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 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 ;
See also:
file: /Techref/scenix/lib/math/sqrt/sqrt32.htm, 3KB, , updated: 2004/6/10 14:40, local time: 2022/5/23 04:31,

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