C Programming - Structures, Unions, Enums - Discussion

I understood how 515 is printed. I don't understand Why does doing u, i print that value?

In the above question:

union a
{
int i;
char ch[2];
};

Here, memory is allocated for int i, i.e 2 bytes now.

u.ch[0]=3;
u.ch[1]=2;

They are stored in memory allocated for int i (as char are of 1 bytes) they are placed in each bytes of integer i, now when we print.

printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i);
return 0;

Then, first u.ch[0] is printed then u.ch[1] then u.i where u.ch[0] then u.ch[1] from right to left is stored in memory allocated for u.i is saved.

Here can we take union u instead of union a u?

Union uses Shared memory.

So, char[] and int i occupy the same memory location.

u.ch[0]='a';

u.ch[1]='b';

Can anyone explain output for this definition instead of u.ch[0]=3;

u.ch[1]=2;?

#include<stdio.h>
int main()
{
union a
{
int i;
char ch[2];
};
union a u;
u.ch[0]=3;
u.ch[1]=2;
printf("%d, %d, %d, %d\n", u.ch[0], u.ch[1], u.ch[2], u.i);
return 0;
}

Getting ans as 3, 2, 110, -143785469.

How that 110 came? Please help

But union takes the memory of maximum size variable. So, in this case, int i has 4 bytes which is greater than value of ch[2] (2 elements * sizeof(char)) = 2 bytes.

#include<stdio.h>

int main()
{
union a
{
short int i;
char ch[2];
};
union a u;
u.ch[0]='a';
u.ch[1]='b';

printf("%d, %d, %d\n", u.ch[0], u.ch[1], u.i);
return 0;
}

Its output is:97, 98, 25185.

if I change 'short int' to 'char'，then the output is; 97, 98, 97.

How to get ui? Please explain.

Please explain me, why use 8 binary bits 00000011?

Anyone explain about it.