8.8 notation....

Discussion in 'C Programming' started by John T., Oct 1, 2003.

  1. John T.

    John T. Guest

    I have a number in 8.8 notation that I wish to convert to a float.......
    8.8 notation is a 16 bit fixed notation, so:
    0xFF11 is the number 255.17
    0x0104 is the number 1.4
    0x2356 is the number 35.86
    and so on....
    Any ideas on how to convert this into a float?

    John
     
    John T., Oct 1, 2003
    #1
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  2. On Wed, 1 Oct 2003, John T. wrote:
    >
    > I have a number in 8.8 notation that I wish to convert to a float.......
    > 8.8 notation is a 16 bit fixed notation, so:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86
    > and so on....
    > Any ideas on how to convert this into a float?


    First of all, you *do* realize that your "8.8 notation"
    as specified is not a one-to-one function, right? For
    example, if 0x0104 is equal to 1.4, then so is 0x008C.
    But here you go, as requested...

    /* the original number */
    unsigned int eight_eight = 0x0104;

    /* extract the fields */
    unsigned int int_part = eight_eight >> 8;
    unsigned int frac_part = eight_eight & 0xFF;

    /* put it back together right-ways */
    double result = int_part + frac_part/100.0;

    (Or in a function:)

    double from_88(unsigned int val)
    {
    /* no error-checking on range of 'val' */
    return (val>>8) + (val & 0xFF)/100.;
    }

    unsigned int to_88(double val)
    {
    /* no error-checking on range of 'val' */
    unsigned int ival = (unsigned int) val;
    unsigned int fval = (unsigned int) (100*(val-ival) + 0.5);
    return (ival << 8) + fval;
    }


    HTH,
    -Arthur
     
    Arthur J. O'Dwyer, Oct 1, 2003
    #2
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  3. On Wed, 1 Oct 2003, Arthur J. O'Dwyer wrote:
    >
    > On Wed, 1 Oct 2003, John T. wrote:
    > >
    > > I have a number in 8.8 notation that I wish to convert to a float.......
    > > 8.8 notation is a 16 bit fixed notation, so:
    > > 0xFF11 is the number 255.17
    > > 0x0104 is the number 1.4
    > > 0x2356 is the number 35.86

    >
    > First of all, you *do* realize that your "8.8 notation"
    > as specified is not a one-to-one function, right? For
    > example, if 0x0104 is equal to 1.4, then so is 0x008C.


    Wait a minute... I realized as soon as I sent that message
    that I can't think of *any* reasonable way to get 0x0104
    from the floating-point value 1.4! If 0x0104 is 1.4, then
    what's 1.04? And what does 0x0140 represent in your
    notation?

    If you really didn't make a typo there, this could be
    a *really* interesting format...

    My solution assumes you meant 0x140 <==> 1.4, BTW.

    -Arthur
     
    Arthur J. O'Dwyer, Oct 1, 2003
    #3
  4. John T.

    Mark Gordon Guest

    On Wed, 1 Oct 2003 08:03:12 +0200
    "John T." <> wrote:

    > I have a number in 8.8 notation that I wish to convert to a
    > float....... 8.8 notation is a 16 bit fixed notation, so:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86
    > and so on....
    > Any ideas on how to convert this into a float?


    So what is your question about C? All I see is a question about
    algorithms which belongs else where, such as comp.programming possibly.

    However, to get you started...

    First check your specification. I would expect 0x0104 to be 1.015625
    (this is the type of fixed point notation I've always found).

    If I am right you just have to do floating point division by 256. If you
    are write you have to do some masking to seperate the two octets, scale
    them then add them together.

    Post here when you have attempted to write the C code and hit problems.
    --
    Mark Gordon
    Paid to be a Geek & a Senior Software Developer
    Although my email address says spamtrap, it is real and I read it.
     
    Mark Gordon, Oct 1, 2003
    #4
  5. John T.

    Richard Bos Guest

    "John T." <> wrote:

    > I have a number in 8.8 notation that I wish to convert to a float.......
    > 8.8 notation is a 16 bit fixed notation, so:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86
    > and so on....
    > Any ideas on how to convert this into a float?


    Yes: by hand. C has no function for this, since fixed-point types are
    not part of C. However, it wouldn't be hard to write one yourself.
    Hints: %, /, 0x100 and/or 0xFF, (float), +.

    Richard
     
    Richard Bos, Oct 1, 2003
    #5
  6. "John T." <> writes:
    > I have a number in 8.8 notation that I wish to convert to a float.......
    > 8.8 notation is a 16 bit fixed notation, so:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86
    > and so on....
    > Any ideas on how to convert this into a float?


    Not without a better specification of your notation.

    I'd normally expect a 16-bit fixed-point notation to have the
    high-order 8 bits represent the integer part and the low-order 8 bits
    represent the fractional part, with 0x0001 representing 1.0/256.0,
    0x00FF representing 255.0/256.0, etc. In that case, you'd simply
    multiply by 256.0.

