Re: Float conversion

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

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. Kazakov
http://www.dmitry-kazakov.de