Andy David says:
Here's my 32 bit routine as written for the 17c43 taken from a mail I sent Scott just after I wrote it, hence the comments about the implementations I used.Looks a lot like Scott's original 16bit sqrt. As the root is going to be a 16 bit number the last subtract is awkward, so the 24bit sqrt method wasn't appropriate. I did actually write this myself rather than automatically converting Scott's code to 32 bit. I DID, however, consciously and unashamedly steal two parts  how to carry out the final 17bit subtraction and how to count iterations  the extra 'counting' bit in the mask was quite a devious idea. This one took a little longer to write than the 24bit, probably because it needs to iterate more times...
Standard disclaimer applies
;========================================================================= ; brSQRT32 ; ; Calculates the square root of a thirtytwo bit number using the ; binary restoring method. ; ; Result in ACCaHI:ACCaLO ; Mask in ACCbHI:ACCbLO ; Input in ACCcHI:ACCcLO:ACCdHI:ACCdLO ; ; Takes between 392 and 439 cycles (incl. call and return). ; Uses 58 words ROM, 8 bytes RAM including 4 holding the input. ; ; brSQRT32: movlw 0x40 ; Initial value for Result is... movwf ACCaHI ; ... 01000000 00000000 clrf ACCaLO,f ; movlw 0xC0 ; Initial value for mask is... movwf ACCbHI ; ... 11000000 00000000 clrf ACCbLO,f ; (second '1' is loop counter). Sub_Cmp:movfp ACCaLO,WREG ; Compare rootsofar with current subwf ACCcLO,f ; ... remainder. movfp ACCaHI,WREG ; subwfb ACCcHI,f ; btfss ALUSTA,C ; goto brstr ; (result is ve, need to restore). In1: movfp ACCbLO,WREG ; set the current bit in the result. iorwf ACCaLO,f ; movfp ACCbHI,WREG ; iorwf ACCaHI,f ; ShftUp: rlcf ACCdLO,f ; rlcf ACCdHI,f ; rlcf ACCcLO,f ; rlcf ACCcHI,f ; rrcf ACCbHI,f ; Shift mask right for next bit, whilst rrcf ACCbLO,f ; ... shifting IN MSB from remainder. btfsc ACCbHI,7 ; If MSB is set, unconditionally set the goto USet1 ; ... next bit. movfp ACCbLO,WREG ; Append '01' to rootsofar xorwf ACCaLO,f ; movfp ACCbHI,WREG ; xorwf ACCaHI,f ; btfss ALUSTA,C ; If second '1' in mask is shifted out, goto Sub_Cmp ; ... then that was the last normal iteration. movfp ACCaLO,WREG ; Last bit Generation. subwf ACCcLO,f ; ... The final subtract is 17bit (15bit root movfp ACCaHI,WREG ; ... plus '01'). Subtract 16bits: if result subwfb ACCcHI,f ; ... generates a carry, last bit is 0. btfss ALUSTA,C ; return ;Peter Harrison says these next two lines are required to set Z correctly movfp ACCcLO,WREG ; Clear zero flag if remainder non zero xorwf ACCcHI,WREG ; ;End section added by Peter Harrison. movlw 1 ; If result is 0 AND msb of is '0', result bit btfsc ALUSTA,Z ; ... is 0, otherwise '1'. btfsc ACCdHI,7 ; xorwf ACCaLO,f ; return USet1: btfsc ALUSTA,C ; If mask has shifted out, leave. final bit return ; ... has been set by iorwf at in1. bcf ACCbHI,7 ; clear bit shifted in from input. movfp ACCbLO,WREG ; Append '01' to rootsofar xorwf ACCaLO,f ; movfp ACCbHI,WREG ; xorwf ACCaHI,f ; movfp ACCaLO,WREG ; This subtraction is guaranteed not to subwf ACCcLO,f ; ... cause a borrow, so subtract and movfp ACCaHI,WREG ; ... jump back to insert a '1' in the subwfb ACCcHI,f ; ... root. goto In1 ; brstr: movfp ACCaLO,WREG ; A subtract above at Sub_Cmp was ve, so addwf ACCcLO,f ; ... restore the remainder by adding. movfp ACCaHI,WREG ; The current bit of the root is zero. addwfc ACCcHI,f ; goto ShftUp ;
Comments:
32 bit Square Root binary restoring routine.
