Re: Float conversion

[Henrique <henriquemagalhaes@gmail.com>]

Thanks for the help (everyone). By using this concept of intervals, I
made the modifications in my code to solve the problem.

On 29 jul, 16:15, "Dmitry A. Kazakov" <mail...@dmitry-kazakov.de>
wrote:
> On Thu, 29 Jul 2010 11:21:52 -0700 (PDT), Henrique wrote:
> > On Jul 29, 12:35 pm, "Dmitry A. Kazakov" <mail...@dmitry-kazakov.de>
> > wrote:
> >> ------------------------
> >> var1 9.99900024414063E+02
> >> var4 9.99900085449219E+02
>
> >> When rounded to 6 decimal digits both are same. But the underlying base
> >> binary type is longer than 6 digits.
>
> >> P.S. It is always useful to think of floating point numbers as intervals
> >> (which they are) rather than numbers.
>
> >> --
> >> Regards,
> >> Dmitry A. Kazakovhttp://www.dmitry-kazakov.de
>
> > As I declared the type with 6 digits, I expected that it would make
> > the comparison only for these digits (this would gave var4 = var1).
>
> No, the underlying base type is more or less free compiler choice. There
> are certain rules, which in essence guarantee you accuracy of the basic
> operations within the precision specified.
>
> > Do I always need to manually truncate the float number to the number
> > of desired digits before making a comparison of them??
>
> You should never use equality or inequality for floating-point types. They
> do not have "physical" sense. As I said, floating-point numbers are
> intervals. Two non-zero length [independent] intervals are *always*
> unequal, even if their bounds are equal.
>
> > So what is the advantage of declaring it as "digits 6"?
>
> That the compiler guarantees you 6 decimal digits accuracy independently on
> whatever hardware you have. The idea behind of Ada numeric types is machine
> independence. You specify your requirements on the precision and range and
> the compiler either gives you what you wanted or else rejects the program.
>
> --
> Regards,
> Dmitry A. Kazakovhttp://www.dmitry-kazakov.de- Ocultar texto das mensagens anteriores -
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