C Programming - Structures, Unions, Enums - Discussion

Is there any restriction to use char in union should contain one int value ? suppose replaced int i to any other (char i, float i. ) what happen ?

@ ashwni
In ashish question the value of the u.i should be 515?

Thanks guys really great.

@Ashish

When I compile it in Turbo c++ IDE

It has given output as: 3 2 0 50

not u.ch[2] as 20

and I think 50 as garbage value, So how u.ch[2] as 0 ?

@Sundar

why 20 comes as value of u.ch[2] in output?
I understand about garbage value. cant understand about u.ch[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;
}
Ans is 3, 2, 20, some garbage value.

Give full information according (2)(3)=512

@Sundar

Your explanation is the best one. THANKS.

@Raju Kumar:

In TurboC it prints 0, 0, 0, 0 (as expected).

But in GCC it prints 0, 0, 0, 134217728 (last value is unexpected).

It is due to the platform dependency of compilers.

Because for 'int' TurboC (under DOS) reserves two bytes only. But in Linux GCC reserves 4 bytes.

You have initialized/assigned only three bytes (u.ch[0], u.ch[1], u.ch[2]) of the reserved bytes to 0 (zero). The remaining one byte (1 out of 4) not yet assigned with any value. So it may have some junk/garbage value.

So while printing the u.i, it prints the garbage value.

If the change your code for union as give below, the above problem will not occur.

union a
{
short i; // change 'int' --> 'short'
char ch[3];
};

Hope this will help you. Have a nice day!

#include<stdio.h>

int main()
{
union a
{
int i;
char ch[3];
};
union a u;
u.ch[0]=0;
u.ch[1]=0;
u.ch[2]=0;
printf("%d, %d, %d, %d\n", u.ch[0], u.ch[1],u.ch[2], u.i);
return 0;
}
output:
0, 0, 0, 134217728


Now, I applied same approach as previous program but last ans should be zero but i got somewhat other than that what's the reason, can anyone help me .

@Sai

The binary value of 2 is 00000010 and dat of 3 is 00000011...

Now (2)(3)---> 0000001000000011 = 2^9 + 2^1 + 2^0 = 515