Why does adding virtual method increase class size in C++?

Question

#include <iostream>
using namespace std;

class a {
  virtual int foo() {
    return 0;
  }
};

class b {
  int foo() {
    return 0;
  }
};

int main() {
  cout << sizeof(b) << endl;
  cout << sizeof(a) << endl;
}

Output (with g++ 4.9, -O3):

1
8

I assume the increase in size is due to adding a vpointer. But I thought the compiler would see that a is not actually deriving or being derived from anything, hence there is no need to add the vpointer?

It is not marked final, so how could the compiler see that it is not being derived from, say in another file? — David Schwartz
– David Schwartz, Commented Dec 5, 2016 at 23:08
I thought the developer would see that it is not actually deriving or being derived from anything, why there is a need to add virtual keyword to class? — icaptan
– icaptan, Commented Dec 5, 2016 at 23:19

owacoder · Accepted Answer · 2016-12-05 23:20:36Z

6

The vpointer is needed because the compiler cannot guarantee an external (e.g. shared) library does not use a derived type. The existence-of-derived-class resolution happens at runtime.

edited Dec 5, 2016 at 23:20

answered Dec 5, 2016 at 23:08

owacoder

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3 Comments

David Schwartz Over a year ago

Imagine if that file contained int j (a& bar) { return bar.foo(); } -- how would the compiler make that work?

owacoder Over a year ago

Is that a rhetorical question?

David Schwartz Over a year ago

Yes. My point is the same as yours -- the compiler has no idea what code it can't see might do.

Edward Strange · Accepted Answer · 2016-12-05 23:05:51Z

1

Run-time type information. Any polymorphic class creates extra meta-data in the program to make things like typeof and dynamic_cast work. This is in addition to the virtual function table.

answered Dec 5, 2016 at 23:05

Edward Strange

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