Hi Bin:
________________________________
From: Bin Huang <[email protected]>
To: Andrew Francis <[email protected]>
Cc: "[email protected]" <[email protected]>
Sent: Friday, June 14, 2013 4:29 PM
Subject: More questions on stackless.run()
>As you suggested, I found thinking of stackless.run() being in an infinite
>loop was helpful. So I >went on to test stackless.run() in a few different
>scenarios. This time, I got confused by different >outputs in different
>scenarios.
Typically you execute stackless.run() once. A common structure for a Stackless
Python
programme is:
your-programme-goes-here
stackless.run()
I ought to look at how stackless.run() is implemented but think of
stackless.run conceptually
as a pump that schedules tasklets as long as there are runnable tasklets.
Again, conceptually
stackless.run() is akin to:
while stackless.getruncount() > 1:
stackless.schedule()
As long as there is more than one tasklet (the main), the next tasklet in the
runnable queue will be
scheduled.
...
>Third, if I don't even call stackless.run(), the stackless runtime is still
>able to schedule some tasklets >to run (e.g., code #2). Why is this?
Good question. I don't have a good answer (without looking at the code) but a
number of operations
will engage the scheduler and get the ball rolling. The scheduler in
non-preemptive mode is a passive entity: it has to be called to do things.
Cheers,
Andrew
_______________________________________________
Stackless mailing list
[email protected]
http://www.stackless.com/mailman/listinfo/stackless
