Posted in Python onMarch 22, 2017
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:
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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