PICkles:

While writing some averaging code, I started realizing my
grounding in the theory of how binary math works is very fuzzy.  It
seems like I worked through binary multiplication some 20 years ago
and then promptly forgot how it worked.

Could one of you kind folks let me know if this is on the right
track?

One of the ways to implement a running averaging algorithm is

(Old average) *256 - (Old Average) + (New data point)

This gives a result that is 256 times the actual average, which
would be quite useful in my case.  Why divide? Or, if I really need
the divided result, just use the high byte of the result.

Multiplication by 256 is just shifting left 8 times, right?  In this
case a one byte number shifted left 8 times becaomes a two byte
number with the low byte being zero.  No actual shifting required.

If (Old average)*256  is stored in AVGHI,AVGLO

the formula becomes

AVGHI,AVGLO  -  AVGHI + (NEW DATA POINT)

It would be a teensy bit more accurate if I could add in a single
digit based on rounding  (AVGLO/256).  This could be computed by
comparing AVGLO to 128 decimal and setting 1 or 0.

No need to call a mupltiply routine, just double precision
addition and subraction routines.


 Does this make sense?


Best Regards,

Lawrence Lile