Question

我通过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()

这项工作，但不是我想要的方式。我希望能够：

def print_test(num):
    print("executing:", num)
    sleep(10)
    print("end of execution:", num)

threads_dict = dict()
thread_range = range(1,20)
for i in thread_range:
    threads_dict[i] = MyThread(target=print_test(i))
    threads_dict[i].start()
for i in thread_range:
    threads_dict[i].join()
    print("joined:", threads_dict[i].name)

显然，一旦启动了5个线程（max_threads），脚本将处于死锁状态，直到达到信号量超时。

我认为start()方法会在新线程中启动run()方法。来自the manual：

开始（）每个线程对象最多只能调用一次。它安排 for对象的run（）方法在一个单独的线程中调用控制。

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)

但是这个代码不是在新线程中执行的。输出是：

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？

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..

Answer 1

关于“奇怪”的输出的问题。这是因为您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))

执行 print_test（i）功能。结果你过去 print_test（i）函数执行的目标结果。

尝试纠正此问题：

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

经过这次修正后，我的输出如下：

('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')
...

如何在执行run（）后自动释放线程中的信号量？

1 个答案: