜＜ endl; 　　cout ＜＜ endl; 　} 　virtual void f1() = 0; 　virtual void f2() = 0; }; class Drive : public Base { 　public: 　Drive() { 　　cout ＜＜ "In Drive" ＜＜ endl; 　　cout ＜＜ "Virtual Pointer = " ＜＜ (int*)this ＜＜ endl; 　　cout ＜＜ "Address of Vtable = " ＜＜ (int*)*(int*)this ＜＜ endl; 　　cout ＜＜ "Value at Vtable 1st entry = " ＜＜ (int*)*((int*)*(int*)this+0) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 2nd entry = " ＜＜ (int*)*((int*)*(int*)this+1) ＜＜ endl; 　　cout ＜＜ endl; 　} 　virtual void f1() { cout ＜＜ "Drive::f1" ＜＜ endl; } 　virtual void f2() { cout ＜＜ "Drive::f2" ＜＜ endl; } }; int main() { 　Drive d; 　return 0; }

　　在debug和release模式下，程序的输出有所不同。下面是debug模式的输出：

In Base Virtual Pointer = 0012FF7C Address of Vtable = 0046C0BC Value at Vtable 1st entry = 00420CB0 Value at Vtable 2nd entry = 00420CB0 In Drive Virtual Pointer = 0012FF7C Address of Vtable = 0046C0A4 Value at Vtable 1st entry = 00401212 Value at Vtable 2nd entry = 0040128F

　　以下则是release模式的输出：

In Base Virtual Pointer = 0012FF80 Address of Vtable = 0042115C Value at Vtable 1st entry = 0041245D Value at Vtable 2nd entry = 0041245D In Drive Virtual Pointer = 0012FF80 Address of Vtable = 00421154 Value at Vtable 1st entry = 00401310 Value at Vtable 2nd entry = 00401380

　　为了更好地弄懂这一原理，我们需要对程序作少许的改动，并尝试使用函数指针来调用虚函数。

　　程序28.

#include ＜iostream＞ using namespace std; typedef void(*Fun)(); class Base { 　public: 　Base() { 　　cout ＜＜ "In Base" ＜＜ endl; 　　cout ＜＜ "Virtual Pointer = " ＜＜ (int*)this ＜＜ endl; 　　cout ＜＜ "Address of Vtable = " ＜＜ (int*)*(int*)this ＜＜ endl; 　　cout ＜＜ "Value at Vtable 1st entry = " ＜＜ (int*)*((int*)*(int*)this+0) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 2nd entry = " ＜＜ (int*)*((int*)*(int*)this+1) ＜＜ endl; 　　// 尝试执行第一个虚函数 　　Fun pFun = (Fun)*((int*)*(int*)this+0); 　　pFun(); 　　cout ＜＜ endl; 　} 　virtual void f1() = 0; 　virtual void f2() = 0; }; class Drive : public Base { 　public: 　Drive() { 　　cout ＜＜ "In Drive" ＜＜ endl; 　　cout ＜＜ "Virtual Pointer = " ＜＜ (int*)this ＜＜ endl; 　　cout ＜＜ "Address of Vtable = " ＜＜ (int*)*(int*)this ＜＜ endl; 　　cout ＜＜ "Value at Vtable 1st entry = " ＜＜ (int*)*((int*)*(int*)this+0) ＜＜ endl; 　　cout ＜＜ "Value at Vtable 2nd entry = " ＜＜ (int*)*((int*)*(int*)this+1) ＜＜ endl; 　　cout ＜＜ endl; 　} 　virtual void f1() { cout ＜＜ "Drive::f1" ＜＜ endl; } 　virtual void f2() { cout ＜＜ "Drive::f2" ＜＜ endl; } }; int main() { 　Drive d; 　return 0; }

　　现在程序的行为在debug和release模式下仍然不同。在debug模式下，它会显示一个运行时错误的对话框：


　　并且，当你按下“忽略”按钮之后，它会显示下面的对话框：


　　而在release模式下运行的话，它只会在控制台窗口中输出错误信息：

In Base Virtual Pointer = 0012FF80 Address of Vtable = 0042115C Value at Vtable 1st entry = 0041245D Value at Vtable 2nd entry = 0041245D runtime error R6025 - pure virtual function call