Question

This is what I offered at an interview today.

int is_little_endian(void)
{
    union {
        long l;
        char c;
    } u;

    u.l = 1;

    return u.c == 1;
}

My interviewer insisted that c and l are not guaranteed to begin at the same address and therefore, the union should be changed to say char c[sizeof(long)] and the return value should be changed to u.c[0] == 1.

Is it correct that members of a union might not begin at the same address?

Answer 1

You are correct in that the “members of a union might begin at the same address”. The relevant part of the Standard is (6.7.2.1 para 13):

The size of a union is sufficient to contain the largest of its members. The value of at most one of the members can be stored in a union object at any time. A pointer to a union object, suitably converted, points to each of its members (or if a member is a bit-field, then to the unit in which it resides), and vice versa.

Basically, a start address of the union is guaranteed to be the same as the start address of each of its members. I believe (still looking for the reference) that a long is guaranteed to be larger than a char. If you assume this, then your solution should* be valid.

* I’m still a little uncertain due to some interesting wording around the representation of integer and, in particular, signed integer types. Take a close read of 6.2.6.2 clauses 1 & 2.