【Python@Thread】Semaphore&糖果机

信号量适用与多线程竞争有限资源的情况。

 from atexit import register
 from time import ctime, sleep
 from threading import Thread, Lock, BoundedSemaphore
 from random import randrange

 lock = Lock()
 MAX = 5                             #信号量大小
 candytray = BoundedSemaphore(MAX)

 def refull():
     with lock:
         print('refulling...')
         try:
             candytray.release()
         except ValueError:
             print('Is Full!')
         else:
             print('OK')

 def buy():
     with lock:
         print('buying...')
         if candytray.acquire(False):    #加入False参数，如果信号量为空，则不阻塞，而是返回错误，感觉类似与C语言中的pthread_mutex_trylock
             print('OK')
         else:
             print('empty')

 def consumer(loops):
     for i in range(loops):
         refull()
         sleep(randrange(3))             #睡眠时间尽量长于creater的概率尽量大，

 def creater(loops):
     for i in range(loops):
         buy()
         sleep(randrange(5))

 def main():
     print('starting...')
     n = randrange(2,6)
     print('the candy mechine full with {0}'.format(MAX))
     Thread(target=creater,args=(randrange(n,n+MAX+2),)).start()
     Thread(target=consumer, args=(randrange(n,n+MAX+2),)).start()

 @register
 def atexitt():
     print('The end!')

 if __name__ == '__main__':
     main()

输出结果：

参考资料：Python核心编程.第四章.Wesley Chun著