How to make a subprocess.call timeout using python 2.7.6?

Question

This has probably been asked but I cannot find anything regarding a subprocess.call timeout when using python 2.7

Answer 1

You could install subprocess32 module mentioned by @gps -- the backport of the subprocess module from Python 3.2/3.3 for use on 2.x. It works on Python 2.7 and it includes timeout support from Python 3.3.

subprocess.call() is just Popen().wait() and therefore to interrupt a long running process in timeout seconds:

#!/usr/bin/env python
import time
from subprocess import Popen

p = Popen(*call_args)
time.sleep(timeout)
try:
    p.kill()
except OSError:
    pass # ignore
p.wait()

If the child process may end sooner then a portable solution is to use Timer() as suggested in @sussudio's answer:

#!/usr/bin/env python
from subprocess import Popen
from threading import Timer

def kill(p):
    try:
        p.kill()
    except OSError:
        pass # ignore

p = Popen(*call_args)
t = Timer(timeout, kill, [p])
t.start()
p.wait()
t.cancel()

On Unix, you could use SIGALRM as suggested in @Alex Martelli's answer:

#!/usr/bin/env python
import signal
from subprocess import Popen

class Alarm(Exception):
    pass

def alarm_handler(signum, frame):
    raise Alarm

signal.signal(signal.SIGALRM, alarm_handler)


p = Popen(*call_args)
signal.alarm(timeout)  # raise Alarm in 5 minutes
try:
    p.wait()
    signal.alarm(0)  # reset the alarm
except Alarm:
    p.kill()
    p.wait()

To avoid using threads and signals here, subprocess module on Python 3 uses a busy loop with waitpid(WNOHANG) calls on Unix and winapi.WaitForSingleObject() on Windows.