C job interview - casting and comparing

I was confronted with a tricky (IMO) question. I needed to compare two MAC addresses, in the most efficient manner.

The only thought that crossed my mind in that moment was the trivial solution - a for loop, and comparing locations, and so I did, but the interviewer was aiming to casting.

The MAC definition:

typedef struct macA {
   char data[6];
} MAC;

And the function is (the one I was asked to implement):

int isEqual(MAC* addr1, MAC* addr2)
{
    int i;

    for(i = 0; i<6; i++)
    {
        if(addr1->data[i] != addr2->data[i]) 
            return 0;
    }
    return 1;
}

But as mentioned, he was aiming for casting.

Meaning, to somehow cast the MAC address given to an int, compare both of the addresses, and return.

But when casting, int int_addr1 = (int)addr1;, only four bytes will be casted, right? Should I check the remaining ones? Meaning locations 4 and 5?

Both char and int are integer types so casting is legal, but what happens in the described situation?

Solution 1:

If he is really dissatisfied with this approach (which is essentially a brain fart, since you aren't comparing megabytes or gigabytes of data, so one shan't really be worrying about "efficiency" and "speed" in this case), just tell him that you trust the quality and speed of the standard library:

int isEqual(MAC* addr1, MAC* addr2)
{
    return memcmp(&addr1->data, &addr2->data, sizeof(addr1->data)) == 0;
}

Solution 2:

If your interviewer demands that you produce undefined behavior, I would probably look for a job elsewhere.

The correct initial approach would be to store the MAC address in something like a uint64_t, at least in-memory. Then comparisons would be trivial, and implementable efficiently.