Question

I could not understand how Union works..

#include <stdio.h>
#include <stdlib.h>

int main()
{
    union {
    int a:4;
    char b[4];
    }abc;
abc.a = 0xF;

    printf(" %d, %d, %d, %d, %d, %d\n", sizeof(abc), abc.a, abc.b[0], abc.b[1], abc.b[2], abc.b[3]);

    return 0;
}

In the above program.
I made int a : 4;
So, a should taking 4 bits.
now I am storing, a = 0xF; //i.e a= 1111(Binary form)
So when I am accessing b[0 0r 1 or 2 or 3] why the outputs are not coming like 1, 1, 1, 1

Answer 1

Your union’s total size will be at least 4 * sizeof(char).

Assuming the compiler you are using handles this as defined behavior, consider the following:

1. abc is never fully initialized, so it contains a random assortment of zeros and ones. Big problem. So, do this first:
   memset(&abc, 0, sizeof(abc));
2. The union should be the size of its largest member, so you should now have 4 zeroed-out bytes: 00000000 00000000 00000000 00000000
3. You are only setting 4 bits high, so your union will become something like this:
   00000000 00000000 00000000 00001111 or
   11110000 00000000 00000000 00000000.

   I’m not sure how your compiler handles this type of alignment, so this is the best I can do.

You might also consider doing a char-to-bits conversion so you can manually inspect the value of each and every bit in binary format:
Access individual bits in a char c++

Best of luck!