Question

For the union declaration below

union a
{
    int i;
    char ch[2];
};
union a u;
u.ch[0] = 3;
u.ch[1] = 2;

Since i and ch store in the same place, i should now be <binary rep. of 3> concatenated with <binary rep. of 2>. However, it is stored in the reverse order.

printf("%d",i);

gives 515.
Why so??

Answer 1

The result depends on the endianness of the machine you are trying your program on.

It is 515 on a little endian machine Intel i686:

$ uname -m
i686
$ ./a.out
515

and is 770 on a bin endian machine like Sparc:

$ uname -m
sun4v
$ ./a.out
770

This happens because the endianess will determine if ch[0] will be higher order byte (in big endian) or lower order byte (in little endian) of i.