Hello,
I want to efficient convert floating point numbers (IEEE754) into a
string. I have no library routines that do the job (like sprintf etc.),
because I work in an embedded environment.
My actual algorithm uses multiplying with 10 to shift the fraction into
an integer value and to aquire the used exponent. But the drawback is
obvious: When I have very small numbers like 3.141E300 I have to make
300 time consuming floating point multiplies to convert this number.
But, since I know the IEEE754 structure and have directly access to the
exponent (of base 2) of a fp number, is there a faster method to convert
fp numbers to ASCII?
Regards
Peter 14 7500
Peter Sprenger wrote:
Hello,
I want to efficient convert floating point numbers (IEEE754) into a
string. I have no library routines that do the job (like sprintf etc.),
because I work in an embedded environment.
My actual algorithm uses multiplying with 10 to shift the fraction into
an integer value and to aquire the used exponent. But the drawback is
obvious: When I have very small numbers like 3.141E300 I have to make
300 time consuming floating point multiplies to convert this number.
But, since I know the IEEE754 structure and have directly access to the
exponent (of base 2) of a fp number, is there a faster method to convert
fp numbers to ASCII?
If you have the frexp() function available, that might
make a good starting point.

Eric Sosman es*****@acmdotorg.invalid
Eric Sosman wrote:
Peter Sprenger wrote:
>Hello,
I want to efficient convert floating point numbers (IEEE754) into a string. I have no library routines that do the job (like sprintf etc.), because I work in an embedded environment.
My actual algorithm uses multiplying with 10 to shift the fraction into an integer value and to aquire the used exponent. But the drawback is obvious: When I have very small numbers like 3.141E300 I have to make 300 time consuming floating point multiplies to convert this number.
But, since I know the IEEE754 structure and have directly access to the exponent (of base 2) of a fp number, is there a faster method to convert fp numbers to ASCII?
If you have the frexp() function available, that might
make a good starting point.
Nope, no frexp() available either.
Peter Sprenger wrote:
Eric Sosman wrote:
>Peter Sprenger wrote:
>>Hello,
I want to efficient convert floating point numbers (IEEE754) into a string. I have no library routines that do the job (like sprintf etc.), because I work in an embedded environment.
My actual algorithm uses multiplying with 10 to shift the fraction into an integer value and to aquire the used exponent. But the drawback is obvious: When I have very small numbers like 3.141E300 I have to make 300 time consuming floating point multiplies to convert this number.
But, since I know the IEEE754 structure and have directly access to the exponent (of base 2) of a fp number, is there a faster method to convert fp numbers to ASCII?
If you have the frexp() function available, that might make a good starting point.
Nope, no frexp() available either.
I'd suggest writing your own frexp() workalike, using
your knowledge of the floatingpoint representation. The
advantage of using Standardlike tools for the task is that
you'll easily be able to move the code to other platforms;
something like a floattostringwithoutsprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little nonportable work to get
yourself up to the frexp()ish baseline, and write portable C
from there upwards.
Good luck!

Eric Sosman es*****@acmdotorg.invalid
>
I'd suggest writing your own frexp() workalike, using
your knowledge of the floatingpoint representation. The
advantage of using Standardlike tools for the task is that
you'll easily be able to move the code to other platforms;
something like a floattostringwithoutsprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little nonportable work to get
yourself up to the frexp()ish baseline, and write portable C
from there upwards.
Good luck!
I was a little bit quick. In fact I can write a frexp() myself, that
separates exponent and mantissa. But then? I have a mantissa and an
exponent of base 2. I have no right idea to transform it from here to
an ascii string.
Regards
Peter
On Thu, 24 May 2007 14:44:39 +0200, Peter Sprenger wrote:
Hello,
I want to efficient convert floating point numbers (IEEE754) into a
string. I have no library routines that do the job (like sprintf etc.),
because I work in an embedded environment.
My actual algorithm uses multiplying with 10 to shift the fraction into
an integer value and to aquire the used exponent. But the drawback is
obvious: When I have very small numbers like 3.141E300 I have to make
300 time consuming floating point multiplies to convert this number.
But, since I know the IEEE754 structure and have directly access to the
exponent (of base 2) of a fp number, is there a faster method to convert
fp numbers to ASCII?
