Can someone please explain the results returned by 'sizeof' in this code

Question

I don't undertand the output shown below.

I know that whenever a virtual function is present it creates a vptr but still the size printed is more than I would expect:

#include<iostream>
using namespace std; 

class Base
{
 int x;
 int y;
 int z;
public:
 virtual void fun(){}
 virtual void fun2(){}
};

class Derived:public Base
{
 public:
  void fun() override {} 
};

int main(int argc, char const *argv[])
{
  cout<<sizeof(Base)<<endl;
  cout<<sizeof(Derived)<<endl;
  cout<<sizeof(int)<<endl; 
}

24
24
4
[Finished in 0.3s]

Possible duplicate: [Why isn't sizeof for a struct equal to the sum of sizeof of each member?](https://stackoverflow.com/questions/119123/why-isnt-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member?noredirect=1&lq=1) — Yksisarvinen, Jun 03 '18 at 08:34
`sizeof(Base) = sizeof(void *) + sizeof(int) * 3` aligned to `sizeof(void *)` which is 8 bytes. — user7860670, Jun 03 '18 at 08:40

Paul Sanders · Accepted Answer · 2018-06-03T09:19:59.163

6

Is this a 64 bit build? If so, sizeof Base would be:

8 (vtable pointer) + (3 * 4 = 12) (member variables) + 4 (pad to multiple of 8 bytes) = 24

Since Derived derives only from Base and adds no member variables, its size is the same.

Why is padding added? To maintain 8-byte alignment in arrays and on the stack. Why is that important? That's a different question.

edited Jun 03 '18 at 09:19

answered Jun 03 '18 at 08:40

Paul Sanders

24,133
4
26
48

Why not? What should it be? – Paul Sanders Jun 03 '18 at 08:49
@Yksisarvinen Motion to strike 'typical' :) – Paul Sanders Jun 03 '18 at 09:20

Can someone please explain the results returned by 'sizeof' in this code

1 Answers1