Testing for Endianness: Why does the following code work?

While I do understand endianness, I am slightly unclear on how the code works below. I guess this question is less about endianness and more about how the char * pointer and int work i.e. type conversion. Also, would it have made any difference if the variable word was not a short but just an int? Thanks!

#define BIG_ENDIAN 0
#define LITTLE_ENDIAN 1

int byteOrder() {
    short int word = 0x0001;
    char * byte = (char *) &word;
    return (byte[0] ? LITTLE_ENDIAN : BIG_ENDIAN);
}