Re: Fractional part of decimal value.

On 08/07/2013 04:38 PM, Wouter Beek wrote:
> Hi Paulo,
>
> Thanks for the response! The functions that you mention have the same
> outcome:
> ~~~
> ?- X is float_fractional_part(1.1).
> X = 0.10000000000000009.
> ~~~
> Even:
> ~~~
> ?- X is 1.1 - 1.
> X = 0.10000000000000009.

Yes.  And they actually use C's modf() :-)  Change your program a little:

#include <stdio.h>
#include <math.h>

int main() {
   double param, fractional_part, integer_part;
   param = 1.1;
   fractional_part = modf(param, &integer_part);
   printf("%.25f = %.25f + %.25f \n", param, integer_part, fractional_part);
   return 0;
}

and we get

hppc823 (~) 8_> gcc -o f f.c
hppc823 (~) 9_> ./f
1.1000000000000000888178420 = 1.0000000000000000000000000 + 
0.1000000000000000888178420

That looks more like Prolog's outcome.  Now,

?- 0.1 =:= 0.1000000000000000888178420.
false.

While

2 ?- 1.1 =:= 1.1000000000000000888178420.
true.

SWI-Prolog's write/1 writes floats with a few as possible digits
that makes =:= succeed with the original value.  So, the claim is
that this is true for any float (where read and write nicely read
and write to/from the same file of course).

	write(X), read(Y), X=:=Y.

Possibly format/3 using ~f can help to accomplish your goal?
Floats are incredibly difficult to understand, especially
binary floats :-(

	Cheers --- Jan






> ~~~
> I understand that most decimal fractions cannot be represented by binary
> fractions, so there is probably a good reason to have the '9' appear at the
> end.
>
> My question is whether this can be circumvented for my simple/naive use.
> (Preferably a solution that does not involve a hack such as using
> format-to-codes and then number_codes/2 :-P.)