Python系统学习第二十五课

• 当多个线程同时访问同一个变量的时候&＃xff0c;会产生共享变量的问题
• 解决变量
  • 锁&＃xff0c;信号灯
  • 锁&＃xff08;Lock&＃xff09;:
    • 是一个标志&＃xff0c;表示一个线程在占用一些资源
    • 使用方法
      • 上锁
      • 使用共享资源&＃xff0c;放心使用
      • 取消锁&＃xff0c;释放锁
    • 锁谁&＃xff1a;哪个资源需要多个线程共享&＃xff0c;锁哪个
    • 理解锁&＃xff1a;锁其实不是锁住谁&＃xff0c;而是一个令牌
  • 线程安全问题&＃xff1a;
    • 如果一个变量/资源&＃xff0c;他对于多线程来讲&＃xff0c;不用加锁也不会引起任何问题&＃xff0c;则成为线程安全
    • 线程不安全变量类型&＃xff1a;list, set, dict
    • 线程安全变量类型&＃xff1a;queue
  • 生产者消费者问题
    • 一个模型&＃xff0c;可以用来搭建消息队列
    • queue是一个用来存放变量的数据结构&＃xff0c;特点是先进先出&＃xff0c;内部元素排队&＃xff0c;可以理解成一个特殊的list

#每次执行都产生不同的值
import threadingsum &＃61; 0
loopSum &＃61; 1000000def myAdd():global sum, loopSumfor i in range(1, loopSum):sum &＃43;&＃61; 1def myMinu():global sum, loopSumfor i in range(1, loopSum):sum -&＃61; 1if __name__ &＃61;&＃61; &＃39;__main__&＃39;:print("Starting ....{0}".format(sum))# 开始多线程的实例&＃xff0c;看执行结果是否一样t1 &＃61; threading.Thread(target&＃61;myAdd, args&＃61;())t2 &＃61; threading.Thread(target&＃61;myMinu, args&＃61;())t1.start()t2.start()t1.join()t2.join()print("Done .... {0}".format(sum))


Starting ....0
Done .... -11539


import threadingsum &＃61; 0
loopSum &＃61; 1000000lock &＃61; threading.Lock()def myAdd():global sum, loopSumfor i in range(1, loopSum):# 上锁&＃xff0c;申请锁lock.acquire()sum &＃43;&＃61; 1# 释放锁lock.release()def myMinu():global sum, loopSumfor i in range(1, loopSum):lock.acquire()sum -&＃61; 1lock.release()if __name__ &＃61;&＃61; &＃39;__main__&＃39;:print("Starting ....{0}".format(sum))# 开始多线程的实例&＃xff0c;看执行结果是否一样t1 &＃61; threading.Thread(target&＃61;myAdd, args&＃61;())t2 &＃61; threading.Thread(target&＃61;myMinu, args&＃61;())t1.start()t2.start()t1.join()t2.join()print("Done .... {0}".format(sum))

Starting ....0
Done .... 0


#生产者消费者模型
import threading
import time# Python2
# from Queue import Queue# Python3
import queueclass Producer(threading.Thread):def run(self):global queuecount &＃61; 0while True:# qsize返回queue内容长度if queue.qsize() < 1000:for i in range(100):count &＃61; count &＃43;1msg &＃61; &＃39;生成产品&＃39;&＃43;str(count)# put是网queue中放入一个值queue.put(msg)print(msg)time.sleep(0.5)class Consumer(threading.Thread):def run(self):global queuewhile True:if queue.qsize() > 100:for i in range(3):# get是从queue中取出一个值msg &＃61; self.name &＃43; &＃39;消费了 &＃39;&＃43;queue.get()print(msg)time.sleep(1)if __name__ &＃61;&＃61; &＃39;__main__&＃39;:queue &＃61; queue.Queue()for i in range(500):queue.put(&＃39;初始产品&＃39;&＃43;str(i))for i in range(2):p &＃61; Producer() #两个生产者p.start()for i in range(5): #五个消费者c &＃61; Consumer()c.start()


