c++ - Testing for Endianness: Why does the following code work? -


while understand endianness, unclear on how code works below. guess question less endianness , more how char * pointer , int work i.e. type conversion. also, have made difference if variable word not short 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); }

a short int made of 2 bytes, in case 0x00 , 0x01. on little endian system, small byte comes first, in memory appears 0x01 followed 0x00. big endian systems are, naturally, reversed. pointers short integers on little endian system:

----------------------- -----------------------  |   0x01   |   0x00   | |          |          |  ----------------------- -----------------------     &word                  &word+1

char pointers, on other hand, incremented sequentially. thus, taking address of first byte of integer , casting char * pointer, may increment through each byte of integer in memory-order. here's corresponding diagram:

------------ ------------ ------------ ------------  |   0x01   | |   0x00   | |          | |          |  ------------ ------------ ------------ ------------     &byte       &byte+1      &byte+2      &byte+3

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