sizeof class with int , function, virtual function in C++?

Question

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

#include
using namespace std; 
class A
{

};
class B
{
int i; 
}; 

class C
{
void         


        

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 |