C Programming - Declarations and Initializations - Discussion

@anonymous..
In that program first we assign u.ch[0]=3 which is array which index 0. so it takes memory in first byte and which is 00000011.

And then u.i=1;
Which REPLACE previous assigned memory of ch[0] and now it become 00000001

And then u.j=2;
Again REPLACE previous assigned memory of i and now it become 00000010

And now u.ch[1]=5 it takes memory in second byte and it is 00000101.

NOW IN MEMORY (00000101)(00000010)
second first byte
AND NOW printf("%d,%d,%d,%d",u.ch[0],u.ch[1],u.i,u.j); statement run.
IN THIS STATEMENT u.ch[0] is print the value which is in first byte. so we convert first byte binary digit in decimal numbers and get 2. so "u.ch[0]=2".

AFTER THAT u.ch[1] RUN AND IT print the value which is in SECOND byte. so we convert SECOND byte binary digit in decimal numbers and get 2. so "u.ch[0]=5".

And u.i run which means the whole memory takes by union so we convert 0000010100000010 into decimal numbers.which you know how convert binary to decimal.

And u.j also means the same

How we got the answer has 515? Try to expalin in detail how to find the union size using same fig used in explaining answer.

Can you tell me why the initialisations for ch[0], i, j are not working?

#include<stdio.h>
int main()
{
union a
{
int i, j;
char ch[2];
};

union a u;
u.ch[0] = 3;
u.i=1;u.j=2;
u.ch[1]=5 ;
printf( "%d, %d, %d, %d\n", u.ch[0], u.ch[1], u.j, u.i );
return 0;
}

O/P:

2, 5, 1282, 1282

It gives different output in gcc or MS vc++.
In gcc the int size is 4 bytes. In the code only 2 bytes are initialized (MSB 2bytes contains uninitialized data). So output is garbage data. To avoid this add

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

Hi nandini good explanation.

I broked my head for understanding. Your explanation helped me.

Thk you.

@nandhini:

I understood wat u told. but how 512 is got?

@nandini.

Thanks a lot. Good explanation.

Hi friends,

Here ch[0]=3 the binary value of 3 = 00000011=011 and ch[1]=2 the binary value of 2=010=00000010.

Now we can call the size of union i.e, 00000010 00000011.

See the fig so 1*1+2*1+512*1=515.

We are taking only 512, 1 and 2 because at these places only we have 1's remaing places we have zero. If we take the remaing also it'll multipled by 0 anyway we can get 0 for that values.

#include<stdio.h>
int main()
{
union a
{
int i;
char ch[2];
};
union a u;
u.ch[0] = 3;
u.ch[1] = 2;

u.i=11;

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

Try output for this program.

Please explain how we get value of i?