PIC Microcontoller Math Method

Divide 24 bit int by 16 bit int to 24 bit int

FXD2416U:
        CLRF REMB0
        CLRF REMB1
        MOVLW 24
        MOVWF LOOPCOUNT
LOOPU2416
        RLF ACCB0, W    ;left shift of accb0's msb to reminder
        RLF REMB1, F
        RLF REMB0, F
        MOVF BARGB1, W  ;REMB -= BARGB
        SUBWF REMB1, F
        MOVF BARGB0, W
        BTFSS _C
        INCFSZ BARGB0,W
        SUBWF REMB0, F

        BTFSC _C
        GOTO UOK46LL    ;if no borrow

        MOVF BARGB1, W  ;REMB += BARGB
        ADDWF REMB1, F
        MOVF BARGB0, W
        BTFSC _C
        INCFSZ BARGB0,W
        ADDWF REMB0, F

        BCF _C
UOK46LL
        RLF AARGB2, F
        RLF AARGB1, F
        RLF AARGB0, F
        DECFSZ LOOPCOUNT, F
        GOTO LOOPU2416

        RETURN

Nikolai Golovchenko says:

The routine above is actually 24 by 15 bits division. Below is a 24 by 16 bits division routine:


;Inputs:
;   Dividend - AARGB0:AARGB1:AARGB2 (0 - most significant!)
;   Divisor  - BARGB0:BARGB1
;Temporary:
;   Counter  - LOOPCOUNT
;   Remainder- REMB0:REMB1
;Output:
;   Quotient - AARGB0:AARGB1:AARGB2
;
;       Size: 28
; Max timing: 4+24*(6+6+4+3+6)-1+3+2=608 cycles (with return)
; Min timing: 4+24*(6+6+5+6)-1+3+2=560 cycles (with return)
;          

FXD2416U:
        CLRF REMB0
        CLRF REMB1
        MOVLW 24
        MOVWF LOOPCOUNT
LOOPU2416
        RLF AARGB2, F           ;shift left divider to pass next bit to remainder
        RLF AARGB1, F           ;and shift in next bit of result
        RLF AARGB0, F

        RLF REMB1, F            ;shift carry into remainder
        RLF REMB0, F

        RLF LOOPCOUNT, F        ;save carry in counter
         
        MOVF BARGB1, W          ;substract divisor from remainder
        SUBWF REMB1, F
        MOVF BARGB0, W
        BTFSS _C
        INCFSZ BARGB0, W
        SUBWF REMB0, W          ;keep that byte in W untill we make sure about borrow

        SKPNC                   ;if no borrow
         BSF LOOPCOUNT, 0       ;set bit 0 of counter (saved carry)

        BTFSC LOOPCOUNT, 0      ;if no borrow
         GOTO UOK46LL           ;jump

        MOVF BARGB1, W          ;restore remainder if borrow
        ADDWF REMB1, F
        MOVF REMB0, W           ;read high byte of remainder to W
                                ;to not change it by next instruction
UOK46LL
        MOVWF REMB0             ;store high byte of remainder
        CLRC                    ;copy bit 0 to carry
        RRF LOOPCOUNT, F        ;and restore counter
        DECFSZ LOOPCOUNT, f     ;decrement counter
         GOTO LOOPU2416         ;and repeat loop if not zero
         
        RLF AARGB2, F           ;shift in last bit of result
        RLF AARGB1, F   
        RLF AARGB0, F
        RETURN

Nikolai Golovchenko shares this code:


 
Here is a slightly optimized version - 1 instruction shorter, 24 cycles faster!

;Inputs:
;   Dividend - AARGB0:AARGB1:AARGB2 (0 - most significant!)
;   Divisor  - BARGB0:BARGB1
;Temporary:
;   Counter  - LOOPCOUNT
;   Remainder- REMB0:REMB1
;Output:
;   Quotient - AARGB0:AARGB1:AARGB2
;
; Size: 27
; Max timing: 4+24*(6+6+4+3+5)-1+3+2=584 cycles (with return)
; Min timing: 4+24*(6+6+5+5)-1+3+2=536 cycles (with return)
;
;25-Sep-2000    Original version
;20-Oct-2001    Made the loop one instruction shorter, comments
;               review.

