Re: understanding floating point types

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

On Sun, 22 Aug 2010 02:11:59 -0700 (PDT), Ada novice wrote:

>     I'm trying to understand the floating point representation in the
> binary and decimal bases. If D is the number of decimal digits in the
> mantissa, then the number if bits B is given by B = D(1og 10/1og 2) +
> 1 = 3.32 * D + 1 and this rounded up to the next integer value.
> 
> Thus for a float type in Ada, mantissa D = 6, then B = 3.32(6) +1 = 21
> to the nearest upwards integer value. GNAT GPL 2010 shows that
> Float'Mantissa is 21 bits.
>
> Now the TOTAL number of bits in a float (Float'Size in GNAT GPL) is B
> = 32 giving the range of exponent as -4B to 4B i.e. -128 to 128. And
> this gives in decimal base a exponent (in Float'Last) of log(2^128)/
> (log 10) = 38.53, which is 38 in GNAT GPL.

Float'Machine_Mantissa = 24 =
23 explicit bits + 1 hidden bit (considering IEEE 754 single precision)

Float'Mantissa gives the model binary mantissa, the least number of bits
required to represent specified decimal mantissa (the "number B"). It is a
model attribute, a view of the real machine's number of bits, which is 24.
[*]

Compare:

   type My_Float is digits 1;

My_Float'Machine_Mantissa = 24 (32-bit IEEE 754 is used to model it)
My_Float'Mantissa = 5 (= 4+1)

   type My_Float is digits 2;

My_Float'Machine_Mantissa = 24 (32-bit IEEE 754 is used to model it)
My_Float'Mantissa = 8 (= 7+1)

etc

-----------
* Excluding denormalized numbers

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de