8.8 notation....

[John T.]

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

 

[Arthur J. O'Dwyer]

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]

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

 

[Mark Gordon]

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.

 

[Richard Bos]

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

 

[Keith Thompson]

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.

 

[pete]

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

 

[Martijn]

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.

 

[Christopher Benson-Manica]

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.

 

[Joona I Palaste]

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 ([email protected]) ---------------------------\
| 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

 

[Darrell Grainger]

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;

 

[Christopher Benson-Manica]

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...

 

[David Rubin]

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


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

 

[Joona I Palaste]

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 ([email protected]) ---------------------------\
| 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

 

[Keith Thompson]

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.

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.