Post by thread:Challenge: 16bit delay loop

Guys & Alice It seems like there's been several questions about delay loops in the last few weeks. So I thought it'd be interesting to see if we could turn this into a challenge: Create a delay loop that can loop for up to 2^16 times while mininmizing the number of cycles per iteration. For example: clrf lo clrf hi movlw 1 loop: addwf lo,f ;increment the low byte skpnc ;if low byte rolls over addwf hi,f ; then increment the high byte skpc ;We're done if the high byte rolls over. goto loop This will loop 2^16 times and each pass through the loop is 6 cycles. I've got a solution that takes 5 cycles per iteration and doesn't use W (oh I shouldn't have said that - now you know the answer!). Any takers? Scott

Hi Scott, How about, 5 cycles per loop, doesn't use W: clrf lo clrf hi loop incfsz lo,f goto $+2 incfsz hi,f goto loop regards, Reggie -- e-mail: spam_OUTrberdinTakeThisOuTbigfoot.com ICQ#: 31651436 URL: http://www.bigfoot.com/~rberdin Scott Dattalo wrote: {Quote hidden}> > Guys & Alice > > It seems like there's been several questions about delay loops in the > last few weeks. So I thought it'd be interesting to see if we could turn > this into a challenge: > > Create a delay loop that can loop for up to 2^16 times while mininmizing > the number of cycles per iteration. For example: > > clrf lo > clrf hi > > movlw 1 > > loop: > > addwf lo,f ;increment the low byte > skpnc ;if low byte rolls over > addwf hi,f ; then increment the high byte > > skpc ;We're done if the high byte rolls over. > goto loop > > This will loop 2^16 times and each pass through the loop is 6 cycles. > > I've got a solution that takes 5 cycles per iteration and doesn't use W > (oh I shouldn't have said that - now you know the answer!). > > Any takers? > > Scott

Actually, this has an AVERAGE number of cycles per loop of very close to 3, since the first goto is only skipped 1/256th of the time. Sean At 09:42 AM 5/7/99 +0800, you wrote: {Quote hidden}>Hi Scott, > >How about, 5 cycles per loop, doesn't use W: > > clrf lo > clrf hi > >loop incfsz lo,f > goto $+2 > incfsz hi,f > goto loop > >regards, >Reggie > >-- >e-mail: .....rberdinKILLspam@spam@bigfoot.com >ICQ#: 31651436 >URL: http://www.bigfoot.com/~rberdin > > >Scott Dattalo wrote: >> >> Guys & Alice >> >> It seems like there's been several questions about delay loops in the >> last few weeks. So I thought it'd be interesting to see if we could turn >> this into a challenge: >> >> Create a delay loop that can loop for up to 2^16 times while mininmizing >> the number of cycles per iteration. For example: >> >> clrf lo >> clrf hi >> >> movlw 1 >> >> loop: >> >> addwf lo,f ;increment the low byte >> skpnc ;if low byte rolls over >> addwf hi,f ; then increment the high byte >> >> skpc ;We're done if the high byte rolls over. >> goto loop >> >> This will loop 2^16 times and each pass through the loop is 6 cycles. >> >> I've got a solution that takes 5 cycles per iteration and doesn't use W >> (oh I shouldn't have said that - now you know the answer!). >> >> Any takers? >> >> Scott > | | Sean Breheny | Amateur Radio Callsign: KA3YXM | Electrical Engineering Student \--------------=---------------- Save lives, please look at http://www.all.org Personal page: http://www.people.cornell.edu/pages/shb7 shb7KILLspamcornell.edu ICQ #: 3329174

