Python多线程,线程死锁及解决,生产者与消费者问题

1.Thread类

普通调用

t = Thread(target=test, args=(i,))	# test为目标函数名, 若函数需要参数将其以元组形									                # 式赋给args, 若无参数可不写

t.start()	# 用start()函数开启线程

例子

import time

from threading import Thread

# 目标函数

def test(i):

    print("hello ", i)

    time.sleep(1)

def main():

    # 循环5次,开起五个线程

    for i in range(5):

        t = Thread(target=test, args=(i,))

        t.start()

if __name__ == '__main__':

    main()

继承Thread类

定义一个自己的类继承自Thread,重写run()方法,即 将原本执行任务的函数内容移植到run()方法中.可通过类的属性传参.

例子

from threading import Thread

import time

class MyThread(Thread):

    def __init__(self, i):

        Thread.__init__(self)   # 初始化父类"构造函数"

        self.i = i  # 初始化,目的将run函数参数作为类的属性

    def run(self):

        time.sleep(1)

        msg = "I'm " + self.name + " @ " + str(self.i)

        print(msg)

def main():

    for i in range(3):	# 开启三个线程

        t = MyThread(i)	# 实例化自己的类

        t.start()

if __name__ == '__main__':

    main()

线程的执行顺序

上面的例子中线程的执行顺序是随机的

2.线程间共享全局变量

下面例子中test1()和test2()共享g_num全局变量.希望test1()执行的结果是1000000,test2()执行的结果是2000000.但是time.sleep()函数会影响结果.

from threading import Thread

import time

g_num = 0

def test1():

    global g_num

    for i in range(1000000):

        g_num += 1

    print('---test1 g_num is %d---' % g_num)

def test2():

    global g_num

    for i in range(1000000):

        g_num += 1

    print('---test2 g_num is %d---' % g_num)

t1 = Thread(target=test1)

t1.start()

# time.sleep(3) # 运行这句话与不运行这句话结果不一样

t2 = Thread(target=test2)

t2.start()

print('-----g_num: %d-----' % g_num)

不执行sleep函数的结果(可能出现多种不同的运行结果)

---test1 g_num is 1312283---

-----g_num: 1312283-----

---test2 g_num is 1341534---

执行sleep()函数的结果

---test1 g_num is 1000000---

-----g_num: 1079982-----

---test2 g_num is 2000000---

其实这也不难理解,sleep之后test1中的任务肯定先完成,而不执行sleep两个函数同能对g_num同时操作

通过轮询的方式解决线程间共享全局变量的问题

from threading import Thread

g_num = 0

g_flag = 1  # 增加一个标识全局变量

def test1():

    global g_num

    global g_flag

    if g_flag == 1:

        for i in range(1000000):

            g_num += 1

    g_flag = 0

    print('---test1 g_num is %d---' % g_num)

def test2():

    global g_num

    # 轮询

    while True:

        if g_flag != 1: # 一旦test1()执行完,即g_flag = 0时,test2()开始执行累加g_num操作

            for i in range(1000000):

                g_num += 1

            break

    print('---test2 g_num is %d---' % g_num)

t1 = Thread(target=test1)

t1.start()

t2 = Thread(target=test2)

t2.start()

print('-----g_num: %d-----' % g_num)

运行结果

-----g_num: 303721-----

---test1 g_num is 1000000---

---test2 g_num is 2000000---

第二个线程一开始并没有执行累加g_num的操作,而是先进行一个死循环,在这个循环中不断的"询问"g_flag的值是否不等于1.一但g_flag不等于1,即test1()结束后便开始干"正事".

通过互斥锁解决线程间共享全局变量的问题

from threading import Thread, Lock  # 导入互斥锁

g_num = 0

def test1():

    global g_num

    for i in range(1000000):

        mutex.acquire()  # 上锁，此时其他的锁会等待  上锁应该遵循最小原则

        g_num += 1

        mutex.release() #　开锁，此时其他的锁会抢着开锁

    print('---test1 g_num is %d---' % g_num)

def test2():

    global g_num

    for i in range(1000000):

        mutex.acquire()

        g_num += 1

        mutex.release()

    print('---test2 g_num is %d---' % g_num)

# 创建一把互斥锁，默认不上锁

mutex = Lock()

t1 = Thread(target=test1)

t1.start()

t2 = Thread(target=test2)

t2.start()

print('-----g_num: %d-----' % g_num)

运行结果

-----g_num: 45012-----

---test1 g_num is 1979942---

---test2 g_num is 2000000---

从结果可以看出test2()的结果是正确的,而test1()的结果很接近test2.这也不难理解.互斥锁会把夹在中间的部分锁定,也就是说,在极短时间内只能有一个线程在执行该代码.一旦开锁了(release),所有线程开始抢这把锁,某个线程抢到之后会把自己的操作锁住,其他线程只能等待,一直反复直至全部任务完成.

