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 |
|
