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..




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Tony Grimer ----  cm1918@scitsc.wlv.ac.uk (Tony D. Grimer)

School of Computing (SCIT)  Room MU406 Tel - 01902-321810.

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