Question

This is an online C++ test question, which has been done.

#include<iostream>
using namespace std; 
class A
{

};
class B
{
int i; 
}; 

class C
{
void foo();
};
class D
{
virtual void foo();
};

class E
{
int i ; 
    virtual void foo();
};
class F
{
int i; 
    void foo();
};
class G
{
    void foo();
    int i;
    void foo1();
};

class H
{
    int i ;
    virtual void foo();
    virtual void foo1();
};
int main()
{
cout <<"sizeof(class A) : " << sizeof(A) << endl ;
cout <<"sizeof(class B) adding the member int i : " << sizeof(B) << endl ;
cout <<"sizeof(class C) adding the member void foo() : " << sizeof(C) << endl ;
cout <<"sizeof(class D) after making foo virtual : " << sizeof(D) << endl ;
cout <<"sizeof(class E) after adding foo virtual , int : " << sizeof(E) << endl ;
cout <<"sizeof(class F) after adding foo  , int : " << sizeof(F) << endl ;
cout <<"sizeof(class G) after adding foo  , int : " << sizeof(G) << endl ;
G g;
cout <<"sizeof(class G) after adding foo  , int : " << sizeof(g) << endl ;
cout <<"sizeof(class H) after adding int 2 virtual " << sizeof(H) << endl ;
return 0; 
}

output:

sizeof(class A) : 1
sizeof(class B) adding the member int i : 4
sizeof(class C) adding the member void foo() : 1
sizeof(class D) after making foo virtual : 8
sizeof(class E) after adding foo virtual , int : 16
sizeof(class F) after adding foo  , int : 4
sizeof(class G) after adding foo   , unsigned int : 4
sizeof(class g) after adding foo  , unsigned int : 4
sizeof(class H) after adding int 2 virtual 16

My questions:

Why siszeof(A) is 1 and sizeof(C) is 1 too ?

Why siszeof(H) is 16 but sizeof(G) is 4 ?

Why siszeof(E) is 16 but sizeof(F) is 4 ?

Why siszeof(D) is 8 but sizeof(E) is 16 ?

My guess:

A virtual function is a pointer with 8 bytes.
But, I do not know why E size is 16 ?
Adding a function to an empty class does not change its size ?

Any help is appreciated.

thanks

Answer 1

First off, a virtual function is not a pointer with 8 bytes. In C++ nothing but sizeof(char) is guaranteed to be any number of bytes.

Second, only the first virtual function in a class increases its size (compiler-dependent, but on most – if not all – it’s like this). All subsequent methods do not. Non-virtual functions do not affect the class’s size.

This happens because a class instance doesn’t hold pointers to methods themselves, but to a virtual function table, which is one per class.

So if you had:

class A
{
   virtual void foo();
}

and

class B
{
   virtual void goo();
   virtual void test();
   static void m();
   void x();
}

you would have sizeof(A) == sizeof(B).

And now:

Why siszeof(A) is 1 and sizeof(C) is 1 too ?

A and C have size 1 just because it’s not allowed for a class to be of size 0. The functions have nothing to do with it. It’s just a dummy byte.

Why siszeof(H) is 16 but sizeof(G) is 4 ?

G has only one member that accounts for memory – the int. And on your platform, sizeof(int) == 4. H, besides the int, also has a pointer to the vftable (virtual function table, see above). The size of this, size of int and allignment are compiler specific.

Why siszeof(E) is 16 but sizeof(F) is 4 ?

Explained above – non virtual methods don’t take up memory in the class.

Why siszeof(D) is 8 but sizeof(E) is 16 ?

D only contains the vftable pointer which is apparently 8 bytes on your platform. E also has an int, and the vftable is aligned to 8 bytes. So it’s something like:

class E

4 bytes for int |  4 padding bytes  |  8 bytes for vftable pointer  | 
| x | x | x | x |    |    |    |    | v | v | v | v | v | v | v | v |