Regards
Peter
If you have log, pow and floor then given a positive x
double log10 = log(x)/log(10.0);
int f = floor( log10);
double y = x*pow( 10.0, f);
gets you y with 1<=y<10 and x = y * pow(10,f);
Duncan
On Thu, 24 May 2007 16:46:06 +0200, Peter Sprenger wrote:
>> I'd suggest writing your own frexp() workalike, using your knowledge of the floatingpoint representation. The advantage of using Standardlike tools for the task is that you'll easily be able to move the code to other platforms; something like a floattostringwithoutsprintf operation seems of sufficiently wide applicability that you may well want it again. So: invest a little nonportable work to get yourself up to the frexp()ish baseline, and write portable C from there upwards.
Good luck!
I was a little bit quick. In fact I can write a frexp() myself, that
separates exponent and mantissa. But then? I have a mantissa and an
exponent of base 2. I have no right idea to transform it from here to
an ascii string.
Regards
Peter
If x is positive
int expon;
double frac = frexp(x, &expon);
int f = floor( expon*log_10_2);
double y = x*pow( 10.0, f);
gets you y with 1<=y<10 and x = y*pow(10,f)
here log_10_2 is log base 10 of 2, ie around 0.301029995663981143
You can get the digits of f by eg iterating
d = (int)y; y = 10.0*(yd);
Note that you're only doing this for the number of digits required.
If you don't have pow (and maybe even if you do) it can be written
fairly easily & efficiently since its second argument is an integer.
Duncan
On Thu, 24 May 2007 16:57:28 +0100, Duncan Muirhead wrote:
On Thu, 24 May 2007 16:46:06 +0200, Peter Sprenger wrote:
>>> I'd suggest writing your own frexp() workalike, using your knowledge of the floatingpoint representation. The advantage of using Standardlike tools for the task is that you'll easily be able to move the code to other platforms; something like a floattostringwithoutsprintf operation seems of sufficiently wide applicability that you may well want it again. So: invest a little nonportable work to get yourself up to the frexp()ish baseline, and write portable C from there upwards.
Good luck! I was a little bit quick. In fact I can write a frexp() myself, that separates exponent and mantissa. But then? I have a mantissa and an exponent of base 2. I have no right idea to transform it from here to an ascii string.
Regards
Peter
If x is positive
int expon;
double frac = frexp(x, &expon);
int f = floor( expon*log_10_2);
double y = x*pow( 10.0, f);
gets you y with 1<=y<10 and x = y*pow(10,f)
here log_10_2 is log base 10 of 2, ie around 0.301029995663981143
You can get the digits of f by eg iterating
d = (int)y; y = 10.0*(yd);
Note that you're only doing this for the number of digits required.
If you don't have pow (and maybe even if you do) it can be written
fairly easily & efficiently since its second argument is an integer.
Duncan
Oops! Sorry, that's not quite right. In fact f above could be one
too small, and so we could have 10<=y<=100. I think the easiest thing
too do is to check for y being at least10, and if so fix it up.
Duncan
Peter Sprenger wrote On 05/24/07 10:46,:
> I'd suggest writing your own frexp() workalike, using your knowledge of the floatingpoint representation. The advantage of using Standardlike tools for the task is that you'll easily be able to move the code to other platforms; something like a floattostringwithoutsprintf operation seems of sufficiently wide applicability that you may well want it again. So: invest a little nonportable work to get yourself up to the frexp()ish baseline, and write portable C from there upwards.
Good luck!
I was a little bit quick. In fact I can write a frexp() myself, that
separates exponent and mantissa. But then? I have a mantissa and an
exponent of base 2. I have no right idea to transform it from here to
an ascii string.
