ATL布幔之下的秘密(2)

核心提示： 现在，你会发现在基类中含有多个虚函数的情况下，ATL布幔之下的秘密(2)(5)，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 <<

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。