Python程序中的线程操作-锁
Python程序中的线程操作-锁
一、同步锁
1.1多个线程抢占资源的情况
from threading import Thread
import os,time
def work():
global n
temp=n
time.sleep(0.1)
n=temp-1
if __name__ == '__main__':
n=100
l=[]
for i in range(100):
p=Thread(target=work)
l.append(p)
p.start()
for p in l:
p.join()
print(n) #结果可能为99
1.1.1对公共数据的操作
import threading
R=threading.Lock()
R.acquire()
'''
对公共数据的操作
'''
R.release()
1.2同步锁的引用
from threading import Thread, Lock
import os, time
def work():
global n
lock.acquire()
temp = n
time.sleep(0.1)
n = temp - 1
lock.release()
if __name__ == '__main__':
lock = Lock()
n = 100
l = []
for i in range(100):
p = Thread(target=work)
l.append(p)
p.start()
for p in l:
p.join()
print(n) # 结果肯定为0，由原来的并发执行变成串行，牺牲了执行效率保证了数据安全
1.3互斥锁与join的区别
# 不加锁:并发执行,速度快,数据不安全
from threading import current_thread, Thread, Lock
import os, time
def task():
global n
print('%s is running' % current_thread().getName())
temp = n
time.sleep(0.5)
n = temp - 1
if __name__ == '__main__':
n = 100
lock = Lock()
threads = []
start_time = time.time()
for i in range(100):
t = Thread(target=task)
threads.append(t)
t.start()
for t in threads:
t.join()
stop_time = time.time()
print('主:%s n:%s' % (stop_time - start_time, n))
'''
Thread-1 is running
Thread-2 is running
......
Thread-100 is running
主:0.5216062068939209 n:99
'''
# 不加锁:未加锁部分并发执行,加锁部分串行执行,速度慢,数据安全
from threading import current_thread, Thread, Lock
import os, time
def task():
# 未加锁的代码并发运行
time.sleep(3)
print('%s start to run' % current_thread().getName())
global n
# 加锁的代码串行运行
lock.acquire()
temp = n
time.sleep(0.5)
n = temp - 1
lock.release()
if __name__ == '__main__':
n = 100
lock = Lock()
threads = []
start_time = time.time()
for i in range(100):
t = Thread(target=task)
threads.append(t)
t.start()
for t in threads:
t.join()
stop_time = time.time()
print('主:%s n:%s' % (stop_time - start_time, n))
'''
Thread-2 start to run
Thread-3 start to run
Thread-1 start to run
Thread-6 start to run
Thread-4 start to run
......
Thread-99 start to run
Thread-96 start to run
Thread-100 start to run
Thread-92 start to run
Thread-93 start to run
主:53.294203758239746 n:0
'''
有的同学可能有疑问:既然加锁会让运行变成串行,那么我在start之后立即使用join,就不用加锁了啊,也是串行的效果啊
没错:在start之后立刻使用jion,肯定会将100个任务的执行变成串行,毫无疑问,最终n的结果也肯定是0,是安全的,但问题是
start后立即join:任务内的所有代码都是串行执行的,而加锁,只是加锁的部分即修改共享数据的部分是串行的
单从保证数据安全方面,二者都可以实现,但很明显是加锁的效率更高.
from threading import current_thread, Thread, Lock
import os, time
def task():
time.sleep(3)
print('%s start to run' % current_thread().getName())
global n
temp = n
time.sleep(0.5)
n = temp - 1
if __name__ == '__main__':
n = 100
lock = Lock()
start_time = time.time()
for i in range(100):
t = Thread(target=task)
t.start()
t.join()
stop_time = time.time()
print('主:%s n:%s' % (stop_time - start_time, n))
'''
Thread-1 start to run
Thread-2 start to run
......