You wrote originally about the time eaten up by the
very many multiplications by ten needed to scale a value
like 3.141E300 to a reasonable range. I'm suggesting
that you use the exponent of two to figure out how many
"decades" of scaling you need, and do them all in one
multiplication. You could use a precomputed array with
the exponents as indices, or multiply the two's exponent
by log10(2) and do a little rounding and/or truncating
to get the ten's exponent.
x = m * 2**e
= m * (10**log10(2))**e
= m * 10**(log10(2)*e)
= m * 10**f
= m * 10**floor(f) * 10**(f  floor(f))
= (m * 10**(f  floor(f))) * 10**floor(f)
 Er*********@sun.com
>
You wrote originally about the time eaten up by the
very many multiplications by ten needed to scale a value
like 3.141E300 to a reasonable range. I'm suggesting
that you use the exponent of two to figure out how many
"decades" of scaling you need, and do them all in one
multiplication. You could use a precomputed array with
the exponents as indices, or multiply the two's exponent
by log10(2) and do a little rounding and/or truncating
to get the ten's exponent.
x = m * 2**e
= m * (10**log10(2))**e
= m * 10**(log10(2)*e)
= m * 10**f
= m * 10**floor(f) * 10**(f  floor(f))
= (m * 10**(f  floor(f))) * 10**floor(f)
Hello Eric,
you are right,
x = m * 2**e
is correct. But if I have use log10 in some way, isn't it easier to
directly get the exponent with directly log10(f) ? (f is my fp number)
And then multiply f to get it in the range 1.0 <= f < 10.0 ?
Your solution will not bring the answer on how many decimal places in
the fraction f has. So I have to multiply it anyway in turns to get the
decimal places after the comma.
Peter
>>Peter
If x is positive int expon; double frac = frexp(x, &expon); int f = floor( expon*log_10_2); double y = x*pow( 10.0, f); gets you y with 1<=y<10 and x = y*pow(10,f) here log_10_2 is log base 10 of 2, ie around 0.301029995663981143
You can get the digits of f by eg iterating d = (int)y; y = 10.0*(yd); Note that you're only doing this for the number of digits required.
If you don't have pow (and maybe even if you do) it can be written fairly easily & efficiently since its second argument is an integer. Duncan
Oops! Sorry, that's not quite right. In fact f above could be one
too small, and so we could have 10<=y<=100. I think the easiest thing
too do is to check for y being at least10, and if so fix it up.
Duncan
Hello Duncan,
you know the IEEE754 floating point format? The exponent says, where in
the mantissa the decimal point is (or should I say the binary point :)
) and the fraction part begins.
Your terms don't use the mantissa, so I think they are not correct. But
like I wrote to Eric, if I can make a log10(x), I can bring x into
1<= x < 10 with one multiplication.
Peter
On Fri, 25 May 2007 09:28:13 +0200, Peter Sprenger wrote:
>
>>>Peter If x is positive int expon; double frac = frexp(x, &expon); int f = floor( expon*log_10_2); double y = x*pow( 10.0, f); gets you y with 1<=y<10 and x = y*pow(10,f) here log_10_2 is log base 10 of 2, ie around 0.301029995663981143
You can get the digits of f by eg iterating d = (int)y; y = 10.0*(yd); Note that you're only doing this for the number of digits required.
If you don't have pow (and maybe even if you do) it can be written fairly easily & efficiently since its second argument is an integer. Duncan
Oops! Sorry, that's not quite right. In fact f above could be one too small, and so we could have 10<=y<=100. I think the easiest thing too do is to check for y being at least10, and if so fix it up. Duncan
Hello Duncan,
you know the IEEE754 floating point format? The exponent says, where in
the mantissa the decimal point is (or should I say the binary point :)
) and the fraction part begins.
Your terms don't use the mantissa, so I think they are not correct. But
like I wrote to Eric, if I can make a log10(x), I can bring x into
1<= x < 10 with one multiplication.
Peter
Indeed, you want a log10 and a pow. However all you really need is the
largest integer less than log10(x). Above I was musing about how you could
do that if you had frexp.
The point is that if
y = frexp( x, &p)
then we have x = pow(2,p)*y and 0.5<=y<=1
Taking logs (base 10) of this
log10(x) = p*log10(2) + log10(y)
but log10(2)<=log10(y) <= 0
so the largest integer less than log10(x) is either the largest integer
less than p*log10(2), or one less than this, You can figure out which
by computing f = floor( p*log10) and then computing y = x*pow(10,p).
if 1<=y<10 we're done; otherwise (ie 0.1<=y<10) multiply y by 0.1
and increase p by one.
