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# 8.8 notation....

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
Nov 13 '05 #1
14 12477

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(unsigne d 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
Nov 13 '05 #2

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
Nov 13 '05 #3
"John T." <jo**@dat.com > 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
Nov 13 '05 #4
On Wed, 1 Oct 2003 08:03:12 +0200
"John T." <jo**@dat.com > 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?

algorithms which belongs else where, such as comp.programmin g 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

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.
Nov 13 '05 #5
"John T." <jo**@dat.com > 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_Keit h) ks*@cts.com <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"
Nov 13 '05 #6
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
Nov 13 '05 #7
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
Nov 13 '05 #8
Arthur J. O'Dwyer <aj*@nospam.and rew.cmu.edu> 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)cybers pace.org |
Nov 13 '05 #9
Christopher Benson-Manica <at***@nospam.c yberspace.org> scribbled the following:
Arthur J. O'Dwyer <aj*@nospam.and rew.cmu.edu> 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 (pa*****@cc.hel sinki.fi) ---------------------------\
| 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
Nov 13 '05 #10

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