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
