jac*_*ack 71 python multithreading
如何对Ctrl + C键事件进行多线程python程序响应？
编辑:代码是这样的:
import threading
current = 0
class MyThread(threading.Thread):
def __init__(self, total):
threading.Thread.__init__(self)
self.total = total
def stop(self):
self._Thread__stop()
def run(self):
global current
while current<self.total:
lock = threading.Lock()
lock.acquire()
current+=1
lock.release()
print current
if __name__=='__main__':
threads = []
thread_count = 10
total = 10000
for i in range(0, thread_count):
t = MyThread(total)
t.setDaemon(True)
threads.append(t)
for i in range(0, thread_count):
threads[i].start()
我试图在所有线程上删除join()但它仍然无效.是因为每个线程的run()过程中的锁段？
编辑:上面的代码应该可以工作但是当当前变量在5,000-6,000范围内并且通过如下错误时它总是被中断
Exception in thread Thread-4 (most likely raised during interpreter shutdown):
Traceback (most recent call last):
File "/usr/lib/python2.5/threading.py", line 486, in __bootstrap_inner
File "test.py", line 20, in run
<type 'exceptions.TypeError'>: unsupported operand type(s) for +=: 'NoneType' and 'int'
Exception in thread Thread-2 (most likely raised during interpreter shutdown):
Traceback (most recent call last):
File "/usr/lib/python2.5/threading.py", line 486, in __bootstrap_inner
File "test.py", line 22, in run
Ale*_*lli 91
除了主线程之外的每个线程都是一个守护进程(t.daemon = True2.6或更好,t.setDaemon(True)在2.6或更少,对于每个线程对象,t然后再启动它).这样,当主线程接收到KeyboardInterrupt时,如果它没有捕获它或捕获它但决定终止,整个过程将终止.查看文档.
编辑:刚刚看到OP的代码(最初没有发布)和"它不起作用"的说法,似乎我必须添加...:
当然,如果你希望你的主线程保持响应(例如控制-c),不要让它陷入阻塞调用,例如join另一个线程 - 尤其不是完全无用的阻塞调用,例如joining 守护程序线程.例如,只需从当前更改主线程中的最后一个循环(无语和破坏):
for i in range(0, thread_count):
threads[i].join()
更合理的事情:
while threading.active_count() > 0:
time.sleep(0.1)
如果你的主要没有比所有线程自己终止,或者接收控制-C(或其他信号)更好的事情.
当然,还有许多其他可用的模式,如果你宁愿让你的线程没有突然终止(作为守护线程) - 除非他们也永远陷入无条件阻塞调用,死锁等等;-) .
Wal*_*ndt 16
有两种主要方式,一种是干净的,一种是简单的.
干净的方法是在你的主线程中捕获KeyboardInterrupt,并设置一个标志,你的后台线程可以检查,以便他们知道退出; 这是一个使用全局的简单/稍微混乱的版本:
exitapp = False
if __name__ == '__main__':
try:
main()
except KeyboardInterrupt:
exitapp = True
raise
def threadCode(...):
while not exitapp:
# do work here, watch for exitapp to be True
凌乱但简单的方法是捕获KeyboardInterrupt并调用os._exit(),它会立即终止所有线程.
小智 5
一个工人可能对你有所帮助:
#!/usr/bin/env python
import sys, time
from threading import *
from collections import deque
class Worker(object):
def __init__(self, concurrent=1):
self.concurrent = concurrent
self.queue = deque([])
self.threads = []
self.keep_interrupt = False
def _retain_threads(self):
while len(self.threads) < self.concurrent:
t = Thread(target=self._run, args=[self])
t.setDaemon(True)
t.start()
self.threads.append(t)
def _run(self, *args):
while self.queue and not self.keep_interrupt:
func, args, kargs = self.queue.popleft()
func(*args, **kargs)
def add_task(self, func, *args, **kargs):
self.queue.append((func, args, kargs))
def start(self, block=False):
self._retain_threads()
if block:
try:
while self.threads:
self.threads = [t.join(1) or t for t in self.threads if t.isAlive()]
if self.queue:
self._retain_threads()
except KeyboardInterrupt:
self.keep_interrupt = True
print "alive threads: %d; outstanding tasks: %d" % (len(self.threads), len(self.queue))
print "terminating..."
# example
print "starting..."
worker = Worker(concurrent=50)
def do_work():
print "item %d done." % len(items)
time.sleep(3)
def main():
for i in xrange(1000):
worker.add_task(do_work)
worker.start(True)
main()
print "done."
# to keep shell alive
sys.stdin.readlines()
您始终可以将线程设置为“守护程序”线程，例如：
t.daemon = True
t.start()
每当主线程死亡时，所有线程都会随之死亡。
http://www.regexprn.com/2010/05/killing-multithreaded-python-programs.html