Re: Float conversion

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

On Thu, 29 Jul 2010 17:08:24 +0200, Georg Bauhaus wrote:

> On 29.07.10 14:46, Dmitry A. Kazakov wrote:
>> On Thu, 29 Jul 2010 05:23:42 -0700 (PDT), Henrique wrote:
>> 
>>> I have some problems in converting float variables in Ada. Look at the
>>> code below.
>>>
>>> Why var4 is not equal var1?
>> 
>> Because they are not. Floating-point operations are inexact.
> 
> Interestingly, a quick rewrite to C translated with gcc gives a different
> impression (of equality),
> 
[...] 
> 
> $ ./a.out
> var3: 999.900024414062500=
> var4: 999.900085449218750=

with Ada.Text_IO;  use Ada.Text_IO;
procedure Test is
  type My_Float is digits 6;

  CONVERSION_CONSTANT  : constant My_Float := 6076.11;
  CONVERSION_CONSTANT2 : constant My_Float := 1.0 / CONVERSION_CONSTANT;

  var1: My_Float := 999.9;
  var2: My_Float := var1 * CONVERSION_CONSTANT;
  var3: My_Float:= var2 / CONVERSION_CONSTANT;
  var4: My_Float := var2 * CONVERSION_CONSTANT2;
begin
  Put_Line ("var1" & Long_Float'Image (Long_Float (Var1)));
  Put_Line ("var4" & Long_Float'Image (Long_Float (Var4)));
end Test;
------------------------
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. Kazakov
http://www.dmitry-kazakov.de