这与之前的许多问题相似，但它提出了一些我无法找到答案的问题。
#include <iostream>
using namespace std;
class Base1 {
public:
int b1_data;
virtual void b1_fn() {cout << "I am b1\n";}
};
class Base2 {
public:
int b2_data;
virtual void b2_fn() {cout << "I am b2\n";}
};
class Derived : public Base1, public Base2 {
public:
int d_data;
void b1_fn() {cout << "I am b1 of d\n";}
void b2_fn() {cout << "I am b2 of d\n";}
};
int main() {
Derived *d = new Derived();
Base1 *b1 = d;
/*My observation mentioned below is implementation dependant, for learning,
I assume, there is vtable for each class containing virtual function and in
case of multiple inheritance, there are multiple vtables based on number of
base classes(hence that many vptr in derived object mem layout)*/
b1->b1_fn(); // invokes b1_fn of Derived because d points to
// start of d's memory layout and hence finds vtpr to
// Derived vtable for Base1(this is understood)
Base2 *b2 = d;
b2->b2_fn(); // invokes b2_fn of Derived but how? I know that it "somehow"
// gets the offset added to d to point to corresponding Base2
// type layout(which has vptr pointing to Derived vtable for
// Base2) present in d's memory layout.
return 0;
}
具体来说，b2 如何指向 vptr for vtable of Derived for Base2 以获取 b2_fn()？我试过从 gcc 看到 memlayout 转储，但想不通。