python线程中超时信号的替代方法
我有一个应用程序,它依赖于一些阻塞操作的超时信号 例如:python线程中超时信号的替代方法,python,multithreading,python-3.x,timeout,signals,Python,Multithreading,Python 3.x,Timeout,Signals,我有一个应用程序,它依赖于一些阻塞操作的超时信号 例如: def wait_timeout(signum, frame): raise Exception("timeout") signal.signal(signal.SIGALRM, wait_timeout) signal.setitimer(signal.ITIMER_REAL, 5) try: while true: print("zzz") sleep(1) except Excep
def wait_timeout(signum, frame):
raise Exception("timeout")
signal.signal(signal.SIGALRM, wait_timeout)
signal.setitimer(signal.ITIMER_REAL, 5)
try:
while true:
print("zzz")
sleep(1)
except Exception as e:
# timeout
print("Time's up")
现在我已经使用相同的方法实现了多线程,但是对于所有线程,我都得到了ValueError:signal只在主线程中起作用
我假设带有信号的超时方法不适用于线程
不幸的是,我不能使用这样的东西:
timeout = 5
start = time.time()
while true:
print("zzz")
sleep(1)
if time.time() <= start+timeout:
print("Time's up)
break
process\u thread
函数通过执行以下操作来处理工具的标准输出:
for line in p.stdout:
# process line of the processes stdout
此过程可能需要很长时间,例如,一旦工具
不产生任何输出。我只想要工具的输出,比如说5秒,所以for循环需要在特定的超时后被中断
这就是我使用信号的目的,但显然它们在线程中不起作用
edit3:我创建了一个更详细、更准确的示例,说明我打算如何在线程中使用信号 你要找的是一个看门狗
def watchdog(queue):
while True:
watch = queue.get()
time.sleep(watch.seconds)
try:
watch = queue.get_nowait()
# No except, got queue message,
# do noting wait for next watch
except queue.Empty:
os.kill(watch.pid, signal.SIGKILL)
def workload_thread(queue):
pid = os.getpid()
queue.put({'pid':pid, 'seconds':5})
# do your work
# Test Watchdog
# time.sleep(6)
queue.put({'pid':pid, 'done':True})
注意:未测试代码,可能有语法错误 这实现了一个类终止符
,
在给定的timeout=5
之后,它向线程进程发送信号.SIG…
。可以使用多个不同的pid
class Terminator(object):
class WObj():
def __init__(self, process, timeout=0, sig=signal.SIGABRT):
self.process = process
self.timeout = timeout
self.sig = sig
def __init__(self):
self.__queue = queue.Queue()
self.__t = Thread(target=self.__sigterm_thread, args=(self.__queue,))
self.__t.start()
time.sleep(0.1)
def __sigterm_thread(self, q):
w = {}
t = 0
while True:
time.sleep(0.1);
t += 1
try:
p = q.get_nowait()
if p.process == 0 and p.sig == signal.SIGTERM:
# Terminate sigterm_thread
return 1
if p.process.pid not in w:
if p.timeout > 0 and p.sig != signal.SIGABRT:
w[p.process.pid] = p
else:
if p.sig == signal.SIGABRT:
del (w[p.process.pid])
else:
w[p.process.pid].timeout = p.timeout
except queue.Empty:
pass
if t == 10:
for key in list(w.keys()):
p = w[key]
p.timeout -= 1
if p.timeout == 0:
""" A None value indicates that the process hasn't terminated yet. """
if p.process.poll() == None:
p.process.send_signal(p.sig)
del (w[p.process.pid])
t = 0
# end if t == 10
# end while True
def signal(self, process, timeout=0, sig=signal.SIGABRT):
self.__queue.put(self.WObj(process, timeout, sig))
time.sleep(0.1)
def close(self, process):
self.__queue.put(self.WObj(process, 0, signal.SIGABRT))
time.sleep(0.1)
def terminate(self):
while not self.__queue.empty():
trash = self.__queue.get()
if self.__t.is_alive():
self.__queue.put(self.WObj(0, 0, signal.SIGTERM))
def __enter__(self):
return self
def __exit__(self, exc_type, exc_val, exc_tb):
self.__del__()
def __del__(self):
self.terminate()
这是工作负载,例如:
def workload(n, sigterm):
print('Start workload(%s)' % n)
arg = str(n)
p = Popen(["tool", "--param", arg], stdin=PIPE, stdout=PIPE, stderr=STDOUT)
sigterm.signal(p, timeout=4, sig=signal.SIGTERM)
while True:
for line in p.stdout:
# process line of the processes stdout
print(line.strip())
time.sleep(1)
if p.poll() != None:
break
sigterm.close(p)
time.sleep(0.1)
print('Exit workload(%s)' % n)
if __name__ == '__main__':
with Terminator() as sigterm:
p1 = Thread(target=workload, args=(1, sigterm)); p1.start(); time.sleep(0.1)
p2 = Thread(target=workload, args=(2, sigterm)); p2.start(); time.sleep(0.1)
p3 = Thread(target=workload, args=(3, sigterm)); p3.start(); time.sleep(0.1)
p1.join(); p2.join(); p3.join()
time.sleep(0.5)
print('EXIT __main__')
使用Python:3.4.2和Python:2.7.9进行测试如何开始线程?您在定义它们时是否设置了daemon=True
?如果是这样,那么当main
线程死亡时,这些线程将被终止。这就是你想要做的吗?是的,我正在以执事的身份启动这些线程,我会在一分钟内编辑OP。不,这不是我要尝试的,我会在OP中更好地解释它。我已经更新了OP@BillyReading你的要点并不能打开我对工作负载过程的理解。但无论如何,我已经更新了我的答案。@stovfl我可以尝试创建一个更清晰的示例代码,但我会查看您的编辑,提前谢谢。我觉得我好像以前见过/使用过这个或类似的东西,但我现在无法理解它。我也发现,在研究看门狗的话题时,我仍然不知道从哪里开始。您能用我发布的代码创建一个示例吗?谢谢您的反馈和示例,但我认为这不适用于多线程,如OP中所述。我在OP中添加了一个更详细的代码片段作为链接。此外,如果我尝试将该解决方案扩展到多线程,我会遇到问题,我所有的线程都报告相同的PID。是的,它不可伸缩。在阅读您更详细的代码片段后,将返回scalabel解决方案。
def workload(n, sigterm):
print('Start workload(%s)' % n)
arg = str(n)
p = Popen(["tool", "--param", arg], stdin=PIPE, stdout=PIPE, stderr=STDOUT)
sigterm.signal(p, timeout=4, sig=signal.SIGTERM)
while True:
for line in p.stdout:
# process line of the processes stdout
print(line.strip())
time.sleep(1)
if p.poll() != None:
break
sigterm.close(p)
time.sleep(0.1)
print('Exit workload(%s)' % n)
if __name__ == '__main__':
with Terminator() as sigterm:
p1 = Thread(target=workload, args=(1, sigterm)); p1.start(); time.sleep(0.1)
p2 = Thread(target=workload, args=(2, sigterm)); p2.start(); time.sleep(0.1)
p3 = Thread(target=workload, args=(3, sigterm)); p3.start(); time.sleep(0.1)
p1.join(); p2.join(); p3.join()
time.sleep(0.5)
print('EXIT __main__')