E’ una semplice domanda: quanto fa 9999999999999999.0 – 9999999999999998.0 nel tuo linguaggio preferito? La risposta potrebbe sorprenderti!
Si sa che spesso le operazioni che riguardano i numeri decimali richiedono molta attenzione. Questo perchè i computer possono rappresentarli fino ad un certo grado di precisione, (semi) arbitrario ma limitato.
Vediamo quindi come si comportano vari linguaggi di programmazione di fronte a questa semplice operazione:
| Ruby: | irb(main):001:0> 9999999999999999.0 – 9999999999999998.0 2.0 |
|---|---|
| Java: | public class Foo{public static void main(String args[]){System.out.println(9999999999999999.0-9999999999999998.0);}} 2.0 |
| Python: | >>> 9999999999999999.0 – 9999999999999998.0 2.0 |
| Rebol: | >> 9999999999999999.0 – 9999999999999998.0 == 2.0 |
| Haskell: | Prelude> 9999999999999999.0 – 9999999999999998.0 2.0 |
| TCL: | % expr “9999999999999999.0-9999999999999998.0” 0.0 |
| Emacs Lisp: | ELISP> (- 9999999999999999.0 9999999999999998.0) 2.0 |
| Common–Lisp: | [1]> (- 9999999999999999.0 9999999999999998.0) 0.0 |
| Maxima: | (%i1) 9999999999999999.0-9999999999999998.0; (%o1) 2.0 |
| Google: | 0 |
| K/Q: | q)9999999999999999.0-9999999999999998.0 2f |
| R: | > 9999999999999999.0-9999999999999998.0 [1] 2 |
| Erlang: | 1> 9999999999999999.0-9999999999999998.0 . 2.0 |
| C: | main(){printf(“%lf\n”,(double)9999999999999999.0-9999999999999998.0);} 2.000000 |
| AWK: | $ awk ‘END{print 9999999999999999.0-9999999999999998.0}'</dev/null 2 |
| GoLang: | var a = 9999999999999999.0; var b = 9999999999999998.0; fmt.Printf(“%f\n”, a-b) 2.000000 |
| Perl: | $ perl -e ‘print 9999999999999999.0-9999999999999998.0;print “\n”;’ 2.0 |
| Perl6: | $ perl6 -e ‘print 9999999999999999.0-9999999999999998.0;print “\n”;’ 1 |
| Wolfram: | 1 |