Python:构建可重入信号量(结合RLock和信号量)
问题描述:
如何结合threading.RLock
和threading.Semaphore
?还是这样的结构已经存在?Python:构建可重入信号量(结合RLock和信号量)
在Python中,有一个可重入锁的原语,threading.RLock(N)
,它允许同一个线程多次获取一个锁,但是没有其他线程可以。还有threading.Semaphore(N)
,它允许在锁定之前获得锁定N
次。如何结合这两种结构?我想最多N
单独的线程能够获得锁,但我希望线程上的每个单独的锁是一个可重入的线程。
答
所以我猜一个可重入信号量不存在。这是我提出的实现,很乐意接受评论。
import threading
import datetime
class ReentrantSemaphore(object):
'''A counting Semaphore which allows threads to reenter.'''
def __init__(self, value = 1):
self.local = threading.local()
self.sem = threading.Semaphore(value)
def acquire(self):
if not getattr(self.local, 'lock_level', 0):
# We do not yet have the lock, acquire it.
start = datetime.datetime.utcnow()
self.sem.acquire()
end = datetime.datetime.utcnow()
if end - start > datetime.timedelta(seconds = 3):
logging.info("Took %d Sec to lock."%((end - start).total_seconds()))
self.local.lock_time = end
self.local.lock_level = 1
else:
# We already have the lock, just increment it due to the recursive call.
self.local.lock_level += 1
def release(self):
if getattr(self.local, 'lock_level', 0) < 1:
raise Exception("Trying to release a released lock.")
self.local.lock_level -= 1
if self.local.lock_level == 0:
self.sem.release()
__enter__ = acquire
def __exit__(self, t, v, tb):
self.release()