Question

#include <iostream>
using namespace std;

class Z
{
public:
    int a;
    virtual void x () {}
};

class Y : public Z
{
public:
    int a;
};

int main() 
{
    cout << "\nZ: "  << sizeof (Z);
    cout << "\nY: "  << sizeof (Y);
} 

Because Y inherits Z, so it will also have a virtual table. Fine. But, it doesn’t have any virtual functions, so what will be the contents of the virtual table of Y?
Will it be empty?

Answer 1

This is entirely compiler-dependent. When I force instantiation of Y and Z, g++ 4.4.5 produces two distinct virtual tables for Y and Z that have the same size.

Both tables point to the same x() but point to different typeinfo structures:

;=== Z's virtual table ===
_ZTV1Z:
        .quad   0
        .quad   _ZTI1Z     ; Z's type info
        .quad   _ZN1Z5xEv  ; x()

_ZTI1Z:
        ; Z's type info (omitted for brevity)

;=== Y's virtual table ===
_ZTV1Y:
        .quad   0
        .quad   _ZTI1Y     ; Y's type info
        .quad   _ZN1Z5xEv  ; x()

_ZTI1Y:
        ; Y's type info (omitted for brevity)