[Tip] If you use union to determine if machine memory is big-endian or small-endian storage @ C

such as int i = 1, if it is small-endian storage, then in memory should be this: 0x01 0x00 0x00 0x00, if it is the big-endian storage, in memory should be: 0x00 0x00 0x00 0x01.

So how to program to determine whether the machine is big-endian storage or small-order storage it?

On the code:

 #include <stdio.h>int   Isbigendian (        {Union { int   I;     char   C;        }test;    test.c  = 1  ;  return  test.i! = 1  ;} int   Main () {printf ( "  %d\n   , Isbigendian ());}  

Since both int and Char in the union are stored from the low address, if it is a small-endian store, the TEST.C is assigned to one, then the Union should be 0x01 0x00 0x00 0x00 0x00 0x00 0x00 0x00, Then the value of TEST.I should also be 1, then test.c==test.i. In the case of big-endian storage, the Union should look like this in memory: 0x00 0x00 0x00 0x01 0x00 0x00 0x00 0x00, then the value of TEST.I should be 2 of 8 square, i.e. test.i==256. That is, if it is a big endian store, test.i! = test.c. If test.i! = test.c, then store for the big endian, or small-endian if equal.

[Tip] If you use union to determine if machine memory is big-endian or small-endian storage @ C