In a message dated 96-05-26 05:38:43 EDT, Clyde Smith-Stubbs wrote:

>msullivan@vax.niobrara.com (Mark K Sullivan) wrote:
>>
>> I needed to divide a 2's complement number by 2 and I just thought of this
>> quickie arithmetic shift.  Stop me if you've heard it before...  Too late
>;)
>>
>>       rlf    MyVar,W     ;copy sign to carry
>>       rrf    MyVar       ;shift and duplicate sign
>
>There is just one slight problem with using shifts to divide -ve
>numbers. To illustrate: what happens to -1 when you do this?
>
>Try this code instead:
>
>        btfsc   MyVar,7
>        incf    MyVar
>        rlf     MyVar,w
>        rrf     MyVar
>
>--

   You bring up an interesting point, Clyde.  It's true that Mark's routine
yields -1 as the result of dividing -1 by 2.  8^(  But that could be seen as
the desired result.  8^).  It leaves an error of 1/2, but then that's true of
*every* odd number in the input.
   Mark's routine also has the advantage of leaving the size of the output
bins the same.  That is, there are *exactly* two input numbers which yield
any given output number.  With your scheme, a result of -40H is given *only*
by an input of -80H, while a result of 0 is given by *three* inputs:  -1, 0,
and 1.
   I'm sure there are applications where either one of these algorithms is
more useful than the other!  Integers are fun, eh wot??

Regards...
Shel Michaels
sbmichaels@aol.com
http://members.aol.com/sbmichaels