BitFields -- is this possible?

[trish]

Hello everyone,
BitFields gives us the liberty to decide how many bits to allot for the particular field.
ex.

struct x
{
   unsigned a:1;
   unsigned b:2;
   unsigned c:3;
};
struct x x1;
x1.a=0; //binary is 0
x1.b=2; // binary is 10
x1.c=5; // binary is 101

so we used only 6 bits.
this means we could accommodate this struct within the size of a char(with char occupying 1 byte i.e 8 bits)

so does this get stored as 010101 in contiguous memory locations ?
if so then
my question here is can we typecast this structure to char and print it with %c or %d as a whole value? i.e decimal of 010101 is 21 and its ASCII equivalent using %c ????

Regards.

[iMalc]

No, there is no guarantee that the individual bitfield values are stored consecutively. In fact, from what I remember it is highly likely that any padding be typically somewhere like just before the second-to-last member.

[quzah]

You can also make padding (force alignment):

struct foo
{
    int a:1;
    int b:2;
    int  :0;
    int c:1;
};

Quzah.

[Subsonics]

Yes you can, but it's a bit of a mess. For example you need to make sure the struct is actually 1 byte in memory, you will need long casts everywhere. It's probably simpler IMO, to just use and unsigned byte, and access the individual bits with | and & bitwise operators.

[trish]

but how is this possible .. i mean as far as i know structure variables are stored in contiguous memory locations .. or if not being in contiguous memory locations is the problem and NOTHING ELSE then we can add #pragma pack(1) so as to store the variables in the addresses that are multiples of 1 i.e. in continguous mem loc. but if i am sure that the variables are stored in contiguous mem loc then how should i typecast it to int .. my compiler is not allowing me to do that ..

[trish]

or is it that we cant typecast structure to integer ???

[Salem]

The problem with bit-fields is that there is no way to know (or control) whether you end up with
abbcccxx
or
xxabbccc
within a single storage unit.

Yes, you can do something like
memcpy( &myint, &myBitfieldStruct, sizeof(int) );
and get all the bit fields into a variable, and from there you can print it in whatever form you choose.

Or the brute-force answer (not guaranteed - google "alignment exception") is

int *p = (int*)&myBitfieldStruct;
// now do something with *p