Re: How to put 200 into an integer sub-type of 16 bits (code included)

[Michael Bode <m.g.bode@web.de>]

ChristopherL <clusardi2k@aol.com> writes:

> The size of floating point number is 32 bits on my system, and my
> float will
> always be positive.
>
> General information about floating point numbers in general:
>
> Floating-point numbers are typically packed into a computer datum as
> the sign bit, the exponent field, and the significand (mantissa).
>
>   Mantissa  Exponent  Value
>   71        0           71
>   71        1          710
>   71        2         7100
>   71        -1           7.1
>   2         2          200
>
> For example, it was shown above that π, rounded to 24 bits of
> precision, has:
>
> sign = 0 ; e = 1 ; s = 110010010000111111011011 (including the hidden
> bit)
> The sum of the exponent bias (127) and the exponent (1) is 128, so
> this is represented in single precision format as
>
> 0 10000000 10010010000111111011011 (excluding the hidden bit).
>
> So, what is the proper way to store a number (never being greater
> than
> 200.5) in a 8 bit short number as outlined above?

Try compressing it with gzip? Your question is how to store 4 bytes in
1 byte. The answer is, you don't. You use 4 8-bit numbers to store 1
32-bit number. In Ada you can do it like this:

with Ada.Unchecked_Conversion;
with Ada.Text_Io;

procedure Convert is
   type SByte is range -128 .. 127;
   for SByte'Size use 8;
   type Float_Bytes is array (0 .. 3) of SByte;

   Arg : Float := 200.5;
   Result : Float_Bytes;

   function To_Bytes is new Ada.Unchecked_Conversion (Source => Float,
                                                      Target => Float_Bytes);
begin
   Result := To_Bytes (Arg);
   for I in Result'Range loop
      Ada.Text_Io.Put (Sbyte'Image (Result(I)) & " ");
   end loop;
   Ada.Text_Io.New_Line;
end Convert;

Output:

 0 -128  72  67