Question

the program is to decide big endian or little endian.

This is the answer given in the book:

int Test(){
    short int word = 0x0001;
    char *byte = (char *) &word;
    return (byte[0] ? BIG:LITTLE);
}

I don’t understand this line: char *byte = (char *) &word; Does it mean “pass the address of word into byte”? So, now byte point to word’s original address? As I known, short int is 2 bytes. So, does “byte” point to higher address or lower address? Why?

How does this work?

Answer 1

It’s just taking the address of word, casting it to a char pointer, and putting it in byte.

At that point, byte will point to the first byte of 2-byte word, and the value of that byte (1 or 0) will tell you if you’re on a big or little endian machine.