Thread: Union problem

Can someone please explain me the output of this program. I searched a lot on the net but didn't find any explanation on this type of question on union.

The output is coming 268 0

Code:

#include<stdio.h>
int main()
{
	union
	{
		struct
		{
			char ax[2];
			char ab[2];
		} s;
		struct
		{
			int a;
			int b;
		} st;
	} u ={12, 1};
	printf("%d %d", u.st.a, u.st.b);
return 0;
}

Thanx

256 * 1 + 12 = 268
This simply means that your initialization pertains to the char portion of the struct. Placing 1 in the upper byte and 12 in the lower. This also indicates little-endian byte order of your particular processor.

Originally Posted by nonoob

256 * 1 + 12 = 268
This simply means that your initialization pertains to the char portion of the struct. Placing 1 in the upper byte and 12 in the lower. This also indicates little-endian byte order of your particular processor.

Thanx. Can you please explain in a little detail that why is the output of u.st.b is 0? and how exactly is the initialization happening in the union?

Code:

Assuming you're on a 32-bit (little endian) machine, the union will be 64 bits long. Initializing with {12, 1} gives:

00110000 10000000 00000000 00000000

Note that the first struct is 4 chars long, which is 32 bits total. Therefore, no matter what you set those chars to, it will never go past the first 32 bits of the 64 bit union, and so the second int will never be touched. Unless of course you explicitly set the u.st.b directly.

Posted in code tags because it wouldn't let me post otherwise.