FXD2416U:
        CLRF REMB0
        CLRF REMB1
        MOVLW 24
        MOVWF LOOPCOUNT
LOOPU2416
        RLF AARGB2, F           ;shift dividend left to move next bit to remainder
        RLF AARGB1, F           ;and shift in next bit of result
        RLF AARGB0, F           ;

        RLF REMB1, F            ;shift carry (next dividend bit) into remainder
        RLF REMB0, F

        RLF LOOPCOUNT, F        ;save carry in counter, since remainder 
                                ;can be 17 bit long in some cases (e.g. 
                                ;0x800000/0xFFFF)
         
        MOVF BARGB1, W          ;substract divisor from 16-bit remainder
        SUBWF REMB1, F          ;
        MOVF BARGB0, W          ;
        BTFSS STATUS, C         ;
        INCFSZ BARGB0, W        ;
        SUBWF REMB0, F          ;

;here we also need to take into account the 17th bit of remainder, which
;is in LOOPCOUNT.0. If we don't have a borrow after subtracting from lower
;16 bits of remainder, then there is no borrow regardless of 17th bit 
;value. But, if we have the borrow, then that will depend on 17th bit 
;value. If it is 1, then no final borrow will occur. If it is 0, borrow
;will occur.

        SKPNC                   ;if no borrow after 16 bit subtraction
         BSF LOOPCOUNT, 0       ;then no no borrow in result. Overwrite
                                ;LOOPCOUNT.0 with 1 to indicate no
                                ;borrow.
                                ;if borrow did occur, LOOPCOUNT.0 will
                                ;hold the eventual borrow value (0-borrow,
                                ;1-no borrow)

        BTFSC LOOPCOUNT, 0      ;if no borrow after 17-bit subtraction
         GOTO UOK46LL           ;skip remainder restoration.

        ADDWF REMB0, F          ;restore higher byte of remainder. (w 
                                ;contains the value subtracted from it
                                ;previously)
        MOVF BARGB1, W          ;restore lower byte of remainder
        ADDWF REMB1, F          ;

UOK46LL
        CLRC                    ;copy bit LOOPCOUNT.0 to carry
        RRF LOOPCOUNT, F        ;and restore counter

        DECFSZ LOOPCOUNT, f     ;decrement counter
         GOTO LOOPU2416         ;and repeat loop if not zero. carry 
                                ;contains next quotient bit (if borrow,
                                ;it is 0, if not, it is 1).
               
        RLF AARGB2, F           ;shift in last bit of quotient
        RLF AARGB1, F   
        RLF AARGB0, F
        RETURN

Nikolai Golovchenko shares this code:

Well, the routine can be made even simpler! This version is 5 instructions shorter and 48 cycles faster. Thanks to Zlatko Petkov for an idea of removing the extra shifts after the loop. Nikolai Golovchenko. 5-Dec-2004.

FXD2416U:
        CLRF REMB0
        CLRF REMB1
        MOVLW .24
        MOVWF LOOPCOUNT
LOOPU2416
        RLF AARGB2, W           ;shift dividend left to move next bit to remainder
        RLF AARGB1, F           ;
        RLF AARGB0, F           ;

        RLF REMB1, F            ;shift carry (next dividend bit) into remainder
        RLF REMB0, F

        RLF AARGB2, F           ;finish shifting the dividend and save  carry in AARGB2.0,
                                ;since remainder can be 17 bit long in some cases
                                ;(e.g. 0x800000/0xFFFF). This bit will also serve
                                ;as the next result bit.
         