Should have included this in the previous message: In case anyone wants to use this loop, the more general form is: movlw x movwf xf movlw y movwf xy delay(A) aloop delay(B) decfsz xf,f goto aloop delay(C) decfsz yf,f goto aloop Which has the timing expression: total cycles = (770+256*B+C)*Y + (B+3) * X + A + 2 In the case given by Reggie, where A=B=C=0: total cycles = 770*Y + 3*X + 2 Sean At 09:42 AM 5/7/99 +0800, you wrote: {Quote hidden}>Hi Scott, > >How about, 5 cycles per loop, doesn't use W: > > clrf lo > clrf hi > >loop incfsz lo,f > goto $+2 > incfsz hi,f > goto loop > >regards, >Reggie > >-- >e-mail: .....rberdinKILLspam.....bigfoot.com >ICQ#: 31651436 >URL: http://www.bigfoot.com/~rberdin > > >Scott Dattalo wrote: >> >> Guys & Alice >> >> It seems like there's been several questions about delay loops in the >> last few weeks. So I thought it'd be interesting to see if we could turn >> this into a challenge: >> >> Create a delay loop that can loop for up to 2^16 times while mininmizing >> the number of cycles per iteration. For example: >> >> clrf lo >> clrf hi >> >> movlw 1 >> >> loop: >> >> addwf lo,f ;increment the low byte >> skpnc ;if low byte rolls over >> addwf hi,f ; then increment the high byte >> >> skpc ;We're done if the high byte rolls over. >> goto loop >> >> This will loop 2^16 times and each pass through the loop is 6 cycles. >> >> I've got a solution that takes 5 cycles per iteration and doesn't use W >> (oh I shouldn't have said that - now you know the answer!). >> >> Any takers? >> >> Scott > | | Sean Breheny | Amateur Radio Callsign: KA3YXM | Electrical Engineering Student \--------------=---------------- Save lives, please look at http://www.all.org Personal page: http://www.people.cornell.edu/pages/shb7 EraseMEshb7spam_OUTTakeThisOuTcornell.edu ICQ #: 3329174

Scott Dattalo wrote: {Quote hidden}> > Guys & Alice > > It seems like there's been several questions about delay loops in the > last few weeks. So I thought it'd be interesting to see if we could > turn > this into a challenge: > > Create a delay loop that can loop for up to 2^16 times while > mininmizing > the number of cycles per iteration. For example: > > clrf lo > clrf hi > > movlw 1 > > loop: > > addwf lo,f ;increment the low byte > skpnc ;if low byte rolls over > addwf hi,f ; then increment the high byte > > skpc ;We're done if the high byte rolls over. > goto loop > > This will loop 2^16 times and each pass through the loop is 6 cycles. > > I've got a solution that takes 5 cycles per iteration and doesn't use > W > (oh I shouldn't have said that - now you know the answer!). > > Any takers? > > Scott clrf lo clrf hi loop: incf lo, f skpnz incfsz hi, f goto loop Bob Fehrenbach Wauwatosa, WI

Hi Sean, I think this is not the case, my routine has a "goto $+2" not a "goto loop" after the first increment (to be isochronous to 5 cycles per loop). The total cycles should be: Cycles = (65536 - (hi * 256 + lo)) * 5 Except when hi and lo is zero which is 65536*5 cycles. regards, Reggie -- e-mail: rberdinspam_OUTbigfoot.com ICQ#: 31651436 URL: http://www.bigfoot.com/~rberdin Sean Breheny wrote: {Quote hidden}> In case anyone wants to use this loop, the more general form is: > > movlw x > movwf xf > movlw y > movwf xy > delay(A) > aloop delay(B) > decfsz xf,f > goto aloop > delay(C) > decfsz yf,f > goto aloop > > Which has the timing expression: > > total cycles = (770+256*B+C)*Y + (B+3) * X + A + 2 > > In the case given by Reggie, where A=B=C=0: > > total cycles = 770*Y + 3*X + 2 > > Sean > > At 09:42 AM 5/7/99 +0800, you wrote: > >Hi Scott, > > > >How about, 5 cycles per loop, doesn't use W: > > > > clrf lo > > clrf hi > > > >loop incfsz lo,f > > goto $+2 > > incfsz hi,f > > goto loop

At 21:58 6/05/99 -0400, you wrote: {Quote hidden}>Should have included this in the previous message: > >In case anyone wants to use this loop, the more general form is: > > > movlw x > movwf xf > movlw y > movwf xy > delay(A) >aloop delay(B) > decfsz xf,f > goto aloop > delay(C) > decfsz yf,f > goto aloop > >Which has the timing expression: > >total cycles = (770+256*B+C)*Y + (B+3) * X + A + 2 > >In the case given by Reggie, where A=B=C=0: > >total cycles = 770*Y + 3*X + 2 > >Sean > Perhaps I am being a bit picky here but I think that the equation for total cycles is wrong 770*7 + 3*X + 2. What happens if X or Y is 0. Dennis

