Post by thread:fast % routine

Hi all, I need a fast 0 - 255 -> % routine Eg Byte value = 89 % value = 34 89 / 255 * 100 I did this routine - any others?? Value in W on entry Result in PcntH on return ToPcnt movwf PCntH movwf tempL clrf PCntL bcf status,carry rrf PCntH rrf PCntL rrf PCntH rrf PCntL rrf PCntH,w movwf tempH rrf PCntL,w call addup clrf tempH bcf status,carry rlf tempL rlf tempH rlf tempL rlf tempH movf tempL,w call addup return addup addwf PCntL btfsc status,carry incf PCntH movf tempH,w addwf PCntH return -- Best regards Tony http://www.picnpoke.com spam_OUTsalesTakeThisOuTpicnpoke.com

On Fri, 25 Feb 2000, Tony Nixon wrote: > Hi all, > > I need a fast 0 - 255 -> % routine > > Eg Byte value = 89 > % value = 34 > > 89 / 255 * 100 > > I did this routine - any others?? > > Value in W on entry > Result in PcntH on return > > ToPcnt movwf PCntH <big snip> You mean like this: ;******************************************************************* ;scale_hex2dec ; The purpose of this routine is to scale a hexadecimal byte to a ;decimal byte. In other words, if 'h' is a hexadecimal byte then ;the scaled decimal equivalent 'd' is: ; d = h * 100/256. ;Note that this can be simplified: ; d = h * 25 / 64 = h * 0x19 / 0x40 ;Multiplication and division can be expressed in terms of shift lefts ;and shift rights: ; d = [ (h<<4) + (h<<3) + h ] >> 6 ;The program divides the shifting as follows so that carries are automatically ;taken care of: ; d = (h + (h + (h>>3)) >> 1) >> 2 ; ;Inputs: W - should contain 'h', the hexadecimal value to be scaled ;Outputs: W - The scaled hexadecimal value is returned in W ;Memory: temp ;Calls: none scale_hex2dec MOVWF temp ;Hex value is in W. CLRC ;Clear the Carry bit so it doesn't affect RRF RRF temp,F CLRC RRF temp,F CLRC RRF temp,F ;temp = h>>3 ADDWF temp,F ;temp = h + (h>>3) RRF temp,F ;temp = (h + (h>>3)) >> 1 ADDWF temp,F ;temp = h + ((h + (h>>3)) >> 1) RRF temp,F CLRC RRF temp,W ;d = W = (h + (h + (h>>3)) >> 1) >> 2 RETURN ? Scott

On Fri, 25 Feb 2000, Tony Nixon wrote: > Scott Dattalo wrote: > > > You mean like this: > [snip] > > > Gee Scott, what took you so long :-)) Well, the first one was a fast mod routine and then it dawned on me you meant percentage :). If you look closely, it doesn't do exactly what you asked. The one I posted divides (implicitly) by 256 while you were asking for 255. The error is quite small, but if necessary can be removed. You're asking for p = x * 100/255 and I gave you a solution for p' = x * 100/256 but you can do a little arithmetic: p = x * 100/256 * (256/255) = p' * (256/255) = p' * (1 + 1/255) = p' + p'/255 and, 1/255 = 1/(-1+256) = 1/256 * ( 1 + 1/256 + 1/256/256 + ...) p = p' + p'/256 + p'/256/256 + ... Now since p' is going to be less than 100, then the first correction is going to be less than 1/2 and there will be no rounding. Hence the answer IS the same. However, if you wanted a fractional result, well here you go. Looks like you get a repeating fraction much like you get when you divide by 9 (10 - 1 instead of 256 - 1). Scott

