Question about waitcnt

Mazzini · 2011-08-12 08:58

Good day,

sry for trivial question ,Can you explain me waitcnt ?
In the code below, it would copy CNT's Value and add 101, then wait for CNT+ t1 cycles and toggle OUTA,1 ?I'm not sure

waitcnt t1,#0 ' zero What does it for?

''***************
''* PASD Test   *
''***************
''

CON

        'Use the following 2 lines if running on a Parallax PropDemo board
        _clkmode        = xtal1 + pll16x
        _xinfreq        = 5_000_000

        'Use the following 2 lines if running on a Hydra board
        '_clkmode        = xtal1 + pll8x
        '_xinfreq        = 10_000_000

VAR

  long  Cog, TestVar

OBJ
  dbg   :       "PASDebug"                '<---- Add for Debugger 


PUB main

  Cog := cognew(@entry, @TestVar) + 1

  dbg.start(31,30,@entry)                 '<---- Add for Debugger


PUB stop

  if Cog
    cogstop(Cog~ -  1)


DAT

                        org     0
entry

'  --------- Debugger Kernel add this at Entry (Addr 0) ---------
   long $34FC1202,$6CE81201,$83C120B,$8BC0E0A,$E87C0E03,$8BC0E0A
   long $EC7C0E05,$A0BC1207,$5C7C0003,$5C7C0003,$7FFC,$7FF8
'  --------------------------------------------------------------
                        mov DIRA,#$FF
loop 
                     
                        mov     t1,CNT
                        add     t1,#101    
                        waitcnt t1,#0
                        xor     OUTA,#1
                        jmp     #loop       
t1                      res 1

Mark_T · 2011-08-12 09:23

waitcnt waits for cnt to reach the value of the dest register and also then adds the source to the dest. The example is unnecessarily having to re-read the CNT register and do an explicit add that could be part of the waitcnt instruction - the usual idiom is

            mov   t1, cnt  ' setup
            add   t1, #101 ' first add
:loop       waitcnt t1, #101 'all subsequent adds use the waitcnt
            xor   outa, #1
            jmp   #:loop

The advantage of this way of doing things is you don't have to count cycles, the wait points are spaced exactly 101 clocks apart as waitcnt auto-adds 101 each time to t1 and never reloads cnt into t1.

Mazzini · 2011-08-12 11:36

Hi,

It works thanks , but my goal is reach 1 second pause and $1FF is the upper limit
Do you know how achieve that ?may be I need an kind of nested loop..
My osc run at 80Mhz --> 20MIPS = 50ns *512 = about 25us, its right ?

Thank so much

ericball · 2011-08-12 11:47

Since you want a 1 second delay, you can simply read long[0] which is CLKFREQ.

            rdlong  delay, #0
            mov     t1, cnt    ' setup
            add     t1, delay  ' first add
:loop       waitcnt t1, delay  'all subsequent adds use the waitcnt
            xor     outa, #1
            jmp     #:loop
delay       long    80_000_000 ' overwritten with clkfreq

Mazzini · 2011-08-12 12:21

ok , so waitcnt wait for delay (80.000.000 cycles) ?
delay*50ns = 4 second(?)
Thanks

StefanL38 · 2011-08-12 13:55

no the systemcounter increments by one every clocktick. So at 80MHz 80_000_000 means exactly 1 second
the waitcnt command makes the cog stop completely until the systemcounter matches the value "t1 + delay"

As long as the cog is stopped as if you would stop the cpu-clock no commands are executed the cog is just frozen.

keep the questions coming
best regards

Stefan

Mazzini · 2011-08-12 15:29

Do you mean every clocktick = 12.5ns ?
from Propeller's datasheet
<<
Note that for a cog clock frequency of 80 MHz, the
***counter update period is a mere 12.5 ns***. This high speed,
combined with 32-bit precision, allows for very dynamic
signal generation and measurement...
>>

Thanks

PJAllen · 2011-08-12 15:39

Ja! Si!
waitcnt (1 + cnt) is 12.5ns
waitcnt (2 + cnt) is 25.0ns
waitcnt (1_000 + cnt) is 12.5us
...

Mazzini · 2011-08-12 15:51

ok

thanks

StefanL38 · 2011-08-12 22:58

there still is a limitation

if you use waitcnt in spin the minimum waittime is 385 clockticks.
This is because of the overhead caused by the SPIN-interpreter.

in PASM the minimum waittime is 9 clockticks. it needs minimum 8 clockticks to load actual value of cnt and then execute the waitcnt-command.

if you try to wait for a shorter timeperiod the systemcounter has already "overrun" the target-value and then it will take 2^32 / 80_000_000 = 53 seconds until
the free running systemcounter again matches the value used in the waitcnt command.
keep the questions coming
best regards

Stefan

Mazzini · 2011-08-13 02:02

Hi,

I don't know SPIN , I just started with Pasm and I read this :

Instruction  Description                           Z     Result      C Result         R  Clocks
WAITCNT D,S  Wait for CNT = D, then add S into D   -                Unsigned Carry  1      5+

Mark_T · 2011-08-13 04:07

PJ Allen wrote: »

Ja! Si!
waitcnt (1 + cnt) is 12.5ns
waitcnt (2 + cnt) is 25.0ns
waitcnt (1_000 + cnt) is 12.5us
...

The first too will wait for a long time - by the time you've calculated cnt+1 or cnt+2 the cnt register will already be 8 greater and you'll be waiting for it to wrap round again...

PJAllen · 2011-08-13 05:51

Figured that I was reinforcing the clockticks-to-12.5ns aspect.

But, true, the time to setup the waitcnt/spin takes up a lot more time ( >> ) than the clockperiod/cnt when you get way, way down there, and that's something to consider.

Question about waitcnt

Comments