    Or you could have the low-order 8 bits represent a multiple of 0.01
    (decimal) rather than of 1.0/256.0; in that case, you'd have to
    extract the high-order and low-order parts and do a little arithmetic.
    In such a notation, though, 0x0104 would be 1.04, not 1.4; was that
    just a typo? Such a notation is a bit inefficient, since you're using
    8 bits to represent any of 100 values rather than 256 values. If
    that's what you're dealing with, though, it's merely odd, not wrong.

    Keep in mind that some fractional values, such as 0.1, cannot be
    represented exactly in binary floating-point. Note also that your
    format has no way of representing negative numbers.

    --
    Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
    San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
    Schroedinger does Shakespeare: "To be *and* not to be"
     
    Keith Thompson, Oct 1, 2003
    #6
  7. John T.

    pete Guest

    John T. wrote:
    >
    > I have a number in 8.8 notation that I wish to convert to a float.......
    > 8.8 notation is a 16 bit fixed notation, so:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86
    > and so on....
    > Any ideas on how to convert this into a float?


    Divide the unsigned short value by 256.0
    If you don't like the resulting decimal fraction,
    then explain it better.

    0xff11 is 255.066406
    0x104 is 1.015625
    0x2356 is 35.335938
    0xffff is 255.996094

    --
    pete
     
    pete, Oct 1, 2003
    #7
  8. John T.

    Martijn Guest

    Keith Thompson wrote:
    > Or you could have the low-order 8 bits represent a multiple of 0.01
    > (decimal) rather than of 1.0/256.0; in that case, you'd have to
    > extract the high-order and low-order parts and do a little arithmetic.
    > In such a notation, though, 0x0104 would be 1.04, not 1.4; was that
    > just a typo? Such a notation is a bit inefficient, since you're using
    > 8 bits to represent any of 100 values rather than 256 values. If
    > that's what you're dealing with, though, it's merely odd, not wrong.
    >
    > Keep in mind that some fractional values, such as 0.1, cannot be
    > represented exactly in binary floating-point. Note also that your
    > format has no way of representing negative numbers.


    In the latter case (the innefficient approach) the "unused" bit can be used
    to denote the sign. Just to make things even weirder :)

    0x0140 = 1.40
    0x01C0 = -1.40

    But given the amount of contraints he as already put on his range, I think
    the fact that it will be able to hold a sign is apparently out of the
    question. :)

    --
    Martijn
    http://www.sereneconcepts.nl
     
    Martijn, Oct 1, 2003
    #8
  9. Arthur J. O'Dwyer <> spoke thus:

    > Wait a minute... I realized as soon as I sent that message
    > that I can't think of *any* reasonable way to get 0x0104
    > from the floating-point value 1.4! If 0x0104 is 1.4, then
    > what's 1.04? And what does 0x0140 represent in your
    > notation?


    If I'm not mistaken, 0x0140 is 1.64 in his notation.

    --
    Christopher Benson-Manica | Jumonji giri, for honour.
    ataru(at)cyberspace.org |
     
    Christopher Benson-Manica, Oct 1, 2003
    #9
  10. Christopher Benson-Manica <> scribbled the following:
    > Arthur J. O'Dwyer <> spoke thus:
    >> Wait a minute... I realized as soon as I sent that message
    >> that I can't think of *any* reasonable way to get 0x0104
    >> from the floating-point value 1.4! If 0x0104 is 1.4, then
    >> what's 1.04? And what does 0x0140 represent in your
    >> notation?


    > If I'm not mistaken, 0x0140 is 1.64 in his notation.


    This would make 0x0101 and 0x010A the same thing, yesno? And there
    would be no way to represent 1.01, yesno?
    Either his "." means a different thing than the fraction separator in
    decimal, he has made a few typos, or he doesn't understand his own
    function.

    --
    /-- Joona Palaste () ---------------------------\
    | Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
    | http://www.helsinki.fi/~palaste W++ B OP+ |
    \----------------------------------------- Finland rules! ------------/
    "All that flower power is no match for my glower power!"
    - Montgomery Burns
     
    Joona I Palaste, Oct 1, 2003
    #10
  11. On Wed, 1 Oct 2003, John T. wrote:

    > I have a number in 8.8 notation that I wish to convert to a float.......
    > 8.8 notation is a 16 bit fixed notation, so:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86
    > and so on....
    > Any ideas on how to convert this into a float?


    Your examples don't tell me how I deal with numbers like:

    0x0064

    Is this 0.100 or is this 1.00?

