In C language, how to determine whether the host is big or small (byte-order)

Http://www.cnblogs.com/52php/p/6114643.html

The so-called big-endian refers to the high value in memory put low address, so-called small end refers to low value in memory put low address. For example, 0x12345678 on the big-endian machine is 12345678, on the small end machine is 78564312, and a host is big or small end to look at the CPU type and running on the operating system. The same CPU in different operating systems use the size of the end case is different. Of course we usually use x86 + windows as the small end.

The test size end typically uses the Union feature. Union is a union in which all variables are common to a piece of memory, but are interpreted differently at different times. It stores in memory the number of bits required by the longest variable to open up memory

#include <stdio.h>

#include <stdbool.h> union {      int number;      char s; } test; bool testBigEndin() {      test.number = 0x01000002;      return (test.s == 0x01); } int main( int argc,  char **argv) {      if (testBigEndin()) {          printf ( "big" );      }  else {          printf ( "small" );      } }

My Mac output: Small

where the actual memory length of union is int, that is, a word, which is 32 bits on a 32-bit machine. Char is a byte (8-bit) and only takes the first low address byte. So its value can be used to determine the size of the end.

Warm tip: When the computer reads the memory data, it is (from left to right) from the low address to the high address

In C language, how to determine whether the host is big or small (byte-order)