我的4个字节在哪里？

Question

#include <iostream>
#include <cstdlib>

using std::cout;

class A
{

public :
    A() { cout << "A()" << this << "\n";}
    ~A() { cout << "~A()" << this << "\n";}
    //void func()  { }
    virtual void debug(int a)  { cout << "A::debug";}
private :
    int a;
};

class A1 : public A
{
public :
    A1() { cout << "A1()"<< this << "\n";}
    ~A1() { cout << "~A1()"<< this << "\n";}
private :
    int a1;
};
class A2 : public A
{
public :
    A2() { cout << "A2()"<< this << "\n";}
    ~A2() { cout << "~A2()"<< this << "\n";}
private :
    int a2;
};

class B : public A1, public A2
{
public :
    B() { cout << "B()"<< this << "\n";}
    ~B() { cout << "~B()"<< this << "\n";}
    void debug() { cout << "B::debug()";  }
private :
    int a3;
};
int main()
{
    cout << "sizeof(int)" << sizeof(int) << "\n";
    cout << "sizeof(void*)" << sizeof(void*) << "\n";
    cout << "sizeof(A): " << sizeof(A) << "\n";
    cout << "sizeof(A1): " << sizeof(A1) << "\n";
    cout << "sizeof(A2): " << sizeof(A2) << "\n";
    cout << "sizeof(B): " << sizeof(B) << "\n";
    B b;
    b.debug();

}

output :

sizeof(int)4
sizeof(void*)4
sizeof(A): 8
sizeof(A1): 12
sizeof(A2): 12
**sizeof(B): 28**
A()0x28fef4
A1()0x28fef4
**A()0x28ff00**
A2()0x28ff00
B()0x28fef4
B::debug()~B()0x28fef4
~A2()0x28ff00
~A()0x28ff00
~A1()0x28fef4
~A()0x28fef4

Both A1 and A2 are 4(vtbl) + 4(A'sint) + 4(respective int) = 12 bytes but B is 28 bytes I know its not guaranteed but what could be the possible use of those 4 bytes...I dont see any padding issues ? Can anyone point out what am I missing ?

Answer 1

sizeof(A): 8

The type A has a member of type int which in your platform is 4 bytes. It also has a virtual function, which means that a vptr (virtual table pointer) is allocated for each object of your class, the size of it is another 4 bytes.

**sizeof(B): 28**

B contains one object of type A1 (12 bytes), and an object of type A2 (another 12 bytes) and it adds another int for a total of 12+12+4 = 28 bytes. This is quite straightforward.

Answer 2

machine word size alignment of data items within structures.

See structure packing for more information.

Answer 3

Multiple inheritance will produce implementation-specific memory layouts of possibly different sizes. Virtual tables and virtual pointers for multiple virtual inheritance and type casting