Andrew David (akdavid at ultronics.co.uk), Software Manager, Ultronics Ltd, Cheltenham says:
This division routine uses the standard binary long-division algorithm. The loop iterates once for each bit in the numerator, so it goes round 24 times. For each loop the numerator (sorry, but I can never remember which is the dividend and the divisor - I guess the divisor is the denominator, but I'd hate to get it wrong) is left shifted 1 bit into the remainder, then the remainder is compared with the denominator. If the remainder is greater than the denominator, the denominator is subtracted from the remainder and a 1 is shifted into the quotient, otherwise a 0 is shifted into the quotient. If you can't see how the routine works, try running through an example on paper, in binary, then think about how you'd write the routine.
;============================================================================= ; DIV24_16u ; ; Divides a 24bit number by a 16bit number. Unsigned. ; ; Inputs: ; 24-but numerator in ACCcLO:ACCdHI:ACCdLO ; 16-bit denominator in ACCbHI:ACCbLO ; ; Outputs: ; 24-bit quotient in ACCcLO:ACCdHI:ACCdLO ; 16-bit rem in ACCaHI:ACCaLO ; ; Locals used: ; R5Hi ; ; ; Inputs are not preserved. ; ; No timing analysis performed. ; Andrew David, Software Manager, Ultronics Ltd, Cheltenham ; akdavid at ultronics.co.uk http://www.ultronics.com ; ;============================================================================= DIV24_16u: movlw .24 ; for 24 shifts movwf R5Hi ; clrf ACCaHI,f ; clear remainder. clrf ACCaLO,f ; d2416lp:rlcf ACCdLO,f ; build up remainder. rlcf ACCdHI,f ; rlcf ACCcLO,f ; rlcf ACCaLO,f ; rlcf ACCaHI,f ; ; remainder is 16-bit, but may have spilled over into carry. btfss ALUSTA,C ; check for remainder spill into carry. goto d2416s ; movfp ACCbLO,WREG ; subwfb ACCaLO,f ; Carry bit is the 17th bit of this subtract! movfp ACCbHI,WREG ; subwfb ACCaHI,f ; bsf ALUSTA,C ; bit is known to be zero here. goto d2416ns ; d2416s: movfp ACCbLO,WREG ; Compare remainder with divisor. subwf ACCaLO,w ; movfp ACCbHI,WREG ; subwfb ACCaHI,w ; btfss ALUSTA,C ; goto d2416ns ; (remainder < divisor), shift in a '0' movfp ACCbLO,WREG ; The remainder is larger, so subtract the subwf ACCaLO,f ; ... divisor FROM the remainder and movfp ACCbHI,WREG ; ... shift a '1' into the quot. subwfb ACCaHI,f ; d2416ns:decfsz R5Hi,f ; check all bits goto d2416lp ; rlcf ACCdLO,f ; shift in final quotient bit. rlcf ACCdHI,f ; rlcf ACCcLO,f ; return
This code does not work if the divisor is smaller than 0x100.+
Replace the five instructions at d2416s with the following to make it work (converted for 16f84):movf ACCbHI,W ; divisor hi subwf ACCaHI,w ; remainder hi (w = remainder - divisor) btfss STATUS, C goto d2416ns ; (remainder < divisor), shift in a '0' movf ACCbLO,W subwf ACCaLO,w btfss STATUS,C
Andy David of Ultronics Ltd Says:
The above (anonymous) poster has not convinced me that the original code doesn't work -- my test cases for divisor < 0x100 are all correct.+
The original code is for the 17c series pics, hence the use of the subwfb instruction. The purpose of the first 5 lines of code at d2416s is to compare the current remainder with the (constant) divisor. Using a 16-bit subtract abd checking the status of carry is standard, I can't get this code to fail. The carry that is subsequently shifted into the quotient is correct as either remainder => divisor (when remainder - divisor is +ve (>= 0) carry = 1) or remainder < divisor (carry = 0).
I've just tried several test cases with divisor < 0x100, I can't see any problem. As a benchmark I've used BaseCalc.
The alternative code looks like it will work but I don't believe there's a reason for using it; you'd normally need a good reason to replace 5 (working) lines of code with 7.
Note that if you want to use this code for 14-bit core you'll need to change more code than suggested.
Now that I've finished saying that the code works, I refer the reader to the standard disclaimer in that I'm not responsible for your use of this code. Clever people use Dimitry's, Nikolai's or Scott's code, anyway.
I have tried to modify and use the code for a PIC 18F2480 (16 bit) but I can't get it to work, even with the modify of mr. jaakko-haakana. This is what I got:+DIV24_16u: movlw d'24' ; for 24 shifts movwf R5Hi ; clrf REMBH ; clear remainder. clrf REMBL ; d2416lp: rlcf DATO1L,f ; build up remainder. rlcf DATO1H,f ; rlcf DATO1U,f ; rlcf REMBL,f ; rlcf REMBH,f ; ; remainder is 16-bit, but may have spilled over into carry. btfss STATUS,C ; check for remainder spill into carry. goto d2416s ; movf DATO2L,W ; subwfb REMBL,f ; Carry bit is the 17th bit of this subtract! movf DATO2H,W ; subwfb REMBH,f ; bsf STATUS,C ; bit is known to be zero here. goto d2416ns ; d2416s: movf DATO2H,W ; divisor hi subwf REMBH,w ; remainder hi (w = remainder - divisor) btfss STATUS, C goto d2416ns ; (remainder < divisor), shift in a '0' movf DATO2L,W subwf REMBL,w btfss STATUS,C goto d2416ns ; (remainder < divisor), shift in a '0' movf DATO2L,W ; The remainder is larger, so subtract subwf REMBL,f ; ... divisor FROM the remainder and movf DATO2H,W ; ... shift a '1' into the quot. subwfb REMBH,f ; d2416ns: decfsz R5Hi,f ; check all bits goto d2416lp ; rlcf DATO1L,f ; shift in final quotient bit. rlcf DATO1H,f ; rlcf DATO1U,f ; returnCan anybody advise me about where is the error? Thanks in advance. Lorenzo. Email: tecnico at innotec.it
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