【Python@Thread】Semaphore&糖果機

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訊號量適用與多線程競爭有限資源的情況。

 1 from atexit import register 2 from time import ctime, sleep 3 from threading import Thread, Lock, BoundedSemaphore 4 from random import randrange 5  6 lock = Lock() 7 MAX = 5                             #訊號量大小 8 candytray = BoundedSemaphore(MAX) 9 10 11 def refull():12     with lock:13         print(‘refulling...‘)14         try:15             candytray.release()16         except ValueError:17             print(‘Is Full!‘)18         else:19             print(‘OK‘)20 21 def buy():22     with lock:23         print(‘buying...‘)24         if candytray.acquire(False):    #加入False參數，如果訊號量為空白，則不阻塞，而是返回錯誤，感覺類似與C語言中的pthread_mutex_trylock25             print(‘OK‘)26         else:27             print(‘empty‘)28 29 def consumer(loops):30     for i in range(loops):31         refull()32         sleep(randrange(3))             #睡眠時間盡量長於creater的機率盡量大，33 34 35 def creater(loops):36     for i in range(loops):37         buy()38         sleep(randrange(5))39 40 41 def main():42     print(‘starting...‘)43     n = randrange(2,6)44     print(‘the candy mechine full with {0}‘.format(MAX))45     Thread(target=creater,args=(randrange(n,n+MAX+2),)).start()46     Thread(target=consumer, args=(randrange(n,n+MAX+2),)).start()47 48 49 @register50 def atexitt():51     print(‘The end!‘)52 53 54 if __name__ == ‘__main__‘:55     main()

輸出結果：

參考資料：Python核心編程.第四章.Wesley Chun著

 

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