C Programming - Declarations and Initializations - Discussion

ch is char type of variable then how can we assign it number?

How we can calculate the value of u(I) at memory block 512?

I read all question and answer but still I can't get answer of why union share common memory to all data elements.

I am not able to understand. And @Nandini you have a good explanation but I have understood it partially can you explain t more deeply?

#include<stdio.h>
int main()
{
union a
{
long int i, j;
int 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;
}

Why value of u?

I means the entire size of union? Can someone help me out?

Please give output with explanation:

int main()
{
union a
{
int i;
float j;
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.j,u.i);
return 0;
}

How does the statement printf("%d %d %d", u.ch[0],u.ch[1],u.i); prints the size of union?

Any variable hold some garbage value without initialization.

Here we din't used sizeof operator too then how its calculating the size and printing. Any one please explain?

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

It prints the value of u.ch[0] = 3, u.ch[1] = 2 and it prints the value of u.i means the value of entire union size." is the explanation by the indiabix.

The last words 'value of entire union size' doesn't means size of the union rather the value represented when entire union is considered as integer of 2 bytes.

Bit confusing. C stores the information in 0s and 1s,

When one say, "Hey! will you please store an int for me" for a machine with int size 2 byte will store the binary equivalent of that number in 2 bytes for example to store decimal 2 it will store 00000000 in 1st byte and 00000010 in second byte, at time of retrieval C will read 2 consecutive bytes change it into decimal and will output it for you.

When one say, "Hey! will you please store a character for me. "Machine will reserve 1 byte of memory and store ascii value of the character in that one byte. For example if character is 'a' ASCII value 65 (Binary:- 01000001) is stored in that byte. At time of retrieval the ASCII code will be changed to corresponding character.

In the case we are discussing the char1 and char0 are stored at memory location say 2000 and 2001, at time of reading (when printf ("%d", u.i) ) , an integer is read at location 2000 (yes, the beauty of union!) and C reads the 2 bytes 2000 and 2001 that are 00000011 and 00000010 c combines it to make 0000000100000011 which is binary for 515.

I think this might not be entirely correct.

Different computers use different orderings for bytes (notice I am not saying bits).
A little endian machine will place the least significant byte first (so ch[0] will come before ch[1]), and a big endian machine will do the reverse.

So the answer cannot be determined without this information.