Duncan
Peter Sprenger wrote:
>> You wrote originally about the time eaten up by the very many multiplications by ten needed to scale a value like 3.141E300 to a reasonable range. I'm suggesting that you use the exponent of two to figure out how many "decades" of scaling you need, and do them all in one multiplication. You could use a precomputed array with the exponents as indices, or multiply the two's exponent by log10(2) and do a little rounding and/or truncating to get the ten's exponent.
x = m * 2**e = m * (10**log10(2))**e = m * 10**(log10(2)*e) = m * 10**f = m * 10**floor(f) * 10**(f  floor(f)) = (m * 10**(f  floor(f))) * 10**floor(f)
Hello Eric,
you are right,
x = m * 2**e
is correct. But if I have use log10 in some way, isn't it easier to
directly get the exponent with directly log10(f) ? (f is my fp number)
And then multiply f to get it in the range 1.0 <= f < 10.0 ?
You only need an approximation to log10(f) to determine
the initial scaling amount. You can get that by extracting
the binary exponent (which is close to lg(f)) and multiplying
by the precomputed constant log10(2), then truncating or
rounding to a nearby integer.
Your solution will not bring the answer on how many decimal places in
the fraction f has. So I have to multiply it anyway in turns to get the
decimal places after the comma.
Yes, but you avoid the nearly three hundred multiplications
by ten that work through the leading zeroes of 3.141E300.
If you get to 0.1 <= abs(f)*scale < 1 in one step, you only
need as many additional multiplications as you want digits.[*]
[*] Optimization: multiply by 1E8 instead of by 10, and
work on groups of eight digits at a time in `long' instead
of one digit at a time in floatingpoint.

Eric Sosman es*****@acmdotorg.invalid
Eric Sosman wrote:
Peter Sprenger wrote: *** and deleted attributions ***
>>> You wrote originally about the time eaten up by the very many multiplications by ten needed to scale a value like 3.141E300
.... snip ...
>> is correct. But if I have use log10 in some way, isn't it easier to directly get the exponent with directly log10(f) ? (f is my fp number) And then multiply f to get it in the range 1.0 <= f < 10.0 ?
.... snip ...
>[*] Optimization: multiply by 1E8 instead of by 10, and work on
groups of eight digits at a time in `long' instead of one digit
at a time in floatingpoint.
To maintain accuracy, that 1E8 should be some value that is
expressed exactly in the FP system in use, and as large as
possible. For division, use the same value, and not it's inverse
(which will not be exact in most systems).

<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
<http://kadaitcha.cx/vista/dogsbreakfast/index.html>
cbfalconer at maineline dot net

Posted via a free Usenet account from http://www.teranews.com
Peter Sprenger wrote:
>>If x is positive int expon; double frac = frexp(x, &expon); int f = floor( expon*log_10_2); double y = x*pow( 10.0, f); gets you y with 1<=y<10 and x = y*pow(10,f) here log_10_2 is log base 10 of 2, ie around 0.301029995663981143
You can get the digits of f by eg iterating d = (int)y; y = 10.0*(yd); Note that you're only doing this for the number of digits required.
If you don't have pow (and maybe even if you do) it can be written fairly easily & efficiently since its second argument is an integer. Duncan
Oops! Sorry, that's not quite right. In fact f above could be one too small, and so we could have 10<=y<=100. I think the easiest thing too do is to check for y being at least10, and if so fix it up.
you know the IEEE754 floating point format? The exponent says, where in
the mantissa the decimal point is (or should I say the binary point :)
) and the fraction part begins.
Your terms don't use the mantissa, so I think they are not correct. But
like I wrote to Eric, if I can make a log10(x), I can bring x into
1<= x < 10 with one multiplication.
Peter,
Duncan has a good approach. It uses an frexp(), which you can write, a
floating multiplication and conversion to integer, then the pow()
function, which can be implemented with some combination of table
lookups and multiplication. And, as Duncan noted later, you may need to
do a final adjustment on the power of 10.
In general, the larger the lookup table, the fewer multiplications that
are needed. For example, there are 81 different powers of 10 from 1e40
to 1e+40. You can use a table of powers of 1e9 from 1e36 to 1e+36,
then powers of 10 from 1e4 to 1e8. With 22 entries and one
multiplication you get any power of 10 from 1e40 to 1e+40. Or you can
use a table with 81 entries and no multiplication. There are other
feasible combinations, as well.

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