question about struct and pointers

Question

let us consider following code

#include <stdio.h>
#include <iostream>
using namespace std;
typedef struct bf_{
    unsigned x:4;
    unsigned y:4;
    unsigned z:4;
    unsigned  w:4;
    }bf;
int main(){
    unsigned short i=8;
    unsigned short j=9;
    bf* bitfields=(bf *)&i;
    bf*bit=(bf*)&j;
    bitfields->w=12;
    printf("%d\n",bitfields->x);
    printf("%d\n",bit->y);
    printf("%d\n",bitfields->w);



     return 0;
}

this fragment

unsigned short j=9;
bf*bit=(bf*)&j;
printf("%d\n",bit->y);

i have add after guess some interesting characteristic of this code for example after this place

bf* bitfields=(bf *)&i;

when we write printf("%d\n",bitfields->x); which prints 8 i understand that using pointers and references value of i will be granted to x and so it prints 8 for instance when we write bitfiled->y it writes 0 so i decided to introduce second element variable j create new instance of bf structer and make reference of j after which statement bit->y should write 9 because as i understand it is order definition but it gives me 0 why? please explain me how this code works?i am not english speaker so please sorry for my bad english

Answer 1

First of all, there are no references used in this code. The & character, when used as a unary operator (instead of as part of a type name), as in this context, means "address-of". So &i is "the address of i", not "a reference to i".

I'm not sure where you're getting the idea that using the value 9 as your bf struct should cause y to get the value 9. When you write this:

bf* bitfields=(bf *)&i;

What happens is that you are telling the compiler "treat the value at memory location &i as if it were a bf structure". The bf structure is defined so that, starting from the least significant (right-most) bits, the first four bits are the x value, the second four bits are the y value, the third four bits are the z value, etc.

So since the value at the location &i is 8, and 8 in 16-bit binary is

0000 0000 0000 1000 

The fields of bitfields are going to be:

0000 0000 0000 1000 
  w    z    y    x

So the value of x (a 4-bit unsigned integer) is 8, while the values of y, z, and w are 0.

In the case where you write

bf*bit=(bf*)&j;

you are doing the same thing as above, except now the value is 9, so the assignments are:

0000 0000 0000 1001 
  w    z    y    x

So x has the value 9, and the other values are still 0.

If you want to assign the value 9 to y instead of x, the assignments would need to look like this:

0000 0000 1001 0000 
  w    z    y    x

So the value you would have to use is the value with the 16-bit binary representation

0000 0000 1001 0000 

Which is 144. So if you let j = 144 instead of j = 9, you would observe that bit->y is 9, and all of the other fields in bit are 0.

Answer 2

I think it's because of the packing that occurs when the struct or class is compiled. See these links:

http://msdn.microsoft.com/en-us/library/83ythb65.aspx#vclrfhowalignworkswithdatapacking.
http://msdn.microsoft.com/en-us/library/2e70t5y1%28VS.80%29.aspx

By default, the size of a struct at compile time, is not the same as the total size of it's members.