Re: Fractional part of decimal value.

Wouter, this is simple floating point error:

1 ?- X is 1.1-1.0.
X = 0.10000000000000009.

You cannot avoid these while you use floating point.  Neither 1.1 nor 0.1 have 
a finite binary representation, and when subtracting as you do, you lose a level 
of precision in your result.

I don't know about the C code, but I'll guess maybe it 
also has the 9 but the number isn't displayed to the same precision.

Probably your solution is to simply reduce the amount of decimals displayed.
Or use whole numbers or fractions.

Alan Baljeu


----- Original Message -----
From: Wouter Beek <[email protected]>
To: SWI-Prolog <[email protected]>
Cc: 
Sent: Wednesday, August 7, 2013 10:14:10 AM
Subject: [SWIPL] Fractional part of decimal value.

Hi all,

I want to extract the integer and fractional part of a decimal value:
~~~{.pl}
decimal_parts(D, I, F):-
  I is floor(D / 1),
  F is D - I * 1.
~~~
What I get is the fractional part of the float value 1.1:
~~~
?- decimal_parts(1.1, _, F).
F = 0.10000000000000009.
~~~
I do not mind the padding zero's (just a notational difference), but I do
not need the 9 at the end.

Is there a way to get the same result as with e.g. C's modf (sample code
below)? Thanks for any suggestions!

---
Cheers,
Wouter.

Sample code in C:
~~~{.c}
#include <stdio.h>
#include <math.h>

int main() {
  double param, fractional_part, integer_part;
  param = 1.1;
  fractional_part = modf(param, &integer_part);
  printf("%f = %f + %f \n", param, integer_part, fractional_part);
  return 0;
}
~~~

Compile and run:
~~~
$ gcc -Wall test.c -o test
$ ./test
1.100000 = 1.000000 + 0.100000
~~~

E-mail: [email protected]
WWW: www.wouterbeek.com
Tel.: 0647674624
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