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

posted @ 2019-03-22 19:10  KainHuck  阅读(...)  评论(...编辑  收藏