This routine has an error in the last bit generation routine  basically the Z bit is incorrectly set after the final subtraction.
I have a routine for the PIC18F with this error corrected  two extra lines of code inserted before the decision to set the final bit. It now works correctly on all the numbers tested so far (several thousand!!)
{ed; here is Peters complete PIC18F routine . His corrections have also been added to the routine above. Thank you Peter!};======================================================================================== ; SQRT32 ; ; Calculates the square root of a thirtytwo bit number using the ; binary restoring method. ; ; Input: acc2_4:acc2_1 ; Mask: acc3_2:acc3_1 ; Result: acc1_2:acc1_1 ; ; Takes between 395 and 457 cycles (incl. call and return). ; Uses 64 words ROM, 8 bytes RAM including 4 holding the input. ;======================================================================================== ; sqrt32 movlw 0x40 ; Initial value for Result is... movwf acc1_2 ; ... 01000000 00000000 clrf acc1_1 ; ; movlw 0xC0 ; Initial value for mask is... movwf acc3_2 ; ... 11000000 00000000 clrf acc3_1 ; (second '1' is loop counter). ; compare movf acc1_1,w ; Compare rootsofar with current subwf acc2_3,f ; ... remainder. movf acc1_2,w ; subwfb acc2_4,f ; bc $+4 ; goto restore ; result is ve, so need to restore ; setcurr movf acc3_1,w ; set the current bit in the result. iorwf acc1_1,f ; movf acc3_2,w ; iorwf acc1_2,f ; ; shftUp rlcf acc2_1,f ; rlcf acc2_2,f ; rlcf acc2_3,f ; rlcf acc2_4,f ; ; rrcf acc3_2,f ; Shift mask right for next bit, whilst rrcf acc3_1,f ; ... shifting IN MSB from remainder. btfsc acc3_2,7 ; If MSB is set, unconditionally set the goto setnext ; ... next bit. ; movf acc3_1,w ; Append '01' to rootsofar. xorwf acc1_1,f ; movf acc3_2,w ; xorwf acc1_2,f ; ; bc $+4 ; If second '1' in mask is shifted out, goto compare ; ... then that was the last normal iteration. ; movf acc1_1,w ; Last bit Generation. subwf acc2_3,f ; ... The final subtract is 17bit (15bit root movf acc1_2,w ; ... plus '01'). Subtract 16bits: if result subwfb acc2_4,f ; ... generates a carry, last bit is 0. bc $+4 ; return ; movf acc2_3,w ; Clear zero flag if remainder non zero xorwf acc2_4,w ; ; movlw 1 ; bnz $+4 ; If result is 0 AND msb of N is '0', result bit btfsc acc2_2,7 ; ... is 0, otherwise '1'. xorwf acc1_1,f ; return ; ; setnext bnc $+4 ; If mask has shifted out, leave. final bit return ; ... has been set by iorwf at in1. bcf acc3_2,7 ; clear bit shifted in from input. ; movf acc3_1,w ; Append '01' to rootsofar xorwf acc1_1,f ; movf acc3_2,w ; xorwf acc1_2,f ; ; movf acc1_1,w ; This subtraction is guaranteed not to subwf acc2_3,f ; ... cause a borrow, so subtract and movf acc1_2,w ; ... jump back to insert a '1' in the subwfb acc2_4,f ; ... root. goto setcurr ; ; restore movf acc1_1,w ; A subtract above at Sub_Cmp was ve, so addwf acc2_3,f ; ... restore the remainder by adding. movf acc1_2,w ; The current bit of the root is zero. addwfc acc2_4,f ; goto shftUp ; ; ;========================================================================================
See also:
file: /Techref/microchip/math/sqrt/sqrt32.htm, 7KB, , updated: 2014/1/9 15:14, local time: 2017/11/23 15:55,

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