Python 之并发编程之线程中

四.线程锁lock(线程的数据安全)

在数据量较大的时候，线程中的数据会被并发，所有数据会不同步，以至于数据会异常。

下面还介绍了两种的上锁方法。

例：

from threading import Thread, Lock

import time

n = 0

def func1(lock):

global n

# time.sleep(0.3)

# print(11)

for i in range(10000):

# 正常上锁

lock.acquire()

print(n)

n -= 1

lock.release()

def func2(lock):

global n

# time.sleep(0.3)

# print(22)

for i in range(10000):

# 用with 自动上锁解锁

with lock:

print(n)

n += 1

if __name__ == "__main__":

# 创建一个锁

lock = Lock()

lst = []

for i in range(10):

t1 = Thread(target=func1, args=(lock,))

t2 = Thread(target=func2, args=(lock,))

t1.start()

t2.start()

lst.append(t1)

lst.append(t2)

for i in lst:

i.join()

print("主线程执行语句结束")

print(n) # n最后得0，如果没有加上锁的话，不会是0

# 程序执行结束

五.线程的信号量

例：效果是5个一打印，5个一打印

from threading import Semaphore, Thread

import time, random

def func(i, sem):

# with简写

with sem:

print(i)

time.sleep(random.uniform(0.1, 1))

"""

# 正常写法

sem.acquire()

print(i)

time.sleep(random.uniform(0.1,1))

sem.release()

"""

if __name__ == "__main__":

sem = Semaphore(5)  # 设置几个线程同时运行几个

for i in range(50):

Thread(target=func, args=(i, sem)).start()

六.线程的锁

1.死锁

例:只有拿到筷子和面才能吃

noodle_lock = Lock()

chopsticks_lock = Lock()

def eat1(name):

noodle_lock.acquire()

print("%s拿到面条" % (name))

chopsticks_lock.acquire()

print("%s拿到筷子" % (name))

print("开始吃")

time.sleep(0.7)

chopsticks_lock.release()

print("%s放下筷子" % (name))

noodle_lock.release()

print("%s放下面条" % (name))

def eat2(name):

chopsticks_lock.acquire()

print("%s拿到筷子" % (name))

noodle_lock.acquire()

print("%s拿到面条" % (name))

print("开始吃")

time.sleep(0.6)

noodle_lock.release()

print("%s放下面条" % (name))

chopsticks_lock.release()

print("%s放下筷子" % (name))

if __name__ == "__main__":

name_list1 = ["one", "two"]

name_list2 = ["three", "four"]

for name in name_list1:

Thread(target=eat1, args=(name,)).start()

for name in name_list2:

Thread(target=eat2, args=(name,)).start()

# 双方都在等待,造成死锁的现象.

2.递归锁RLock

递归锁专门用来解决死锁现象

临时用于快速解决服务器崩溃的异常现象,用递归锁应急

解决应急问题的

(1)基本用法

from threading import Thread,RLock

# 递归锁如果3个,就对于释放3分锁,忽略上锁过程,进行解锁

rlock = RLock()

def func(name):

rlock.acquire()

print(name,1)

rlock.acquire()

print(name,2)

rlock.acquire()

print(name,3)

rlock.release()

rlock.release()

rlock.release()

lst = []

for i in range(10):

t1 = Thread(target=func,args=("name%s" % (i), ))

t1.start()

lst.append(t1)

for i in lst:

i.join()

print("程序结束了")

(2)用递归锁应急解决死锁现象

# 用递归锁应急解决死锁现象

noodle_lock = chopsticks_lock = RLock()

def eat1(name):

noodle_lock.acquire()

print("%s拿到面条" % (name))

chopsticks_lock.acquire()

print("%s拿到筷子" % (name))

print("开始吃")

time.sleep(0.7)

chopsticks_lock.release()

print("%s放下筷子" % (name))

noodle_lock.release()

print("%s放下面条" % (name))

def eat2(name):

chopsticks_lock.acquire()

print("%s拿到筷子" % (name))

noodle_lock.acquire()

print("%s拿到面条" % (name))

print("开始吃")

time.sleep(0.6)

noodle_lock.release()

print("%s放下面条" % (name))

chopsticks_lock.release()

print("%s放下筷子" % (name))

if __name__ == "__main__":

name_list1 = ["one", "two"]

name_list2 = ["three", "four"]

for name in name_list1:

Thread(target=eat1, args=(name,)).start()

for name in name_list2:

Thread(target=eat2, args=(name,)).start()

3.互斥锁

    从语法上来看,锁是可以互相嵌套的,但是不要使用

    上一次锁,就对应解开一把锁,形成互斥锁

    吃面条和拿筷子是同时的,上一次锁就够了,不要分别上锁

    尽量不要形成锁的嵌套,容易死锁

例：

from threading import Thread,RLock

mylock = Lock()

def eat1(name):

mylock.acquire()

print("%s拿到面条" % (name))

print("%s拿到筷子" % (name))

print("开始吃")

time.sleep(0.7)

print("%s放下筷子" % (name))

print("%s放下面条" % (name))

mylock.release()

def eat2(name):

mylock.acquire()

print("%s拿到筷子" % (name))

print("%s拿到面条" % (name))

print("开始吃")

time.sleep(0.6)

print("%s放下面条" % (name))

print("%s放下筷子" % (name))

mylock.release()

if __name__ == "__main__":

name_list1 = ["one", "two"]

name_list2 = ["three", "four"]

for name in name_list1:

Thread(target=eat1, args=(name,)).start()

for name in name_list2:

Thread(target=eat2, args=(name,)).start()