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.
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:
James Ashley Hillman of Industrial Interface Research Ltd shares this code:
The archive on this page links to 2 divide 24 by 8 routines.
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:
| CODE |
;*********************************************************** ;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 |