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

@Andrey: I was about to ask the same, but looking at ecatmur's answer I can see this being a valid question. The muddy waters between UD and OK are a little hard to test with "just running it and see". — Benjamin Bannier
– Benjamin Bannier, Commented Aug 23, 2012 at 13:43
It's funny how people use these weird features, and then go on to complain C++ is a minefield... — enobayram
– enobayram, Commented Aug 23, 2012 at 14:04

ecatmur · Accepted Answer · 2012-08-23 14:12:28Z

12

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.

edited Aug 23, 2012 at 14:12

answered Aug 23, 2012 at 13:38

ecatmur

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6 Comments

Aesthete Over a year ago

Would it be advised to use size_t if one were needing to do this?

David Rodríguez - dribeas Over a year ago

Beware: this is guaranteed only for standard layout types. Add a virtual function and you will be in undefined behavior land.

Kerrek SB Over a year ago

Only if the type is standard-layout.

ecatmur Over a year ago

@Aesthete I don't see how size_t is relevant.

G. Martinek Over a year ago

This violates class visibilty too. x and y are both private.

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