PIC Microcontoller Math Method

Divide 24 bit int by 8 bit int to 24 bit int and remainder

James Ashley Hillman of Industrial Interface Research Ltd shares this code:

The archive on this page links to 2 divide 24 by 8 routines. 

-PICList post "24bit by 16bit Division" (actually 24 / 8)
-PICList post "Divide 24/8 routine" ASM embedded in C  

During testing I discovered that high numbers cause the wrong 
answers to be calculated (eg 0xF00000 / 0xFD = wrong answer)

I wrote my own routine which is Nikolai's 24 bits by 16, 
modified to divide by 8 bits, remainder is also 8 bits:

;***********************************************************
;Unsigned 24 bit by 8 bit divide routine
;
; Inputs:
;   Dividend  - x,x+1,x+2 (x+2 - most significant!)
;   Divisor   - y
; Temporary:
;   Counter   - counter
; Output:
;   Quotient  - x,x+1,x+2 (x+2 - most significant!)
;   Remainder - x+3
;
; Size: 17
; Timing: 342 cycles (including call and return)
;
; This is basically Nikolai Golovchenko's 24 by 16 bit 
; divide routine, with some instructions removed to 
; optimise it for an 8 bit divide. 
; Thanks to Nikolai for the original post.
;
; James Hillman, 2 December 2005 
;***********************************************************

FXD248U:
    CLRF    x+3          ;remainder
    MOVLW   d'24'
    MOVWF   counter

LOOPU248
    RLF     x,W      ;shift dividend left to move next bit to remainder
    RLF     x+1,F    ;
    RLF     x+2,F    ;
    RLF     x+3,F    ;shift carry (next dividend bit) into remainder
    RLF     x,F      ;finish shifting the dividend and save carry in x.0,
                     ;since remainder can be 9 bit long in some cases
                     ;This bit will also serve as the next result bit.
         
    MOVF    y,W      ;substract divisor from 8-bit remainder
    SUBWF   x+3,F        

;here we also need to take into account the 9th bit of remainder, which
;is in x.0. If we don't have a borrow after subtracting from 
;8 bits of remainder, then there is no borrow regardless of 9th bit 
;value. But, if we have the borrow, then that will depend on 9th 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.

    BTFSC   STATUS,0  ;if no borrow after 8 bit subtraction
    BSF     x,0       ;then there is no borrow in result. Overwrite
                      ;x.0 with 1 to indicate no borrow.
                      ;if borrow did occur, x.0 already
                      ;holds the final borrow value (0-borrow, 
                      ;1-no borrow)

    BTFSS   x,0       ;if no borrow after 9-bit subtraction
    ADDWF   x+3,F     ;restore remainder. (w contains the value 
                      ;subtracted from it previously)
    DECFSZ  counter,F    
    GOTO    LOOPU248    
    RETURN

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