Here... I'll contribute a try-to-be-brief summary (to the best of my
recollection):
There is a special: the "main" tasklet.
There is a doubly-linked (circular) list of "runnable" tasklets.
The head of the runnables is the "current" tasklet (unless "main" is
current...)
Calling tasklet.insert() inserts (or re-inserts) at the END of the
runnables.
Calling tasklet.remove() removes from runnables.
Calling stackless.schedule() rotates the head of runnables to the NEXT
one after current
and switches to it.
Calling chan.receive/send() can remove the current tasklet from
runnables
if nothing is "waiting at the other end" (the NEXT tasklet in
runnables
is continued).
OR (depending on the channel preference) continue in the caller
(inserting the sender/receiver at the "other end" onto runnables)
or rotate the head of runnables to the sender/receiver and pick up
there.
"Preemptive" Scheduling:
t = stackless.run() pauses the "main" tasklet and begins running from
runnables.
If the timeout argument is used, whatever tasklet is current after
some number
of eval loops is paused and the "main" tasklet returned to. (A simple
scheduler
at this point could call say "t.insert()" to rotate runnables and call
stackless.run()
again causing "preemptive round robin" behavior.
Note that the scheduling order is not necessarily strictly a fair
"round robin" --
since the head of the queue is "rotated" at some points, theoretically
such a
scheduler could starve some tasklets under certain circumstances.
On Nov 11, 2008, at 9:44 AM, Andrew Francis wrote:
Hi Folks:
I understand that the Stackless documentation says that tasklets are
scheduled in a round-robin fashion. However I am interested in a
more detailed description of the scheduling algorithm. To keep thing
simple, a description of the scheduler in non-preemptive mode would
be nice. In particular, I am interested in how channel preferences
come into the picture. I know the code is out there, but I would
like to start with a high level description.
Cheers,
Andrew
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