Thread-100 start to run
主:350.6937336921692 n:0 #耗时是多么的恐怖
'''
二、死锁与递归锁
所谓死锁：是指两个或两个以上的进程或线程在执行过程中，因争夺资源而造成的一种互相等待的现象，若无外力作用，它们都将无法推进下去。此时称系统处于死锁状态或系统产生了死锁，这些永远在互相等待的进程称为死锁进程，如下就是死锁
2.1死锁
from threading import Thread,Lock
import time
mutexA=Lock()
mutexB=Lock()
class MyThread(Thread):
def run(self):
self.func1()
self.func2()
def func1(self):
mutexA.acquire()
print('\033[41m%s 拿到A锁\033[0m' %self.name)
mutexB.acquire()
print('\033[42m%s 拿到B锁\033[0m' %self.name)
mutexB.release()
mutexA.release()
def func2(self):
mutexB.acquire()
print('\033[43m%s 拿到B锁\033[0m' %self.name)
time.sleep(2)
mutexA.acquire()
print('\033[44m%s 拿到A锁\033[0m' %self.name)
mutexA.release()
mutexB.release()
if __name__ == '__main__':
for i in range(10):
t=MyThread()
t.start()
'''
Thread-1 拿到A锁
Thread-1 拿到B锁
Thread-1 拿到B锁
Thread-2 拿到A锁
然后就卡住，死锁了
'''
解决方法：递归锁，在Python中为了支持在同一线程中多次请求同一资源，python提供了可重入锁RLock。
这个RLock内部维护着一个Lock和一个counter变量，counter记录了acquire的次数，从而使得资源可以被多次require。直到一个线程所有的acquire都被release，其他的线程才能获得资源。上面的例子如果使用RLock代替Lock，则不会发生死锁。
mutexA=mutexB=threading.RLock() #一个线程拿到锁，counter加1,该线程内又碰到加锁的情况，则counter继续加1，这期间所有其他线程都只能等待，等待该线程释放所有锁，即counter递减到0为止
三、典型问题：科学家吃面
3.1死锁问题
import time
from threading import Thread,Lock
noodle_lock = Lock()
fork_lock = Lock()
def eat1(name):
noodle_lock.acquire()
print('%s 抢到了面条'%name)
fork_lock.acquire()
print('%s 抢到了叉子'%name)
print('%s 吃面'%name)
fork_lock.release()
noodle_lock.release()
def eat2(name):
fork_lock.acquire()
print('%s 抢到了叉子' % name)
time.sleep(1)
noodle_lock.acquire()
print('%s 抢到了面条' % name)
print('%s 吃面' % name)
noodle_lock.release()
fork_lock.release()
for name in ['哪吒','nick','tank']:
t1 = Thread(target=eat1,args=(name,))
t2 = Thread(target=eat2,args=(name,))
t1.start()
t2.start()
3.2递归锁解决死锁问题
import time
from threading import Thread, RLock
fork_lock = noodle_lock = RLock()
def eat1(name):
noodle_lock.acquire()
print('%s 抢到了面条' % name)
fork_lock.acquire()
print('%s 抢到了叉子' % name)
print('%s 吃面' % name)
fork_lock.release()
noodle_lock.release()
def eat2(name):
fork_lock.acquire()
print('%s 抢到了叉子' % name)
time.sleep(1)
noodle_lock.acquire()
print('%s 抢到了面条' % name)
print('%s 吃面' % name)
noodle_lock.release()
fork_lock.release()
for name in ['哪吒', 'nick', 'tank']:
t1 = Thread(target=eat1, args=(name,))
t2 = Thread(target=eat2, args=(name,))
t1.start()
t2.start()
四、信号量Semaphore
同进程的一样
Semaphore管理一个内置的计数器，
每当调用acquire()时内置计数器-1；
调用release() 时内置计数器+1；
计数器不能小于0；当计数器为0时，acquire()将阻塞线程直到其他线程调用release()。
实例：(同时只有5个线程可以获得semaphore,即可以限制最大连接数为5)：
from threading import Thread,Semaphore
import threading
import time
def task():
sm.acquire()
print(f"{threading.current_thread().name} get sm")
time.sleep(3)
sm.release()
if __name__ == '__main__':
sm = Semaphore(5) # 同一时间只有5个进程可以执行。
for i in range(20):
t = Thread(target=task)
t.start()
与进程池是完全不同的概念，进程池Pool(4)，最大只能产生4个进程，而且从头到尾都只是这四个进程，不会产生新的，而信号量是产生一堆线程/进程。
