ATL布幔之下的秘密(2)

核心提示： 那么现在，让我邀请纯虚函数来加入这一游戏，ATL布幔之下的秘密(2)(6)，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 <&l

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。