只有对上述代码稍微修改便可以实现我们想要的结果

修改后的代码

from threading import Thread, Lock  # 导入互斥锁

g_num = 0

def test1():

    global g_num

    mutex.acquire()  # 上锁，此时其他的锁会等待  上锁应该遵循最小原则

    for i in range(1000000):

        g_num += 1

    mutex.release() #　开锁，此时其他的锁会抢着开锁

    print('---test1 g_num is %d---' % g_num)

def test2():

    global g_num

    mutex.acquire()

    for i in range(1000000):

        g_num += 1

    mutex.release()

    print('---test2 g_num is %d---' % g_num)

# 创建一把互斥锁，默认不上锁

mutex = Lock()

t1 = Thread(target=test1)

t1.start()

t2 = Thread(target=test2)

t2.start()

print('-----g_num: %d-----' % g_num)

结果

-----g_num: 220254-----

---test1 g_num is 1000000---

---test2 g_num is 2000000---

值得注意的是,互斥锁上的范围太大就失去了线程的意义,别的线程都把时间浪费在了等待上.轮询同理.

3.线程间使用非全局变量

from threading import Thread

import threading

import time

def test1():

    name = threading.current_thread().name  # 获取当前线程名字

    print('----thread name is %s----' % name)

    g_num = 100

    if name == 'Thread-1':

        g_num += 1

    else:

        time.sleep(2)

    print('---thread is %s | g_num is %d---' % (name, g_num))

t1 = Thread(target=test1)

t1.start()

t2 = Thread(target=test1)

t2.start()

运行结果

----thread name is Thread-1----

---thread is Thread-1 | g_num is 101---

----thread name is Thread-2----

---thread is Thread-2 | g_num is 100---

非全局对于同一个函数来说.可以通过线程的名字来区分.

4.线程死锁

import threading

import time

class MyThread1(threading.Thread):

    def run(self):

        if mutexA.acquire():

            print(self.name + '---do1---up---')

            time.sleep(1)

            if mutexB.acquire():

                print(self.name + '---do1---down---')

                mutexB.release()

            mutexA.release()

class MyThread2(threading.Thread):

    def run(self):

        if mutexB.acquire():

            print(self.name + '---do2---up---')

            time.sleep(1)

            if mutexA.acquire():

                print(self.name + '---do2---down---')

                mutexA.release()

            mutexB.release()

if __name__ == '__main__':

    mutexA = threading.Lock()

    mutexB = threading.Lock()

    t1 = MyThread1()

    t2 = MyThread2()

    t1.start()

    t2.start()

运行结果(卡在了这两句,未结束)

Thread-1---do1---up---

Thread-2---do2---up---

分析代码,t1的代码在等待mutexB解锁的时候t2在等待mutexA解锁.而t1必须先执行完mutexB锁中的代码执行完才能释放mutexA,t2必须先执行完mutexA锁中的代码执行完才能释放mutexB,这就导致两个线程一直等待下去形成死锁,会浪费CPU资源.

解决死锁的办法

设置超时时间 mutexA.acquire(2)

当然也可以从算法上避免死锁

5.使用ThreadLocal

import threading

# 创建全局ThreadLocal对象

local_school = threading.local()

def process_student():

    # 获取当前线程相关联的student

    std = local_school.student

    print('Hello, %s in %s' % (std, threading.current_thread().name))

def process_thread(name):

    # 绑定ThreadLocal的student

    local_school.student = name

    process_student()

t1 = threading.Thread(target=process_thread, args=('kain',), name='Thread-A')

t2 = threading.Thread(target=process_thread, args=('huck',), name='Thread-B')

t1.start()

t2.start()

t1.join()

t2.join()

运行结果

Hello, kain in Thread-A

Hello, huck in Thread-B

6.生产者与消费者问题

import threading

import time

# Python2

# from Queue import Queue

# Python3

from queue import Queue

class Producer(threading.Thread):

    def run(self):

        global queue

        count = 0

        while True:

            if queue.qsize() < 1000:

                for i in range(100):

                    count += 1

                    msg = '生成产品' + str(count)

                    queue.put(msg)

                    print(msg)

            time.sleep(0.5)

class Consumer(threading.Thread):

    def run(self):

        global queue

        while True:

            if queue.qsize() > 100:

                for i in range(3):

                    msg = self.name + '消费了' + queue.get()

                    print(msg)

            time.sleep(1)

if __name__ == '__main__':

    queue = Queue()

    for i in range(500):

        queue.put('初始产品'+str(i))    # 向队列中塞内容

    for i in range(2):

        p = Producer()

        p.start()

    for i in range(5):

        c = Consumer()

        c.start()

运行结果过长不予展示

巴特西

Python多线程,线程死锁及解决,生产者与消费者问题

1.Thread类

2.线程间共享全局变量

通过轮询的方式解决线程间共享全局变量的问题

通过互斥锁解决线程间共享全局变量的问题

3.线程间使用非全局变量

4.线程死锁

5.使用ThreadLocal

6.生产者与消费者问题

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