I'm in the middle of a project in which I need to compare
a 12-bit value from an A/D with a set-value.

Let's say that the A/D-value is stored in two registers, the low eight bits
are stored in AD_LOW, and the four high bits are stored in AD_HIGH.

The remaining four bits in AD_HIGH can be considered to be 0.

The same criterion is applicable for the set-value, stored in SET_LOW
and SET_HIGH.

Bit 2 in a register named STAT holds the last compare status.

The A/D-value shall be compared to the set-value, and if it
is greater, then bit 2 in the STAT register will be set.
If it is smaller (or equal) the bit shall be cleared.

The W register can be thrashed, but the other registers
must be preserved

How fast and/or small code can be written to do this?

Here's one way :

  movf AD_HIGH,0   ;Copy the high bits of the A/D value to W
  xorwf SET_HIGH,0   ;Compare with the high bits of the set-value
  skpz
  goto label1
  movf AD_LOW,0   ;Copy the low bits of the A/D value to W
  subwf SET_LOW,0   ;Subtract from the SET_LOW value
  skpc
  bsf STAT,2   ;If it is greater than SET_LOW bit 2 of STAT is set
  skpnc
  bcf STAT,2   ;If it is less than SET_LOW (or equal to) bit 2 of STAT is 
cleared
  goto label2
label1   ; Compare the high parts
  movf AD_HIGH,0
  subwf SET_HIGH,0
  skpc
  bsf STAT,2
  skpnc
  bcf STAT,2
label2   ;Compare finished

Points will be scored as follows:

8 points for each instruction word used.
20 points for each (extra) RAM location used.
6 points for each instruction cycle. (worst case)

-8 points for isosynchronous execution.

My rather sluggish example scores (if I've counted correctly):

17*8+0+11*6 = 202 points..

It should be possible to shave a lot of points here...

There is no award to be won, but the winner will get all the glory...


-Oyvind          oyvind.kaurstad@nofac.abb.no