Hi Reggie, Doh! Yes, of course, you are right, I was a bit over zealous in my analysis there! However, both methods are valid loops (although, mine is not isochronous). Yours gives a greater maximum number of cycles than mine,too. Dennis also pointed out that my expression is wrong for the case where Y=0 or X=0. Yes, although you can count Y=0 as Y=256 and X=0 as X=256,and then it works. I should have mentioned that. Thanks for the corrections. Sean At 10:22 AM 5/7/99 +0800, you wrote: {Quote hidden}>Hi Sean, > >I think this is not the case, my routine has a "goto $+2" not a "goto >loop" after the first increment (to be isochronous to 5 cycles per >loop). The total cycles should be: > >Cycles = (65536 - (hi * 256 + lo)) * 5 > >Except when hi and lo is zero which is 65536*5 cycles. > >regards, >Reggie > >-- >e-mail: @spam@rberdinKILLspambigfoot.com >ICQ#: 31651436 >URL: http://www.bigfoot.com/~rberdin > > >Sean Breheny wrote: >> In case anyone wants to use this loop, the more general form is: >> >> movlw x >> movwf xf >> movlw y >> movwf xy >> delay(A) >> aloop delay(B) >> decfsz xf,f >> goto aloop >> delay(C) >> decfsz yf,f >> goto aloop >> >> Which has the timing expression: >> >> total cycles = (770+256*B+C)*Y + (B+3) * X + A + 2 >> >> In the case given by Reggie, where A=B=C=0: >> >> total cycles = 770*Y + 3*X + 2 >> >> Sean >> >> At 09:42 AM 5/7/99 +0800, you wrote: >> >Hi Scott, >> > >> >How about, 5 cycles per loop, doesn't use W: >> > >> > clrf lo >> > clrf hi >> > >> >loop incfsz lo,f >> > goto $+2 >> > incfsz hi,f >> > goto loop > | | Sean Breheny | Amateur Radio Callsign: KA3YXM | Electrical Engineering Student \--------------=---------------- Save lives, please look at http://www.all.org Personal page: http://www.people.cornell.edu/pages/shb7 KILLspamshb7KILLspamcornell.edu ICQ #: 3329174

On Thu, 6 May 1999, Bob Fehrenbach wrote: > clrf lo > clrf hi > loop: > incf lo, f > skpnz > incfsz hi, f > goto loop That's exactly my solution. And On Fri, 7 May 1999, Regulus Berdin wrote: {Quote hidden}> > How about, 5 cycles per loop, doesn't use W: > > clrf lo > clrf hi > > loop incfsz lo,f > goto $+2 > incfsz hi,f > goto loop > Yep! Here's another solution that gives two extra cycles for other stuff (like CLRWDT or testing an I/O bit) clrf lo clrf hi loop1 nop ;2 free cycles executed 255 out of 256 nop ;iterations loop2 decfsz lo,f ;or incfsz goto loop1 decfsz hi,f ;if low byte rolls over, dec hi byte, goto loop2 ;but don't branch to the beginning.

On Fri, 7 May 1999, Scott Dattalo wrote: > Here's another solution that gives two extra cycles for other stuff > (like CLRWDT or testing an I/O bit) > > clrf lo > clrf hi > > loop1 > > nop ;2 free cycles executed 255 out of 256 > nop ;iterations > > loop2 > > decfsz lo,f ;or incfsz > goto loop1 > decfsz hi,f ;if low byte rolls over, dec hi byte, > goto loop2 ;but don't branch to the beginning. After a little churning, I found a 3 cycle solution: clrf lo clrf hi loop decfsz lo,f goto loop ;lo is zero right now. decf hi,f skpnz goto done ;or perhaps return ;okay, we've gotta lot of choices here ; ;we could do this: ; incf lo,f ; nop ;or clrwdt ; incf lo,f ; goto loop ; ;But I like this: bsf lo,1 ;Equivalent to 2 incf's nop ;Two free cycles nop ;everytime lo rolls over goto loop ;Or if you wanted more free cycles: ; bsf lo,2 ;Equivalent to 4 incf's ; goto $+1 ;Eight free cycles ; goto $+1 ;everytime lo rolls ; goto $+1 ;over ; goto $+1 ; ; goto loop ; ;Of course, you could preset the lo byte ;to any value and then have the remaining ;cycles evailable for other stuff. ; Like everything I comment, this untested :). The time at which the lo byte is incremented does not happen exactly every third cycle after it rolls over. In fact, some values of lo are skipped altogether. However, the total number of cycles between the beginning of the loop and the 'goto done' instruction is 2^16 * 3. This kind of loop would be useful if you wanted to have a fixed delay AND still be able to do other things too. Scott