why output is not zero

Question

Hello

Below is sample code:
https://code.sololearn.com/cUgCQw4rxsLg

I thought output is zero as uninitialized array should not be used but it might give zero value.
Here, structure has two element and hence a[2] is third element which should be unassigned I guess.

Please help me understand output

Answer 1

The reason is:
virtual void f() {}

The virtual method causes a function pointer address to get stored with your struct. That function pointer is the size of 2 int's. If the method wasn't virtual, there wouldn't be an address of f stored in each instance of the struct.

Notice that the following will print 22 just the same. I commented out the virtual method and changed the array index from 2 to 0.
#include <iostream>
using namespace std;

struct A
{
int data[2];

A(int x, int y) : data{x, y} {}
//virtual void f() {}
};

int main(int argc, char **argv)
{
A a(22, 33);

int *arr = (int *) &a;
cout << arr[0] << endl;
cout << "Number of bytes in struct A: " << sizeof(struct A) << endl;

return 0;
}

Note that you don't need the std:: prefix since you indicated that you're "using namespace std" above.

The sizeof returns 8 in that example but 16 when f is declared as a virtual method because the function pointer is 8 bytes.

Answer 2

I couldn't find anything saying that c++ standardizes that the virtual function pointers are always stored before the usual data members so I don't know.

That is how it works with Sololearn's tools at least. I wouldn't assume that in any non-educational programming situation.

This will obviously work the same way regardless of the virtual methods:
int *arr = (int *) &a.data; // the (int *) casting is optional since data is an array.

Answer 3

Oh...I see... I could observe virtual... Vptr is always on top of the allocated members?