piclist 1996\05\30\141006a >
www.piclist.com/techref/method/math.htm?key=divide
BY : myke predko email (remove spam text)

>I can save 600bytes in a lookup table if I can figure out a good way to
>divide a 16 bit number by 10.
>
>I have the app notes with the general 16 math, but I need a smaller and
>faster routine.
>
>

As everybody was probably expecting, I have to put in my two cents worth and
drop in some code...

Here is a 16 bit unsigned division algorithm that I have used in the past.
Initially, it shifts the Divisor until bit 14 is set and then begins
shifting down.  If the divisor can be taken from the dividend, it is.  Next,
the divisor is shifted down until you get the initial divisor.  The results
(quotient and remainder) are all 16 bits long and an additional 8 bit
register is used in the example code below.

Here is the psuedo-code:

Count = 0                       - Count keeps track of where the Divisor is
Quotient = 0                    - Quotient is actually a 16 bit sum

while ( Divisor & 0x04000 ) != 0  - Find where to shift the Divisor up to
Count = Count + 1             - Record How Many Bits Shifted
Divisor = Divisor << 1        - Shift up the Divisor

while Count != 0                - Now, do the Shifting Subtraction (Division)
if Dividend >= Divisor        - Can Subtract from the Result
Quotient = Quotient + ( 2 ^ Count )
Divident = Divident - Divisor
Count = Count - 1
Divisor = Divisor >> 1

That's it.  Divident contains the remainder and Quotient contains the
quotient of the value.  Note that this algorithm will go into an endless
loop if the divisor is equal to zero.  I stop the shifting up with bit 14 of
the Divisor so that signed values can be supported (even though this
algorithm won't work with negative values).

The PIC code for doing this is below.  Note, that I have changed the code in
two places.  The first is with regards to Count.  Rather than using a
counter, I shift a "1" up and down (ending the division when the "1" ends up
in the Carry Flag).  This means that I can add Count to the Quotient
directly.  The second area that I have changed is in regard to the
comparison of the Dividend to the Divisor, note that I save the contents of
the subtraction (compare) and use it later, rather than subtracting twice.

clrf  Quotient                ;  Initialize the Variables
clrf  Quotient + 1

movlw 1                       ;  Instead of a Counter, Count is a shifted
movwf Count                   ;  Value for adding to the Quotient
clrf  Count + 1

StartLoop                       ;  Find the Top Value for the Divisor
btfsc Dividend, 6             ;  If Bit 14 Set, then we have the value
goto DivLoop

bcf   STATUS, C               ;  Shift over the Carry and the Divisor
rlf   Count
rlf   Count + 1
rlf   Divisor                 ;  Note, Carry will be Zero from Count
rlf   Divisor + 1

goto  StartLoop               ;  Now, see if we can shift again

DivLoop                         ;  Do the Shifted Subtraction
movf  Divisor + 1, w          ;  Compare Values, High First
subwf Dividend + 1, w
movwf Temp                    ;  Save Result for later (just in case)
movf  Divisor, w
subwf Dividend
btfss STATUS, C               ;  Make Sure Carry is accounted for
decf Temp

btfsc Temp, 7                 ;  Do we have a Negative Number from Subtract?
goto DivSkip                 ;  Yes, Don't Subtract this value

movwf Dividend                ;  Else, save the result for the Next
movf  Temp, w                 ;   Subtract
movwf Dividend

movf  Count + 1, w            ;  Add the Bit Offset to the Quotient
movf  Count, w

DivSkip                         ;  Now, Shift the Values Down

bcf   STATUS, C               ;  Shift down the Divisor
rrf   Divisor + 1
rrf   Divisor

rrf   Count + 1               ;  Now see if the Count is finished
rrf   Count

btfss STATUS, C               ;  Finished if Carry is Set
goto DivLoop

Note that with this code, "Dividend" now contains the Remainder and
"Divisor" is the original "Divisor" >> 1.

In terms of space and execution speed, I think you'll find this to be a
pretty good improvement from the code in the ECBK (although not as good as
some of the algorithms other people have put in for a direct divide by 10).

Myke
Myke

"We're Starfleet officers, weird is part of the job."

Capt. Catherine Janeway
<m0uPC9A-001CROC@passport.ca>