如何在执行run()之后自动释放线程中的信号量?

2024-05-23 16:29:46 发布

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我用这种方式跟踪信号量的线程数:

#!/usr/bin/python3
import threading

class MyThread(threading.Thread):

    max_threads           = 5
    max_threads_semaphore = threading.BoundedSemaphore(value=max_threads)
    semaphore_timeout     = 60 

    def __init__(self, target=None, name=None, args=(), kwargs={}):
        super().__init__(target=target, name=name, args=args, kwargs=kwargs)

    def start(self):
        semaphore_aquired = self.max_threads_semaphore.acquire(blocking=True, timeout=self.semaphore_timeout)
        if semaphore_aquired:
            print("Sempahore acquired by:", self.name)
            super().start()
        else:
            raise OSError("Time out aquiring max threads semaphore to start new thread")

    def join(self):
        super().join()
        semaphore_released = self.max_threads_semaphore.release()

这是工作,但不是我想要的方式。我希望能够:

^{pr2}$

显然,一旦启动了5个线程(max_threads),脚本就会处于死锁状态,直到信号量超时为止。在

我认为start()方法将在一个新线程中启动run()方法。来自the manual

start() It must be called at most once per thread object. It arranges for the object’s run() method to be invoked in a separate thread of control.

所以在MyThread中,我将join()部分替换为:

def run(self):
    super().run()
    semaphore_released = self.max_threads_semaphore.release()
    print("Sempahore released by:", self.name)

但这段代码不会在新线程中执行。输出为:

executing: 1
end of execution: 1
Sempahore acquired by: Thread-1
Sempahore released by: Thread-1
executing: 2
end of execution: 2
Sempahore acquired by: Thread-2
Sempahore released by: Thread-2
executing: 3
end of execution: 3
Sempahore acquired by: Thread-3
Sempahore released by: Thread-3
executing: 4
end of execution: 4
Sempahore acquired by: Thread-4
Sempahore released by: Thread-4
executing: 5
end of execution: 5
Sempahore acquired by: Thread-5
Sempahore released by: Thread-5
executing: 6
end of execution: 6
Sempahore acquired by: Thread-6
executing: 7
Sempahore released by: Thread-6
end of execution: 7
Sempahore acquired by: Thread-7
executing: 8
Sempahore released by: Thread-7
end of execution: 8
Sempahore acquired by: Thread-8
executing: 9
Sempahore released by: Thread-8
end of execution: 9
Sempahore acquired by: Thread-9
joined: Thread-1
joined: Thread-2
joined: Thread-3
joined: Thread-4
joined: Thread-5
joined: Thread-6
joined: Thread-7
joined: Thread-8
Sempahore released by: Thread-9
joined: Thread-9

有没有办法重载threading.Thread函数,以便在目标函数结束时自动释放信号量?在

我不明白为什么输出不是:

Sempahore acquired by: Thread-1
executing: 1
Sempahore acquired by: Thread-2
executing: 2
Sempahore acquired by: Thread-3
executing: 3
Sempahore acquired by: Thread-4
executing: 4
Sempahore acquired by: Thread-5
executing: 5
(sleep 10 secondes)
end of execution: 1
Sempahore released by: Thread-1
Sempahore acquired by: Thread-6
executing: 6
etc..

Tags: ofnameselfbythreadmaxendexecution
1条回答
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1楼 · 发布于 2024-05-23 16:29:46

关于“奇怪”输出的问题。这是意料之中的,因为在创建新线程之前,在主线程中调用print_test(i)

threads_dict[i] = MyThread(target=print_test(i))

这是执行打印测试(i)函数。结果是,您超过了目标结果打印测试(i)函数执行。在

请尝试更正:

^{pr2}$

修正后,我得到如下输出:

('Sempahore acquired by:', 'Thread-1')
('Sempahore acquired by:', 'Thread-2')
('executing:', 1)
('Sempahore acquired by:', 'Thread-3')
('executing:', 2)
('Sempahore acquired by:', 'Thread-4')
('executing:', 3)
('Sempahore acquired by:', 'Thread-5')
('executing:', 4)
('executing:', 5)
('end of execution:', 1)
('Sempahore released by:', 'Thread-1')
('Sempahore acquired by:', 'Thread-6')
...

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