    Obviously, the first step is seperating the number into the whole number
    and the fraction:

    /* let n be the number we are dealing with */
    float result;
    int whole = (n >> 8) & 0xff;
    int fraction = n & 0xff;

    If the first method is the desired result you would need:

    if(fraction > 99)
    result = whole + fraction/100.0;
    else
    result = whole + fraction/10.0;

    If the second method is the desired result you would need:

    result = whole + fraction/10.0;

    --
    Send e-mail to: darrell at cs dot toronto dot edu
    Don't send e-mail to
     
    Darrell Grainger, Oct 1, 2003
    #11
  12. Joona I Palaste <> spoke thus:

    > This would make 0x0101 and 0x010A the same thing, yesno?


    Hm, yes, if he's trying to convert these to floats.

    > And there
    > would be no way to represent 1.01, yesno?


    Again, yes, although I'm not sure he wants to. The format itself makes sense
    (although I don't know what one would use it for).

    > Either his "." means a different thing than the fraction separator in
    > decimal, he has made a few typos, or he doesn't understand his own
    > function.


    I'm betting on the first one, but I'm not sure...

    --
    Christopher Benson-Manica | Jumonji giri, for honour.
    ataru(at)cyberspace.org |
     
    Christopher Benson-Manica, Oct 1, 2003
    #12
  13. John T.

    David Rubin Guest

    Christopher Benson-Manica wrote:
    >
    > Joona I Palaste <> spoke thus:
    >
    > > This would make 0x0101 and 0x010A the same thing, yesno?

    >
    > Hm, yes, if he's trying to convert these to floats.
    >
    > > And there
    > > would be no way to represent 1.01, yesno?

    >
    > Again, yes, although I'm not sure he wants to. The format itself makes sense
    > (although I don't know what one would use it for).


    The format is kind of lame IMO. Why not use 0x[0-9A-Z]{2}[0-9]{2} to
    represent 000.00 to 255.99? This is only slightly less efficient, but
    much more straightforward.

    /david

    --
    Andre, a simple peasant, had only one thing on his mind as he crept
    along the East wall: 'Andre, creep... Andre, creep... Andre, creep.'
    -- unknown
     
    David Rubin, Oct 1, 2003
    #13
  14. David Rubin <> scribbled the following:
    > Christopher Benson-Manica wrote:
    >> Joona I Palaste <> spoke thus:
    >> > This would make 0x0101 and 0x010A the same thing, yesno?

    >>
    >> Hm, yes, if he's trying to convert these to floats.
    >>
    >> > And there
    >> > would be no way to represent 1.01, yesno?

    >>
    >> Again, yes, although I'm not sure he wants to. The format itself makes sense
    >> (although I don't know what one would use it for).


    > The format is kind of lame IMO. Why not use 0x[0-9A-Z]{2}[0-9]{2} to
    > represent 000.00 to 255.99? This is only slightly less efficient, but
    > much more straightforward.


    Because (10+26) squared is 1296, not 256?

    --
    /-- Joona Palaste () ---------------------------\
    | Kingpriest of "The Flying Lemon Tree" G++ FR FW+ M- #108 D+ ADA N+++|
    | http://www.helsinki.fi/~palaste W++ B OP+ |
    \----------------------------------------- Finland rules! ------------/
    "A computer program does what you tell it to do, not what you want it to do."
    - Anon
     
    Joona I Palaste, Oct 1, 2003
    #14
  15. Joona I Palaste <> writes:
    > Christopher Benson-Manica <> scribbled
    > the following:
    > > Arthur J. O'Dwyer <> spoke thus:
    > >> Wait a minute... I realized as soon as I sent that message
    > >> that I can't think of *any* reasonable way to get 0x0104
    > >> from the floating-point value 1.4! If 0x0104 is 1.4, then
    > >> what's 1.04? And what does 0x0140 represent in your
    > >> notation?

    >
    > > If I'm not mistaken, 0x0140 is 1.64 in his notation.

    >
    > Either his "." means a different thing than the fraction separator in
    > decimal, he has made a few typos, or he doesn't understand his own
    > function.


    The OP wrote:
    > 0xFF11 is the number 255.17
    > 0x0104 is the number 1.4
    > 0x2356 is the number 35.86


    Assuming that 1.4 was a typo, and it should be 1.04, the notation
    looks odd but consistent. The high-order 8 bits are the integer part;
    the low-order 8 bits are the fractional part multiplied by 100.0
    decimal. One advantage of such a notation is that it can represent
    decimal fractions such as 0.1 or 0.01 exactly; of course, converting
    to binary floating-point loses that property.

    > This would make 0x0101 and 0x010A the same thing, yesno? And there
    > would be no way to represent 1.01, yesno?


    Nononono. 0x0101 is 1.01, and 0x010A is 1.10.

    It would mean that 0x0164 and 0x0200 would both represent 2.00, but
    presumably 0x0164 would be the canonical normalized representation.

    I suggest we wait for the OP to come back and tell us what he really
    meant.

    --
    Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
    San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
    Schroedinger does Shakespeare: "To be *and* not to be"
     
    Keith Thompson, Oct 1, 2003
    #15
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