Re: Fractional part of decimal value.

Hi Wouter


when I had to compute numbers with fixed precisions (for instance,
presenting prices after taxes or discounts) I resorted to 'hardcode' the
precision in this way (here 2 decimal digits - actually € cents)

...
TwoDec is round(NoIva * 100) / 100,
...

or, using lambda:

...
FmtNum = \I^O^(T is round(I * 100) / 100, atom_number(A, T),
replace_word('.',',',A,O)),
...
   call(FmtNum, DeltaG, DeltaF),
   call(FmtNum, Rate, RateF),
   call(FmtNum, Ivato, IvatoF),
   call(FmtNum, Cost, CostF),
...

HTH



2013/8/7 Wouter Beek <[email protected]>

> 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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