C Programming - Declarations and Initializations - Discussion

Why we are assigning the 515 to the int i?

Why 'i' is not having any other value(like 0 or 1)?

@Suman sagar.

Consider it has a memory cell.

These are 2 memory cell(1byte =8 bits) ,size of the union is 2memory cells u.ch[0] occupies 1st memory cell ie 1-128, u.ch[1] occupies 2nd memory cell ie 256-65536. value assigned to u.ch[0] is 3 indicate the in 1st memory cell, u.ch[1] is assigned 2 indicate that is 2nd memory cell look for u.ch[1] it is like starting 1-128 but 1-65536 is one memory cell as a whole for the union.

u.ch[2] view   128 64 32   ....  2   1    |  128  ..  4   2   1   

u[i] view     65536  .. 1024   512    256 |  128   64..  2    1  

                               1     0                   1    1   

So according to u.ch[1] its value is 2.
But according to u[i] its value is 512 + 2+ 1.
Int occupies two memory cells whereas char occupies 1 memory cell.

Actually union size is the biggest size of its variable. But by using the instance of union if we access u.i it will indicate the value of union member/variable. But why it is giving the answer as 515?

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

I got these output...not the mentioned in the explanation.

3, 2, 1075053059.......why?

I have a doubt that structure and union is almost same then why do we need union in first place if structure is already there? I know the difference between them but don't know the purpose of using union over structure? please anyone?

%d is a format specifier for integer. So when u use %d in printf() statement for a character it will print the ascii value of that character? Explain in detail how could the solution gain those output value.

union has two variables indeed. the one is an integer and a character sequence of 2.

So memory is allocated either for int variable or for character sequence. but could you access the all the variables of the union at a time?

I tried out this

#include<stdio.h>
int main()
{
union a
{
int i;
char ch[2];
};
union a u;
u.ch[0] = 3;
u.ch[1] = 2;
u.ch[2] = 7;
printf("%d, %d, %d, %d\n", u.ch[0], u.ch[1], u.ch[2], u.i);
return 0;
}

I got the ans as 3,2,7,459267 and I also verified manually

But can u explain me why when char[2] declaration changed to char[3] gives result as 3,2,7,134676995

I mean I get the above result for the following program

#include<stdio.h>
int main()
{
union a
{
int i;
char ch[3];
};
union a u;
u.ch[0] = 3;
u.ch[1] = 2;
u.ch[2] = 7;
printf("%d, %d, %d, %d\n", u.ch[0], u.ch[1], u.ch[2], u.i);
return 0;
}

What does char[2] and char[3] declarations actually do to the program?
Pls explain me.

Thanks Soma.

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.

@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