Bitfield

This is a discussion on Bitfield within the C Programming forums, part of the General Programming Boards category; Hi Guys, Question on a BitField. How will I check the sixth bit in a six bits storage? For example ...

  1. #1
    Registered User ssharish2005's Avatar
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    Bitfield

    Hi Guys,

    Question on a BitField. How will I check the sixth bit in a six bits storage? For example

    Code:
    struct BITF
    {
       unsigned Bits:6; 
       unsigned :2;
    };
    struct BITF Buffer;
    unsigned D;
    D = (unsigned *)&Buffer;
    *D = 0x04;
    if( Buffer.Bits & ??? )
      ...
    I know, i will have to use bit mask to check the sixth bit.

    0x04
    0000 0100
    Buffer.Bits ==> 0000 01

    how will i check the sixth bit ( LSB == 1 )?

    thanks guys!

    ~ssharish
    Last edited by ssharish2005; 10-03-2009 at 12:28 PM.
    Life is like riding a bicycle. To keep your balance you must keep moving - Einstein

  2. #2
    ATH0 quzah's Avatar
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    Check it the same way you check the 6th bit of any integer.
    Code:
    struct BITF b;
    ...
    if( b.Bits & (1<<5) )

    Quzah.
    Hope is the first step on the road to disappointment.

  3. #3
    Registered User ssharish2005's Avatar
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    This is a bit confusing. When I do the following way.

    Code:
    struct BITF
    {
       unsigned Bits:6; 
       unsigned Bits2:6;
       unsigned :4;
    };
    struct BITF Buffer;
    unsigned D;
    D = (unsigned *)&Buffer;
    *D = 0x0421;
    if( Buffer.Bits & (8>>2)) // This would give me the right results; Thanks Quzah
      ...
    But when i try o do this
    Code:
    if( Buffer.Bits2 & 0x08) // This is not giving me the right answer
    0x0421

    0000 01 | 00 0010 | 0001
    Bits = 0000 01
    Bits2 = 00 0010

    What am I doing wrong?

    Thanks a lot!

    ~ssharish
    Life is like riding a bicycle. To keep your balance you must keep moving - Einstein

  4. #4
    ATH0 quzah's Avatar
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    0x0421 doesn't mean "set the first three bits. And 8 right shifted twice is just 2. I'm not sure you know what you're trying to do. Or at least, you aren't explaining what you're trying to do very well.


    Quzah.
    Hope is the first step on the road to disappointment.

  5. #5
    Registered User ssharish2005's Avatar
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    Apologies, for not being clear.

    Think 0x0421 is some buffer value which we got over the network. From the buffer value extract the first 6 bits.

    So for example 0x0421 in binary would look as follow

    0000 0100 0010 0001

    Now I need the first 6 bits to be int Buffer.Bits and the next Six bits in Buffer.Bits2 and hence the bitfield along the structure.

    Now then, assuming all bits set and initialised to the right values. I can now access the first six bits by just Buffer.Bits.

    I need to check if the LSBit in the 6 bits is set. depending on that I'll have to do some further processing.

    I tried my best. Please let me know if i missed out something.

    Thanks Quzah!

    ~ssharish
    Life is like riding a bicycle. To keep your balance you must keep moving - Einstein

  6. #6
    and the Hat of Guessing tabstop's Avatar
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    Why isn't checking LSB just
    Code:
    (whatever & 1)
    ?

  7. #7
    ATH0 quzah's Avatar
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    Code:
    union u
    {
        unsigned int big;
        struct
        {
            unsigned int firstsix:6;
            unsigned int nextsix:6;
            :4;
        } s;
    };
    You could always use a union. Or just mask off the six you need:
    Code:
    unsigned int foo = 0x0124;
    ...
    y = (foo & ((1<<6)-1); /* or just foo & 63 */

    Quzah.
    Hope is the first step on the road to disappointment.

  8. #8
    Registered User ssharish2005's Avatar
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    Yeah masking is the probably very ideal. But the complication here is the buffer value is huge instead of just 0x0421.

    Really its something like 0x04215435432534435435432 and hence i decided to go for structure representation.

    ssharish
    Life is like riding a bicycle. To keep your balance you must keep moving - Einstein

  9. #9
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by tabstop View Post
    Why isn't checking LSB just
    Code:
    (whatever & 1)
    ?
    Seconded!
    My homepage
    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

  10. #10
    Malum in se abachler's Avatar
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    Code:
    bit6 = (whatever & 32) >> 5;
    Until you can build a working general purpose reprogrammable computer out of basic components from radio shack, you are not fit to call yourself a programmer in my presence. This is cwhizard, signing off.

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