C job interview - casting and comparing - C job interview - casting and comparing

problem ：

I was confronted with a tricky (IMO) question. I encountered a difficult problem (IMO) problem .I needed to compare two MAC addresses , in the most efficient manner. I need to compare the two in the most effective way MAC Address .

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. At that moment , The only thing I can think of is a trivial solution —— One for loop , Compare positions , So I did , But the interviewer's goal is to cast .

The MAC definition:MAC Definition ：

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

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

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. But as mentioned before , His goal is to cast .

Meaning, to somehow cast the MAC address given to an int, compare both of the addresses, and return. intend , In some way, the given MAC Address conversion to int, Compare two addresses , Then return .

But when casting, int int_addr1 = (int)addr1; But when converting , int int_addr1 = (int)addr1;, only four bytes will be casted, right?, Only four bytes will be projected , Am I right? ？Should I check the remaining ones? Should I check the rest ？Meaning locations 4 and 5? It means location 4 and 5？

Both char and int are integer types so casting is legal, but what happens in the described situation?char and int They're all integer types , Therefore, coercion is legal , But what happens in the described situation ？

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Solution ：

Reference resources ： https://stackoom.com/en/question/1PWLU