[Stackless] How tasklets are scheduled

[Christian Tismer]

Stephan Diehl wrote:

> I played around with stackless a little bit more. I made some observations 
> about scheduling which I like to share here. It's probably stuff, everybody 
> knows, but me :-)
> Please correct me, if the description is wrong.
> 
> Stackless has an internal list of running tasklets. The main tasklet (the 
> program that is executed) is always on that list.

No. The main program can play all games the others
can, and if it is waiting in a channel or remove()d,
it is not in the list.
But if an exception is not caught, main will be
forcibly inserted to handle this.

> In order to get a new tasklet on that list, its 'run' method must be called.

Not really. You can do so, but it suffices to schedule(),
or to call stackless.run(), see its doc.

> The tasklet is removed from that list, if it calls 'schedule()', i.e. that it 
> will only continue to run, if its 'run' method is called again.

All not necessary.
t=tasklet(func)
# creates a tasklet.
t(param1,...)
# initialized func, and *inserts* the tasklet
...

> Instead of the 'run' method, 'insert' could be used as well. But this gives a 
> different runtime behaviour.
> (maybe 'insert' should be called 'append' since it appends the tasklet to the 
> queue)

Insert was used, because the queue is a circle.
I was also planning (but forgot) to give it
a parameter which gives the position in the list.
(Default = -1)

cheers - chris

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