In the last two lines of below program, static_cast<void*>
and dynamic_cast<void *>
behave differently. From what I understand, The result of a dynamic_cast<void*>
always resolves to the address of the complete object. So it uses RTTI in some way. Could anyone explain how compilers uses RTTI to differentiate between the two.
#include <iostream>
using namespace std;
class Top {
protected:
int x;
public:
Top(int n) { x = n; }
virtual ~Top() {}
friend ostream& operator<<(ostream& os, const Top& t) {
return os << t.x;
}
};
class Left : virtual public Top {
protected:
int y;
public:
Left(int m, int n) : Top(m) { y = n; }
};
class Right : virtual public Top {
protected:
int z;
public:
Right(int m, int n) : Top(m) { z = n; }
};
class Bottom : public Left, public Right {
int w;
public:
Bottom(int i, int j, int k, int m): Top(i), Left(0, j), Right(0, k) { w = m; }
friend ostream& operator<<(ostream& os, const Bottom& b) {
return os << b.x << ',' << b.y << ',' << b.z<< ',' << b.w;
}
};
int main() {
Bottom b(1, 2, 3, 4);
cout << sizeof b << endl;
cout << b << endl;
cout << static_cast<void*>(&b) << endl;
Top* p = static_cast<Top*>(&b);
cout << *p << endl;
cout << p << endl;
cout << static_cast<void*>(p) << endl;
cout << dynamic_cast<void*>(p) << endl;
return 0;
}
Possible output: https://ideone.com/WoX5DI
28
1,2,3,4
0xbfcce604
1
0xbfcce618
0xbfcce618
0xbfcce604