我在Python中有这个例子,它演示了条件变量的使用。
import logging
import threading
import time
logging.basicConfig(level=logging.DEBUG, format='%(asctime)s (%(threadName)-2s) %(message)s',)
def consumer(cond):
# wait for the condition and use the resource
logging.debug('Starting consumer thread')
t = threading.currentThread()
cond.wait()
logging.debug('Resource is available to consumer')
def producer(cond):
# set up the resource to be used by the consumer
logging.debug('Starting producer thread')
logging.debug('Making resource available')
cond.notifyAll()
condition = threading.Condition()
# pass each thread a 'condition'
c1 = threading.Thread(name='c1', target=consumer, args=(condition,))
c2 = threading.Thread(name='c2', target=consumer, args=(condition,))
p = threading.Thread(name='p', target=producer, args=(condition,))
# start two threads and put them into 'wait' state
c1.start()
c2.start()
# after two seconds or after some operation notify them to free or step over the wait() function
time.sleep(2)
p.start()
但是,它会在线程上引发运行时错误un-acquired lock
。我知道我需要使用acquire
和release
函数,但我不确定它们的用法以及它们的用途。
答案 0 :(得分:9)
条件是提供等待/通知功能的基础Lock
的包装器。你需要先acquire
一个锁才能释放它 - wait
做了什么。值得注意的是,一旦它被重新唤醒,重新获取锁定。因此,在获得和释放之间确保互斥,wait
"产生"控制锁,如果这是有道理的。
不要手动获取/释放,只需使用Condition
作为上下文管理器:
def consumer(cond):
with cond:
cond.wait()
logging.debug('Resource is available to consumer')
如果由于某种原因你在没有上下文管理器的情况下卡在了python版本上,这相当于:
def consumer(cond):
try:
cond.acquire()
cond.wait()
finally:
cond.release()
logging.debug('Resource is available to consumer')
通常你想确保只有一个消费者被唤醒,所以经常使用以下习语:
with cond:
while some_queue.isEmpty():
cond.wait()
#get one from queue
因此,你可以notify
任意数量的消费者和额外的消费者只要在队列为空时立即重新入睡。