Tony Grimer wrote:
>
> At 10:38 11/03/97 +0200, you wrote:
> >I have been trying to figure out the way how Microchip has
> >adopted the IEEE floating point math to their format they use in the
> >pic 16 range.
> >I have got the floating point math application notes which is fine if you
> >know how to convert the answer back to legable values without
> >using the float to int convertion routines.
> >Can anyone give me help how to convert the 32 bit fp. to a
> >understandable value and the other way arround aswell.
>
> IEEE 32 numbers are easy really if you understand the following
>
> The 32 bits are stored as
>
> 1 bit sign
> 8 bits exponent
> 23 Mantissa
>
> The mantissa is held in the range 1.0000--- 23 zeros in total
>                                   1.1111--- 23 ones in total
>
> Notice the 1. is not stored because it is always present so you have to put
> it back
>
> The exponent is held in biased 127 form.. this is to ensure it monotonic and
> makes manipulation simpler. It is NOT a 2's complement number which is one
> of the most common mistakes.
>
> The sign is the sign of the Number. So we can assume a Signed Magnitude
 overall.
>
> Thats a lot of words lets look at a simple example.
>
> If the 32 bits held
>
>         1 01111111  11000000000000000000000
>
> Sign = 1 so number is negative
>
> Mantissa is   1.11000000000000000000000   We have added the 1. back...
>
> Biased Exponent is as unsigned value 127
>
> So putting this together  1.11 x 2^ 127 -127  Note: remove the bias
>
>                           1.11 x 2^0
>                           1.11 x 1
>
>         In decimal this would be 1. 0.5+0.25 for the two '1's
>                                   1.75 x 1
> The whole thing is negative so ANSWER is -1.75.
>
> Taking this the other way if we started with say the number
>
>                                 0.5 decimal
>
> Sign is positive so sign bit is a 0
>
> Mantissa is 0.1 x 2^0
>
> Normalise this to 1.0 x 2^-1
>
> Bias the exponent so 2^-1 + 127 = 2^126
>
> Mantissa 23 bits = 0000000000000000000000  the 1. is not stored.
>
> Exponent is 126 in binary 8 bits unsigned = 01111110
>
> Final 32 bit are
>      0 01111110 0000000000000000000000
>
> Special cases ..  Zero  this is represented by 32 bits all at zero
>
> Since bias is 127 then value 255 in exponent can't occur
>
> Use this pattern i.e 11111111 to indicate limit conditions
>
> Negative Infinity is stored as Sign 1 - Exponent 11111111 mantissa 23 zero's
>
> Positive Infinity is stored as Sign 0 - Exponent 11111111 mantissa 23 ones's
>
> Hope this helps the algorithm is very straight forward really..
>
> ----------------------------------------------------------------------------
> ----------
> Tony Grimer ----  cm1918@scitsc.wlv.ac.uk (Tony D. Grimer)
>
> School of Computing (SCIT)  Room MU406 Tel - 01902-321810.
>
> ----------------------------------------------------------------------------
> ----------


DOH !!!!!!!!! system overload.... system overload..... system overload


I'm just gona stick to INT's I think :)))))))

--

Cheers Peter ..........

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