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 3.141E-300 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  7972 
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 3.141E-300 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*****@acm-dot-org.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 3.141E-300 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 3.141E-300 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() work-alike, using
your knowledge of the floating-point representation. The
advantage of using Standard-like tools for the task is that
you'll easily be able to move the code to other platforms;
something like a float-to-string-without-sprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little non-portable work to get
yourself up to the frexp()-ish baseline, and write portable C
from there upwards.
Good luck!
--
Eric Sosman es*****@acm-dot-org.invalid
>
I'd suggest writing your own frexp() work-alike, using
your knowledge of the floating-point representation. The
advantage of using Standard-like tools for the task is that
you'll easily be able to move the code to other platforms;
something like a float-to-string-without-sprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little non-portable 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 3.141E-300 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() work-alike, using
your knowledge of the floating-point representation. The
advantage of using Standard-like tools for the task is that
you'll easily be able to move the code to other platforms;
something like a float-to-string-without-sprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little non-portable 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 0.3010299956639 81143
You can get the digits of f by eg iterating
d = (int)y; y = 10.0*(y-d);
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() work-alike, using
your knowledge of the floating-point representation. The
advantage of using Standard-like tools for the task is that
you'll easily be able to move the code to other platforms;
something like a float-to-string-without-sprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little non-portable 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 0.3010299956639 81143
You can get the digits of f by eg iterating
d = (int)y; y = 10.0*(y-d);
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() work-alike, using
your knowledge of the floating-point representation. The
advantage of using Standard-like tools for the task is that
you'll easily be able to move the code to other platforms;
something like a float-to-string-without-sprintf operation
seems of sufficiently wide applicability that you may well
want it again. So: invest a little non-portable 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 3.141E-300 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 3.141E-300 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.
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