python - BoundedSemaphore 在 KeyboardInterrupt 上的线程中挂起

Question

如果在尝试获取信号量时引发 KeyboardInterrupt，则也尝试释放相同信号量对象的线程将无限期挂起。

代码:

import threading
import time

def worker(i, sema):
    time.sleep(2)
    print i, "finished"
    sema.release()


sema = threading.BoundedSemaphore(value=5)
threads = []
for x in xrange(100):
    sema.acquire()
    t = threading.Thread(target=worker, args=(x, sema))
    t.start()
    threads.append(t)

启动它，然后 ^C 在它运行时启动。它会挂起并且永远不会退出。

0 finished
3 finished
1 finished
2 finished
4 finished
^C5 finished
Traceback (most recent call last):
  File "/tmp/proof.py", line 15, in <module>
    sema.acquire()
  File "/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/threading.py", line 290, in acquire
    self.__cond.wait()
  File "/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/threading.py", line 214, in wait
    waiter.acquire()
KeyboardInterrupt
6 finished
7 finished
8 finished
9 finished

我怎样才能让它让最后几个线程自然死亡然后正常退出？ (如果你不尝试打断它，它会这样做)

Answer 1

您可以使用信号模块设置一个标志，告诉主线程停止处理:

import threading
import time
import signal
import sys

sigint = False

def sighandler(num, frame):
  global sigint
  sigint = True

def worker(i, sema):
  time.sleep(2)
  print i, "finished"
  sema.release()

signal.signal(signal.SIGINT, sighandler)
sema = threading.BoundedSemaphore(value=5)
threads = []
for x in xrange(100):
  sema.acquire()
  if sigint:
    sys.exit()
  t = threading.Thread(target=worker, args=(x, sema))
  t.start()
  t.join() 
  threads.append(t)