Size of virtual pointer-C++

Question

What is the size of virtual pointer(VPTR) for a virtual table in C++? Also this is not a homework question...just a question that came to my mind while I was reading a C++ b

Answer 1

Note that in order to gracefully handle multiple inheritance, there can be more than one VPTR in an object, but in general each is likely to be a simple architecture dependent pointer.

Try running something like this to see how your compiler lays things out:

#include 
using namespace std;

struct Base
{
    int B;
    virtual ~Base() {} 
};

struct Base2
{
    int B2;
    virtual ~Base2() {} 
};

struct Derived : public Base, public Base2
{
    int D;
};

int main(int argc, char* argv[])
{
    cout << "Base:" << sizeof (Base) << endl;
    cout << "Base2:" << sizeof (Base2) << endl;
    cout << "Derived:" << sizeof (Derived) << endl;

    Derived *d = new Derived();
    cout << d << endl;
    cout << static_cast(d) << endl;
    cout << &(d->B) << endl;
    cout << static_cast(d) << endl;
    cout << &(d->B2) << endl;
    cout << &(d->D) << endl;
    delete d;
    return 0;
}

On my 32-bit compiler, this give 8 bytes for both Base classes, and 20 bytes for the Derived class (and double those values when compiled for 64 bits):

4 bytes Derived/Base VPTR
4 bytes int B
4 bytes Derived/Base2 VPTR
4 bytes int B2
4 bytes int D

You can see how by looking at the first 8 bytes, you can treat a Derived as a Base, and how by looking at the second 8 bytes, you can treat it as a Base2.