Question

I got the output 0 2 for this program…..
but don’t know why?
Please explain i think only int i is initialized with 512.
But how ch[1] got the value 2.

#include <stdio.h>
int main()
{
    union a /* declared */
    {
        int i;   char ch[2];
    };
    union a z = { 512 };   

    printf("%d%d", z.ch[0], z.ch[1]);
    return 0;
}

Answer 1

Union declaration means that all its members are allocated the same memory. So your int i and char ch[2] are referencing the same memory space — in other words, they are aliased. Whenever you change one, you will change the other as well.

Now, assuming your ints are 32-bit wide and you’re on a little-endian system like x86, i = 512 (512 == 0x00000200) actually looks like this in memory:

0x00  0x02  0x00  0x00.

with the first two values corresponding directly to the 2-character array:

ch[0] ch[1]

So you get ch[0] == 0x0 and ch[1] == 0x02.

Try setting your i = 0x1234 and see what effect it will have on your character array.

Based on your question, it’s possible that you may want to use a struct instead of union — then its members would be allocated in memory sequentially (one after the other).