Python多线程经典问题之乘客做公交车算法实例_Python

本文实例讲述了Python多线程经典问题之乘客做公交车算法。分享给大家供大家参考，具体如下：

问题描述：

乘客乘坐公交车问题，司机，乘客，售票员协同工作,通过多线程模拟三者的工作。
司机：开车，停车
售票员：打开车门，关闭车门
乘客：上车，下车

用Python的Event做线程同步通信，代码如下：

				?

									# *-* coding:gb2312 *-*

									import threading

									import time

									stationName=("车站0","车站1","车站2","车站3","车站4","车站5","车站6")

									currentStationIndex = -1

									eventBusStop = threading.Event()

									eventClosedDoor = threading.Event()

									eventOpenedDoor = threading.Event()

									stationCount = len(stationName)

									class Passenger(threading.Thread):

									  def __init__(self,no,getonStation,getoffStation):

									    self.no =no

									    self.getonStation=getonStation

									    self.getoffStation=getoffStation

									    threading.Thread.__init__(self)

									  def run(self):

									    bExit= False

									    global currentStationIndex

									    global stationCount

									    bAlreadyGetOnStation = False

									    while not bExit:

									      eventOpenedDoor.wait()

									      if self.getonStation == currentStationIndex and bAlreadyGetOnStation == False:

									        print "乘客%d在%s上车" %(self.no,stationName[currentStationIndex])

									        bAlreadyGetOnStation =True

									      elif self.getoffStation == currentStationIndex:

									        print "乘客%d在%s下车" %(self.no,stationName[currentStationIndex])

									        bExit = True

									      time.sleep(1)

									class Driver(threading.Thread):

									  def run(self):

									    bExit= False

									    global currentStationIndex

									    global stationCount

									    while not bExit:

									      print "司机: 公交车开始行驶....."

									      time.sleep(5)

									      currentStationIndex += 1

									      print "司机: 到站 ",stationName[currentStationIndex]

									      eventBusStop.set()

									      eventClosedDoor.wait()

									      eventClosedDoor.clear()

									      if currentStationIndex == stationCount-1:

									        bExit= True

									class Conductor(threading.Thread):

									  def run(self):

									    bExit= False

									    global currentStationIndex

									    global stationCount

									    while not bExit:

									      eventBusStop.wait()

									      eventBusStop.clear()

									      print "售票员打开车门：%s到了" %(stationName[currentStationIndex])

									      eventOpenedDoor.set()

									      time.sleep(5)

									      print "售票员关闭车门"

									      eventOpenedDoor.clear()

									      eventClosedDoor.set()

									      if currentStationIndex == stationCount-1:

									        bExit = True

									def test():

									  passPool=[]

									  passPool.append(Passenger(0,0,3))

									  passPool.append(Passenger(1,1,3))

									  passPool.append(Passenger(2,2,4))

									  passPool.append(Passenger(3,0,5))

									  passPool.append(Passenger(4,1,3))

									  passPool.append(Passenger(5,2,4))

									  passPool.append(Passenger(6,4,5))

									  passPool.append(Passenger(7,0,2))

									  passPool.append(Passenger(8,1,3))

									  passPool.append(Conductor())

									  passPool.append(Driver())

									  leng = len(passPool)

									  for i in range(leng):

									    passPool[i].start()

									if __name__=='__main__':

									  test()

输出结果如下：

Python多线程经典问题之乘客做公交车算法实例

希望本文所述对大家Python程序设计有所帮助。

Python多线程经典问题之乘客做公交车算法实例

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