Are pointers to members of an anonymous union equal?

Question

C++11 allows an anonymous union to be defined in a function, and its members can be accessed as variables of the function. If I examine the pointers to the different members, they are the same, but the == operator says they're unequal.

Such strange behavior is typically the result of undefined behavior, but I don't see anything undefined in my code (I made sure to that both members are of the same type).

The behavior is different if I use a named variable of an unnamed union type. In this case, the pointers compare equal.

This program demonstrates both cases:

#include <iostream>
using namespace std;

#ifdef NAMED
// Create a named object of a union type
#define NAME n
#define ADDR(mem) &(NAME.mem)
#else
// Create an anonymous union in main()
#define NAME
#define ADDR(mem) &mem
#endif

int main()
{
    union {
        int a;
        int b;
    } NAME;

    cout << "&a = " << ADDR(a) << endl;
    cout << "&b = " << ADDR(b) << endl;
    cout << "(&a==&b) = " << (ADDR(a) == ADDR(b)) << endl;

    return 0;
}

When compiled with -DNAMED it prints to identical pointers, and 1 (equal pointers). Without -DNAMED, it again prints identical pointers, but then 0 (unequal pointers).

Tested with g++ 5.4.0, Ubuntu 16.04, x86_64.

Answer 1

Your address checking is well defined (as YSC pointed out) and the standard guarantees that all members shall have the same address (cfr. [class.union]/3).

You must have stumbled upon a now fixed compiler bug.