• 死锁问题

#死锁案例
import threading
import timelock_1 &＃61; threading.Lock()
lock_2 &＃61; threading.Lock()def func_1():print("func_1 starting.........")lock_1.acquire()print("func_1 申请了 lock_1....")time.sleep(2)print("func_1 等待 lock_2.......")lock_2.acquire()print("func_1 申请了 lock_2.......")lock_2.release()print("func_1 释放了 lock_2")lock_1.release()print("func_1 释放了 lock_1")print("func_1 done..........")def func_2():print("func_2 starting.........")lock_2.acquire()print("func_2 申请了 lock_2....")time.sleep(4)print("func_2 等待 lock_1.......")lock_1.acquire()print("func_2 申请了 lock_1.......")lock_1.release()print("func_2 释放了 lock_1")lock_2.release()print("func_2 释放了 lock_2")print("func_2 done..........")if __name__ &＃61;&＃61; "__main__":print("主程序启动..............")t1 &＃61; threading.Thread(target&＃61;func_1, args&＃61;())t2 &＃61; threading.Thread(target&＃61;func_2, args&＃61;())t1.start()t2.start()
# 锁1先醒了之后&＃xff0c;等待锁2 的醒来&＃xff0c;但是&＃xff0c;锁2 一直在函数2中沉睡&＃xff0c;所以就是说&＃xff0c;没有释放锁2。形成死锁&＃xff0c;所以使用的时候要注意t1.join()t2.join()print("主程序启动..............")


主程序启动..............
func_1 starting.........
func_2 starting.........
func_2 申请了 lock_2....
func_1 申请了 lock_1....
func_1 等待 lock_2.......
func_2 等待 lock_1.......


# 锁的等待时间问题import threading
import timelock_1 &＃61; threading.Lock()
lock_2 &＃61; threading.Lock()def func_1():print("func_1 starting.........")lock_1.acquire(timeout&＃61;4) #超时时间&＃xff0c;等待时间是四秒&＃xff0c;不能超过四秒print("func_1 申请了 lock_1....")time.sleep(2)print("func_1 等待 lock_2.......")rst &＃61; lock_2.acquire(timeout&＃61;2)if rst:print("func_1 已经得到锁 lock_2")lock_2.release()print("func_1 释放了锁 lock_2")else:print("func_1 注定没申请到lock_2.....")lock_1.release()print("func_1 释放了 lock_1")print("func_1 done..........")def func_2():print("func_2 starting.........")lock_2.acquire()print("func_2 申请了 lock_2....")time.sleep(4)print("func_2 等待 lock_1.......")lock_1.acquire()print("func_2 申请了 lock_1.......")lock_1.release()print("func_2 释放了 lock_1")lock_2.release()print("func_2 释放了 lock_2")print("func_2 done..........")if __name__ &＃61;&＃61; "__main__":print("主程序启动..............")t1 &＃61; threading.Thread(target&＃61;func_1, args&＃61;())t2 &＃61; threading.Thread(target&＃61;func_2, args&＃61;())t1.start()t2.start()t1.join()t2.join()print("主程序结束..............")

• semphore
  • 允许一个资源由多个线程使用

import threading
import time# 参数定义最多几个线程同时使用资源
semaphore &＃61; threading.Semaphore(3)def func():if semaphore.acquire():for i in range(5):print(threading.currentThread().getName() &＃43; &＃39; get semaphore&＃39;)time.sleep(15)semaphore.release()print(threading.currentThread().getName() &＃43; &＃39; release semaphore&＃39;)for i in range(8):t1 &＃61; threading.Thread(target&＃61;func)t1.start()


• threading.Timer
  • timer是利用多线程&＃xff0c;在指定时间后启动一个功能

#案例
import threading
import timedef func():print("I am running.........")time.sleep(4)print("I am done......")if __name__ &＃61;&＃61; "__main__":t &＃61; threading.Timer(6, func)t.start()i &＃61; 0while True:print("{0}***************".format(i))time.sleep(3)i &＃43;&＃61; 1

• 可重入锁
  • 一个锁&＃xff0c;可以被一个线程多次申请
  • 主要解决递归调用的时候&＃xff0c;需要申请锁的情况

import threading
import timeclass MyThread(threading.Thread):def run(self):global numtime.sleep(1)if mutex.acquire(1):num &＃61; num&＃43;1msg &＃61; self.name&＃43;&＃39; set num to &＃39;&＃43;str(num)print(msg)mutex.acquire()mutex.release()mutex.release()num &＃61; 0mutex &＃61; threading.RLock()def testTh():for i in range(5):t &＃61; MyThread()t.start()if __name__ &＃61;&＃61; &＃39;__main__&＃39;:testTh()