How can a C++ base class determine at runtime if a method has been overridden?

Question

The sample method below is intended to detect whether or not it has been overridden in a derived class. The error I get from MSVC implies that it is simply wrong to try to g

Answer 1

You can actually find this out. We encountered the same problem and we found a hack to do this.

#include
#include
#include

using namespace std;

class A {
public:
    virtual void hi(int i) {}
    virtual void an(int i) {}
};

class B : public A {
public:
    void hi(int i) {
        cout << i << " Hello World!" << endl;
    }
};

We have two classes A and B and B uses A as base class.

The following functions can be used to test if the B has overridden something in A

int function_address(void *obj, int n) {
    int *vptr = *(int **)&obj;
    uintptr_t vtbl = (uintptr_t)*vptr;

    // It should be 8 for 64-bit, 4 for 32-bit 
    for (int i=0; i(p);
}

bool overridden(void *base, void* super, int n) {
    return (function_address(super, n) != function_address(base, n));
}

The int n is the number given to method as they are stored in vtable. Generally, it's the order you define the methods.

int main() {
    A *a = new A();
    A *b = new B();

    for (int i=0; i<2; i++) {
        if (overridden(a, b, i)) {
            cout << "Function " << i << " is overridden" << endl;
        }
    }

    return 0;
}

The output will be

Function 0 is overridden

EDIT: We get the pointers to vtables for each class instance and then compare the pointer to the methods. Whenever a function is overridden, there will be a different value for super object.