Python 简单的线程问题 - V2EX
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shenyansycn
V2EX    Python

Python 简单的线程问题

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  •   shenyansycn 2021-03-18 10:19:01 +08:00 2055 次点击
    这是一个创建于 1673 天前的主题,其中的信息可能已经有所发展或是发生改变。

    代码如下:

    import _thread as thread stdoutmutex = thread.allocate_lock() exitmutexes = [thread.allocate_lock() for i in range(10)] def counter(myId, count): for i in range(count): print('1 %s -> %s' % (myId, stdoutmutex.locked())) stdoutmutex.acquire() print('2 %s -> %s' % (myId, stdoutmutex.locked())) print(stdoutmutex.locked()) print('[%s] => %s' % (myId, i)) stdoutmutex.release() exitmutexes[myId].acquire() for i in range(10): thread.start_new_thread(counter, (i, 10)) for mutex in exitmutexes: while not mutex.locked(): pass print('Main thread exiting') 

    问题: exitmutexes[myId].acquire()获得了锁的状态,while not mutex.locked(): pass 判断没有锁的话就继续执行,但我不知道什么时候释放的锁。 谢谢大家了!

    4 条回复    2021-03-18 16:04:49 +08:00
    linw1995
        1
    linw1995  
       2021-03-18 11:31:06 +08:00
    想感知线程中函数的执行情况,可以用 Thread.join 。或者更细粒度的可以用 threading.Event
    no1xsyzy
        2
    no1xsyzy  
       2021-03-18 13:20:21 +08:00
    while not mutex.locked(): pass

    这是判断没有锁就卡住,有锁继续执行。
    shenyansycn
        3
    shenyansycn  
    OP
       2021-03-18 15:53:46 +08:00
    @no1xsyzy 应该是有锁就卡住,没锁才执行 pass 吧
    shenyansycn
        4
    shenyansycn  
    OP
       2021-03-18 16:04:49 +08:00
    @no1xsyzy 我明白了,谢谢!
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