I\'m working on a renderfarm, and I need my clients to be able to launch multiple instances of a renderer, without blocking so the client can receive new commands. I\'ve got
Found the answer to my own question. The primary problem was that I was calling a third-party application rather than a function. When I call the subprocess [either using call() or Popen()] it creates a new instance of python whose only purpose is to call the new application. However when python exits, it will kill this new instance of python and leave the application running.
The solution is to do it the hard way, by finding the pid of the python process that is created, getting the children of that pid, and killing them. This code is specific for osx; there is simpler code (that doesn't rely on grep) available for linux.
for process in pool:
processId = process.pid
print "attempting to terminate "+str(processId)
command = " ps -o pid,ppid -ax | grep "+str(processId)+" | cut -f 1 -d \" \" | tail -1"
ps_command = Popen(command, shell=True, stdout=PIPE)
ps_output = ps_command.stdout.read()
retcode = ps_command.wait()
assert retcode == 0, "ps command returned %d" % retcode
print "child process pid: "+ str(ps_output)
os.kill(int(ps_output), signal.SIGTERM)
os.kill(int(processId), signal.SIGTERM)
I found solution: stop pool in separate thread, like this:
def close_pool():
global pool
pool.close()
pool.terminate()
pool.join()
def term(*args,**kwargs):
sys.stderr.write('\nStopping...')
# httpd.shutdown()
stophttp = threading.Thread(target=httpd.shutdown)
stophttp.start()
stoppool=threading.Thread(target=close_pool)
stoppool.daemon=True
stoppool.start()
signal.signal(signal.SIGTERM, term)
signal.signal(signal.SIGINT, term)
signal.signal(signal.SIGQUIT, term)
Works fine and always i tested.
signal.SIGINT
Interrupt from keyboard (CTRL + C). Default action is to raise KeyboardInterrupt.
signal.SIGKILL
Kill signal. It cannot be caught, blocked, or ignored.
signal.SIGTERM
Termination signal.
signal.SIGQUIT
Quit with core dump.
If you're still experiencing this issue, you could try simulating a Pool
with daemonic processes (assuming you are starting the pool/processes from a non-daemonic process). I doubt this is the best solution since it seems like your Pool
processes should be exiting, but this is all I could come up with. I don't know what your callback does so I'm not sure where to put it in my example below.
I also suggest trying to create your Pool
in __main__
due to my experience (and the docs) with weirdness occurring when processes are spawned globally. This is especially true if you're on Windows: http://docs.python.org/2/library/multiprocessing.html#windows
from multiprocessing import Process, JoinableQueue
# the function for each process in our pool
def pool_func(q):
while True:
allRenderArg, otherArg = q.get() # blocks until the queue has an item
try:
render(allRenderArg, otherArg)
finally: q.task_done()
# best practice to go through main for multiprocessing
if __name__=='__main__':
# create the pool
pool_size = 2
pool = []
q = JoinableQueue()
for x in range(pool_size):
pool.append(Process(target=pool_func, args=(q,)))
# start the pool, making it "daemonic" (the pool should exit when this proc exits)
for p in pool:
p.daemon = True
p.start()
# submit jobs to the queue
for i in range(totalInstances):
q.put((allRenderArgs[i], args[2]))
# wait for all tasks to complete, then exit
q.join()
# -*- coding:utf-8 -*-
import multiprocessing
import time
import sys
import threading
from functools import partial
#> work func
def f(a,b,c,d,e):
print('start')
time.sleep(4)
print(a,b,c,d,e)
###########> subProcess func
#1. start a thead for work func
#2. waiting thead with a timeout
#3. exit the subProcess
###########
def mulPro(f, *args, **kwargs):
timeout = kwargs.get('timeout',None)
#1.
t = threading.Thread(target=f, args=args)
t.setDaemon(True)
t.start()
#2.
t.join(timeout)
#3.
sys.exit()
if __name__ == "__main__":
p = multiprocessing.Pool(5)
for i in range(5):
#1. process the work func with "subProcess func"
new_f = partial(mulPro, f, timeout=8)
#2. fire on
p.apply_async(new_f, args=(1,2,3,4,5),)
# p.apply_async(f, args=(1,2,3,4,5), timeout=2)
for i in range(10):
time.sleep(1)
print(i+1,"s")
p.close()
# p.join()