Re: sizeof bit-fields

[Eljay Love-Jensen <eljay at adobe dot com>]

Hi Massimiliano,

>How can I obtain a 5 bytes long structure (if it is possible)? 

The bit-fields are going to pack like this in memory (probably), big-endian representation on a 32-bit/word machine:
  +--3---------2---------1---------+
0 +..ccccccccccbbbbbbbbbbaaaaaaaaaa+
4 +......................dddddddddd+
  +--------------------------------+

So the 32-bit word at offset 0 has a, b, and c and 2 fallow bits.  The 32-bit word at offset 4 has d and 22 fallow bits.

Note that bit-fields (typically) do not span across words.

If you really want the structure to be 5 bytes AND have byte alignment, you'll have to do more of the work yourself.  For example, in C++, I'd do it this way:

typedef char byte;
struct rec
{
    byte m[5]; // 40-bits to pack our 4 10-bit unsigned values.

    void setA(unsigned i_10);
    void setB(unsigned i_10);
    void setC(unsigned i_10);
    void setD(unsigned i_10);
    void setE(unsigned i_10);

    unsigned getA() const;
    unsigned getB() const;
    unsigned getC() const;
    unsigned getD() const;
    unsigned getE() const;
};

void rec::setA(unsigned i_10)
{
    m[0] = i_10 & 0xFF;
    m[1] = (m[1] & 0xFC) | ((i_10 >> 8) & 0x03);
}

unsigned rec::getA() const
{
    return (m[1] & 0x03) | (m[0] & 0xFF);
}

I leave get/set B, C, D as an exercise.  Note that the compiler's optimizer will optimize the seemingly superfluous masking where it's unnecessary, which makes this stuff fairly platform agnostic.

Another advantage is that you can read/write this structure directly to file or over a socket, on any 8-bit/byte platform, and it'll work.

HTH,
--Eljay