Question

我的反对是采取行动,然后在再次采取行动之前睡觉第二次。然而,该物体有各种变数,无论在第二个睡眠期,均须随时取用。当目标执行被中止时,其变量是否无法使用?

这是一种 Python。

感谢。

EDIT 1: Clarifying inaccessible :

Object A has variable x that Object B needs to access repeatedly. Object A sleeps for 1 second. Will there be a problem if Object B tries to use x while A is sleeping?

EDIT 2: Forgot to mention the two objects are running as individual processes (I m using processes to avoid the GIL)

EDIT:

class QueueController(Process):
    def __init__(self):
        Process.__init__(self)
        self.queue_stream    = Queue()
        self.queue_language  = Queue()
        self.queue_expander  = Queue()
        self.queue_feature   = Queue()
        self.queue_classify  = Queue()
        self.db = connect_to_db()

    def run(self):
        # Every second, allow 10 more tweets from the raw_tweets db
        # enter the queue_stream to be processed
        value = 0
        while True:
            for i in db.raw_tweets.find().skip(value).limit(30):
                self.queue_stream.put(i)
                value+=30
                sleep(1)

是否还有另一个使用这一奎埃勒舱位的物体仍然能够使用电线——变量,即使它睡了二次吗? 恐怕睡觉会停止处决,但也会接触这些格言——作为一方影响的变量。

Answer 1

<代码>时间.sleep(>> 确实释放了GIL,因此允许其他校对运行。

我之所以作出这一回答,而不是回答你提出的问题,是因为你提出的问题毫无意义。除了其他透镜的运行能力之外,没有其他途径。

Answer 2

Never touched much on python but still I think I do know the answer. The thing is: there are any threads other than main? If the answer is yes, then they are accessible (watch out for racing conditions here), since a thread can go to sleep but the others are still active. If you don t have any other threads, then you won t be able to do much, since the main thread is sleeping, so nothing much will happen.

EDIT:

目标A有变数x,目标B需要反复接触。反对A 睡觉1秒。如果目标B试图在A睡觉时使用x,是否会出现问题?

正如有人已经说过的那样,物体不去睡觉,read子就行了。因此,如果你对B有不同的视线和A要睡觉,那么你就可以从B处获得A所认为的(如果不是的话)主线(我假定A在主线上运行)将睡觉,而不会发生任何情况。如果你怀疑底线是什么,请查看:http://en.wikipedia.org/wiki/Thread_%28computer_science%29。

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