C Programming - Structures, Unions, Enums - Discussion

Here we came to know that, union allocates memory of 2 bytes for an integer.

In 1st prg we have assigned as 20 and in 2nd prg we have assigned as 10. i.e., when you print the address of both we. A and we. B the address are same, because union allocates the largest memory size for 2bytes only.

Dear All,

#include<stdio.h>
int main()
{
union var
{
int a, b;
};
union var v;
v.a=10;
v.b=20;
printf("%d\n", v.a);
return 0;
}

OutPut : 20
-----------------------------------

#include<stdio.h>
int main()
{
union var
{
int a, b;
};
union var v;
//v.a=10;
v.b=20;
v.a=10;
printf("%d\n", v.a);
return 0;
}

OutPut : 10

Here we came to know that, union allocates memory of 2 bytes for an integer.
In 1st prg we have assigned as 20 and in 2nd prg we have assigned as 10. i.e., when you print the address of both v.a and v.b the address are same, because union allocates the largest memory size for 2bytes only.

#include<stdio.h>

int main()
{
union var
{
int a, b;
};
union var v;
//v.a=10;
v.b=20;
printf("%d\n", v.a);

return 0;
}
output for the above program is :20.
can u guess? v.a is not assigned any value instead the value is printed...

Good Explanation...

I think it will give compile error...
cause
v.a=10;
v.b=20;
printf("%d\n", v.a);
since v.b=20 overwrites the value v.a=10
but program is trying to print the value which doesnot exist.
...
..
.
correct me if i m wrong...

The difference between union and structure is

Structure used as prototype in creating objects with same memory, it can not compress the memory.

struct a{ int a; float b; }obj;
Here obj is having 6 bytes in memory.

Where as for union, it allocates largest memory among all datatypes defined in the union.

union a{ int a; float b;}obj;
Here obj is having 4 bytes instead of 6 Bytes.

So, union utilizes largest memory assigned to variable among all variables and it will be used for all its operations.

Good Logic Bro.
But it also in structure?

If in above question a is initiallised with 20 and b with 10 then
o/p becomes 10(the last updated value is 10)..... ..!!!!

If in above question a is initiallised with 20 and b with 10 then what will be the o/p?

All variable inside in the union share the same memory area so all modification done in the same memory hence answer is the last updated value i.e. 20.