Re: Fractional part of decimal value.

On 07/08/2013, at 15:14, Wouter Beek <[email protected]> wrote:

> 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
> ~~~


I assume that you're aware of the float_integer_part/1 and float_fractional_part/1 ISO Prolog standard functions but for some reason you found them not adequate for your task?

Cheers,

Paulo


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Paulo Moura
Logtalk developer

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Web:   <http://logtalk.org/>
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