Peter Brooks writes:
> I'm not sure if this'll interest anybody, but I expect that I'm
> going to get some mutual recursion in my simulation, so I needed to
...
> returned, then this solution won't help you. Often, though, you're
> not interested in what's returned and would just like the routine to
> exit once it's called itself, or another process, recursively.
>
> If that's the case, this solution, using threads, allows you to have
> functions call themselves, or each other, indefinitely. It works OK
> on a macbook pro and the thread count varies from 2-3, so it handles
... hard to resist ... but somehow I manage ... whew ...
A light-weighter way is to have each task end by assigning the next
task and returning, instead of calling the next task directly. When a
task returns, a driver loop will call the assigned task, which again
does a bounded amount of work, assigns the next task, and returns.
Tasks can even pass parameters in the same way.
Like so, Dr. Fred keeps adding to a pile as long as there is a pile,
and Mr. Jim keeps taking from it as long as it's there to take from:
from random import choice
def fred():
global fun, arg
print('Fred adds 1')
fun, arg = choice((fred, jim)), arg + 1
def jim():
global fun, arg
print('Jim takes 1')
fun, arg = choice((fred, jim)), arg - 1
if __name__ == '__main__':
fun, arg = choice((fred, jim)), 3
while arg:
print('Pile is', arg, end = '\t')
fun()
else:
print('Pile is gone')
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