Explicit type casting of object to int *

Question

What is the output of the following c++ code ?

#include<iostream> 
using namespace std;
class IndiaBix
{
    int x, y; 
    public:
    IndiaBix(int xx)
    {
        x = ++xx;
    } 
    ~IndiaBix()
    {
        cout<< x - 1 << " ";
    }
    void Display()
    {
        cout<< --x + 1 << " ";
    } 
};
int main()
{
    IndiaBix objBix(5);
    objBix.Display();
    int *p = (int*) &objBix;
    *p = 40;
    objBix.Display();
    return 0; 
}

I did n't understand the following line ::

 int *p = (int*) &objBix;//Explicit type cast of a class object to integer pointer type

Answer 1

It is possible to cast an object pointer (of a standard-layout type) to a pointer to its first member. This is because it is guaranteed that the first member of a standard-layout object has the same address as the overall object:

c++11

9.2 Class members [class.mem]

20 - A pointer to a standard-layout struct object, suitably converted using a reinterpret_cast, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa.

Thus int *p = (int*) &objBix; is a pointer to objBix.x, since objBix is standard-layout; both its data members x and y are private, and the class has no virtual methods or base classes.