I am trying to run a simple multiple processes application in Python. The main thread spawns 1 to N processes and waits until they all done processing. The processes each run an infinite loop, so they can potentially run forever without some user interruption, so I put in some code to handle a KeyboardInterrupt:
#!/usr/bin/env python
import sys
import time
from multiprocessing import Process
def main():
# Set up inputs..
# Spawn processes
Proc( 1).start()
Proc( 2).start()
class Proc ( Process ):
def __init__ ( self, procNum):
self.id = procNum
Process.__init__(self)
def run ( self ):
doneWork = False
while True:
try:
# Do work...
time.sleep(1)
sys.stdout.write('.')
if doneWork:
print "PROC#" + str(self.id) + " Done."
break
except KeyboardInterrupt:
print "User aborted."
sys.exit()
# Main Entry
if __name__=="__main__":
main()
The problem is that when using CTRL-C to exit, I get an additional error even though the processes seem to exit immediately:
......User aborted.
Error in atexit._run_exitfuncs:
Traceback (most recent call last):
File "C:\Python26\lib\atexit.py", line 24, in _run_exitfuncs
func(*targs, **kargs)
File "C:\Python26\lib\multiprocessing\util.py", line 281, in _exit_function
p.join()
File "C:\Python26\lib\multiprocessing\process.py", line 119, in join
res = self._popen.wait(timeout)
File "C:\Python26\lib\multiprocessing\forking.py", line 259, in wait
res = _subprocess.WaitForSingleObject(int(self._handle), msecs)
KeyboardInterrupt
Error in sys.exitfunc:
Traceback (most recent call last):
File "C:\Python26\lib\atexit.py", line 24, in _run_exitfuncs
func(*targs, **kargs)
File "C:\Python26\lib\multiprocessing\util.py", line 281, in _exit_function
p.join()
File "C:\Python26\lib\multiprocessing\process.py", line 119, in join
res = self._popen.wait(timeout)
File "C:\Python26\lib\multiprocessing\forking.py", line 259, in wait
res = _subprocess.WaitForSingleObject(int(self._handle), msecs)
KeyboardInterrupt
I am running Python 2.6 on Windows. If there is a better way to handle multiprocessing in Python, please let me know.