Nice done, Scott! To add, if Tony needs a little better precision (especially for higher values of input byte), one more term should be added, so that d = (((h >> 3 + h) >> 3 + h) >> 1 + h) >> 2 = h * (1/4 + 1/8 + 1/64 + 1/512) This adds only 6 cycles, and reduces errors from 51 to 11 cases per all combinations of input (in both routines absolute errors are small, +-1). > MOVWF temp ;Hex value is in W. > CLRC ;Clear the Carry bit so it doesn't affect RRF RRF temp,F CLRC RRF temp,F CLRC RRF temp,F ADDWF temp, F {Quote hidden}> RRF temp,F > CLRC > RRF temp,F > CLRC > RRF temp,F ;temp = h>>3 > ADDWF temp,F ;temp = h + (h>>3) > RRF temp,F ;temp = (h + (h>>3)) >> 1 > ADDWF temp,F ;temp = h + ((h + (h>>3)) >> 1) > RRF temp,F > CLRC > RRF temp,W ;d = W = (h + (h + (h>>3)) >> 1) >> 2 > RETURN Nikolai On Friday, February 25, 2000 Scott Dattalo wrote: {Quote hidden}> On Fri, 25 Feb 2000, Tony Nixon wrote: >> Hi all, >> >> I need a fast 0 - 255 -> % routine >> >> Eg Byte value = 89 >> % value = 34 >> >> 89 / 255 * 100 >> >> I did this routine - any others?? >> >> Value in W on entry >> Result in PcntH on return >> >> ToPcnt movwf PCntH > <big snip> > You mean like this: > ;******************************************************************* > ;scale_hex2dec > ; The purpose of this routine is to scale a hexadecimal byte to a > ;decimal byte. In other words, if 'h' is a hexadecimal byte then > ;the scaled decimal equivalent 'd' is: > ; d = h * 100/256. > ;Note that this can be simplified: > ; d = h * 25 / 64 = h * 0x19 / 0x40 > ;Multiplication and division can be expressed in terms of shift lefts > ;and shift rights: > ; d = [ (h<<4) + (h<<3) + h ] >> 6 > ;The program divides the shifting as follows so that carries are > automatically > ;taken care of: > ; d = (h + (h + (h>>3)) >> 1) >> 2 > ; > ;Inputs: W - should contain 'h', the hexadecimal value to be scaled > ;Outputs: W - The scaled hexadecimal value is returned in W > ;Memory: temp > ;Calls: none > scale_hex2dec > MOVWF temp ;Hex value is in W. > CLRC ;Clear the Carry bit so it doesn't affect RRF > RRF temp,F > CLRC > RRF temp,F > CLRC > RRF temp,F ;temp = h>>3 > ADDWF temp,F ;temp = h + (h>>3) > RRF temp,F ;temp = (h + (h>>3)) >> 1 > ADDWF temp,F ;temp = h + ((h + (h>>3)) >> 1) > RRF temp,F > CLRC > RRF temp,W ;d = W = (h + (h + (h>>3)) >> 1) >> 2 > RETURN > ? > Scott

> I need a fast 0 - 255 -> % routine > > Eg Byte value = 89 > % value = 34 > > 89 / 255 * 100 The fastest routine, of course, would be a 256 byte lookup table... Assuming that code space is also a concern, try this: it's a bit shorter than the other solutions I've seen posted. Note that it actually calculates input * 102 / 256, but due to roundoff errors it produces 100 for an input of 255, so it should do the job for you. percentify ; input in W, output in temp movwf temp bcf status,c rrf temp,f ;temp = input * 128 / 256 addwf temp,f ;carry/temp = input * 384 / 256 rrf temp,f ;temp = input * 192 / 256 bcf status,c rrf temp,f ;temp = input * 96 / 256 swapf temp,w andlw 0x0F ;w = input * 6 / 256 addwf temp,f ;temp = input * 102 / 256 ;or: addwf temp,w for result in w return Jason Harper

Hi Scott. Nice code ;-) {Quote hidden}> scale_hex2dec > > MOVWF temp ;Hex value is in W. > CLRC ;Clear the Carry bit so it doesn't affect RRF > RRF temp,F > CLRC > RRF temp,F > CLRC > RRF temp,F ;temp = h>>3 > ADDWF temp,F ;temp = h + (h>>3) > RRF temp,F ;temp = (h + (h>>3)) >> 1 > ADDWF temp,F ;temp = h + ((h + (h>>3)) >> 1) > RRF temp,F > CLRC > RRF temp,W ;d = W = (h + (h + (h>>3)) >> 1) >> 2 > > RETURN > > ? Possible realization of 1 + 1/256 + ... correction. ;1/256 * (1 + 1/256) =~ 0.999985 / 255 ; scale255to100: ;25 = 16+8+1 clrf temp ;int clrf temp1 ;frac addwf temp,F ;1 rrf temp,F ;/2 rrf temp1,F rrf temp,F ;/4 rrf temp1,F rrf temp,F ;/8 rrf temp1,F addwf temp,F ;8 rrf temp,F ;/16 rrf temp1,F addwf temp,F ;16 rrf temp,F ;/32 rrf temp1,F rrf temp,W ;/64 rrf temp1,F ; ; movwf temp2 ;1/256/256 if longer tail ; movwf temp3 ;1/256/256/256 is required ; etc... ; addwf temp1,F ;1 + 1/256 correction btfss temp1,7 ;either round if >= 0.5 skpnc ;or take carry addlw 1 ;temp1 holds frac part ; ; movwf temp ;answer is either W or temp return If >=0.5 correction isn't required <btfss temp1,7> line should be commented. WBR Dmitry.