Time delay & CNT roll over ??

twm47099

I'm trying to learn Spin, and have a question about the WAITCNT command.

According to the Prop Manual states that at 80MHz clock, it takes about 53 seconds for the System Counter to wrap around.

If I wrote:
 


delay := clkfreq * 30
waitcnt(delay + cnt)

The cog should wait for 30 seconds. But that would also result in a high probability that cnt would roll over to 0, and the wait would not be satisfied. Using smaller delays would have a smaller but finite probability of having that occur.

Is that correct? What am mixing up? How does this work, and is there a way to specify long deterministic delays?

Thanks
Tom

Phil Pilgrim (PhiPi)

It doesn't matter if cnt rolls over during the waitcnt. The only time your scheme won't work is if delay is more than about 53 seconds. That's how long it takes for cnt to make one full cycle at 80 MHz.
-Phil

Ariba

waitcnt compares to equality of cnt and the value you pass.(delay + cnt) is calculated with 32bit arithmetic, so the overflow behaves exactly the same as the rollover of the 32bit cnt register.
Andy

Sam_sneak

The Propeller Manual explains all this, but the language might be a bit technical for some.

If you take a look at your actual code, you retrieve the value of CNT and add it too your DELAY to make up a target.

Since that is all in 32 bit binary math in the actual machine, the addition compensates for the CNT roll over at Zero if and only if your total delay is less than the 53 seconds.

The 53 seconds limit is bases on going higher than what CNT originally read.  That is a second and importantly different roll over.

The 53 second is roughly based on 80Mhz clock being very near to the where you pass the same value that you got in CNT.