        MOVF BARGB1, W          ;substract divisor from 16-bit remainder
        SUBWF REMB1, F          ;
        MOVF BARGB0, W          ;
        BTFSS STATUS, C         ;
        INCFSZ BARGB0, W        ;
        SUBWF REMB0, F          ;

;here we also need to take into account the 17th bit of remainder, which
;is in AARGB2.0. If we don't have a borrow after subtracting from lower
;16 bits of remainder, then there is no borrow regardless of 17th bit 
;value. But, if we have the borrow, then that will depend on 17th bit 
;value. If it is 1, then no final borrow will occur. If it is 0, borrow
;will occur. These values match the borrow flag polarity.

        SKPNC                   ;if no borrow after 16 bit subtraction
         BSF AARGB2, 0          ;then there is no borrow in result. Overwrite
                                ;AARGB2.0 with 1 to indicate no
                                ;borrow.
                                ;if borrow did occur, AARGB2.0 already
                                ;holds the final borrow value (0-borrow,
                                ;1-no borrow)

        BTFSC AARGB2, 0         ;if no borrow after 17-bit subtraction
         GOTO UOK46LL           ;skip remainder restoration.

        ADDWF REMB0, F          ;restore higher byte of remainder. (w 
                                ;contains the value subtracted from it
                                ;previously)
        MOVF BARGB1, W          ;restore lower byte of remainder
        ADDWF REMB1, F          ;

UOK46LL

        DECFSZ LOOPCOUNT, f     ;decrement counter
         GOTO LOOPU2416         ;and repeat the loop if not zero.

        RETURN

Fathy Lotfy Samaha of Freelancer Says:

Thanks so much, this is a much simple routine,
but it did not work with me until I changed
the first line to :
	movlw .24 ,
I am using MPASM and MPLAB, it recogonize
the decimal numbers precedded with a dot .

Questions:

Tori Torin asks:
Dear Piclist People:

Thanks to all for sharing your knowledge. I was wondering if some of you have had a problem like the one I describe below. Any help will be welcomed!
Iâ¬™m dealing with integer unsigned binary division using a PIC. I implemented 24/24 bit division and everything went ok. But when performing 24/16 bit division, something goes wrong.
The dividend (numerator) is a 24 bit constant, but the divisor (denominator) is a variable number. When this denominator takes a low value (but still 16 bit value), say 10700 i.e., the error in the result is quite a lot, and it becomes bigger as the difference between both operands does too. Is there a solution to this?
Thanks again for your help. Greetings!
or 16 bits divided by 16 bits to 24 bits floating point? thanks!+
anne_lauren_a at yahoo.com.ph asks: " How about dividing 24 bits by 16 bits resulting into a floating point? Thanks a lot! =)" +

Comments:

rick at silver-willows.co.uk

reference the LAST code segment.
Using a PIC18F4520, please note the code changes.
THANKS YOU all for saving me a lot of work.
LOOPU2416
	bcf	CARRY			; addition of code line
	rlcf	BSR1_Larger24Lo,W 	; shift dividend left to move next bit to remainder
	rlcf	BSR1_Larger24Mi,F
	rlcf	BSR1_Larger24Hi,F
	rlcf	BSR1_Remain16Lo,F	; shift carry (next dividend bit) into remainder
	rlcf	BSR1_Remain16Hi,F
	rlcf	BSR1_Larger24Lo,F	; finish shifting the dividend and save carry in
					;  BSR1_Larger24Lo.0,
					; since remainder can be 17 bit long in some cases
					; (e.g. 0x800000/0xFFFF). This bit will also serve
					; as the next result bit.
      
I apologise for spiling the convention in MY labels, but you can work backwards from the origonal code.
Rick

rick at silver-willows.co.uk

Sorry, I forgot...
Also, for PIC184520...
replace SKPNC with btfsc CARRY
;	SKPNC	 	;if no borrow after 16 bit subtraction
      btfsc CARRY	; if no borrow after 16 bit subtraction

file: /Techref/microchip/math/div/24by16.htm, 12KB, , updated: 2021/6/1 08:34, local time: 2024/11/24 11:24,

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PIC Microcontoller Math Method

Divide 24 bit int by 16 bit int to 24 bit int

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