死锁(哲学家就餐问题)
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
import time
from threading import Thread, Lock,RLock
lock1 = Lock()
lock2 = Lock()
# lock1 = RLock() # 递归锁, 可重入锁
# lock2 = RLock()
def eat1( name):
lock1.acquire()
print("%s抢到了面条" % name)
time.sleep(1)
lock2.acquire()
print("%s抢到了筷子" % name)
time.sleep(1)
lock2.release()
print("%s放下了筷子" % name)
time.sleep(1)
print("%s放下了面条" % name)
lock1.release()
def eat2(name):
lock2.acquire()
print("%s抢到了筷子" % name)
time.sleep(1)
lock1.acquire()
print("%s抢到了面条" % name)
time.sleep(1)
lock1.release()
print("%s放下了面条" % name)
time.sleep(1)
print("%s放下了筷子" % name)
lock2.release()
if __name__ == '__main__':
for name in ['egon', 'jason', 'ly']:
t = Thread(target=eat1, args=( name,))
t.start()
for name in ['qq', 'tom', 'kevin']:
t1 = Thread(target=eat2, args=(name,))
t1.start()
递归锁RLock/可重入锁
死锁解决方法: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。
线程队列
import queue
# q = queue.Queue()
#
# q.put('first')
# q.put('second')
# q.put('third')
#
# print(q.get())
# print(q.get())
# print(q.get())
'''
结果(先进先出):
first
second
third
'''
q = queue.LifoQueue()
q.put('first')
q.put('second')
q.put('third')
print(q.get())
print(q.get())
print(q.get())
'''
结果(后进先出):
third
second
first
'''
q = queue.PriorityQueue()
# put进入一个元组,元组的第一个元素是优先级(通常是数字,也可以是非数字之间的比较),数字越小优先级越高
q.put((20, 'a'))
q.put((10, 'b'))
q.put((10, 'b1'))
q.put((30, 'c'))
print(q.get())
print(q.get())
print(q.get())
print(q.get())
进程锁和线程锁
from multiprocessing import Process
from concurrent.futures import ProcessPoolExecutor, ThreadPoolExecutor
def task(n, m):
return n + m
# 回调函数:
def xxx(res):
print(res.result())
if __name__ == '__main__':
# p = Process(target=task)
# p.start()
# p_pool = ProcessPoolExecutor(3) #
# p_pool.submit(task, n=1, m=2)
#
# p_pool.shutdown() # join
# print("end...")
p_pool = ProcessPoolExecutor(3)
p_pool.submit(task, n=1, m=2).add_done_callback(xxx)
import os, time, random
def task(n):
print('%s is runing' % os.getpid())
time.sleep(random.randint(1, 3))
return n ** 2
if __name__ == '__main__':
executor = ThreadPoolExecutor(max_workers=3)
# for i in range(11):
# future=executor.submit(task,i)
# executor.map(task, range(1, 12)) # map取代了for+submit
for i in range(1, 12):
executor.submit(task, i)
协程
是单线程下的并发
协程是一种用户态的轻量级线程,即协程是由用户程序自己控制调度的。
需要强调的是:
python的线程属于内核级别的,即由操作系统控制调度(如单线程遇到io或执行时间过长
就会被迫交出cpu执行权限,切换其他线程运行)
**单线程内开启协程,一旦遇到io,就会从应用程序级别(而非操作系统)控制切换**,
以此来提升效率(!!!非io操作的切换与效率无关)
开线程的开销比协程的开销大太多了
对比操作系统控制线程的切换,用户在单线程内控制协程的切换。
优点如下:
协程的切换开销更小,属于程序级别的切换,操作系统完全感知不到,因而更加轻量级
单线程内就可以实现并发的效果,最大限度地利用cpu
缺点如下:
协程的本质是单线程下,无法利用多核
可以是一个程序开启多个进程,每个进程内开启
多个线程,每个线程内开启协程
协程指的是单个线程,因而一旦协程出现阻塞,将会阻塞整个线程
总结协程特点:
必须在只有一个单线程里实现并发
修改共享数据不需加锁
用户程序里自己保存多个控制流的上下文栈
附加:一个协程遇到IO操作自动切换到其它协程(如何实现检测IO,yield、greenlet都
无法实现,就用到了gevent模块(select机制))
greenlet实现状态切换
需安装第三方模块:pip3 install greenlet
def eat(name):
print('%s eat 1' %name)
g2.switch('nick')
print('%s eat 2' %name)
g2.switch()
def play(name):
print('%s play 1' %name)
g1.switch()
print('%s play 2' %name)
g1=greenlet(eat)
g2=greenlet(play)
g1.switch('nick')#可以在第一次switch时传入参数,以后都不需要
运行结果:
nick eat 1
nick play 1
nick eat 2
nick play 2
gevent模块
需安装第三方模块:gevent
例1:
import gevent
def eat(name):
print('%s eat 1' %name)
gevent.sleep(2)
print('%s eat 2' %name)
def play(name):
print('%s play 1' %name)
gevent.sleep(1)
print('%s play 2' %name)
g1=gevent.spawn(eat,'lqz')
g2=gevent.spawn(play,name='lqz')
g1.join()
g2.join()
#或者gevent.joinall([g1,g2])
print('主')
运行结果:
lqz eat 1
lqz play 1
lqz play 2
lqz eat 2
主
需安装第三方模块:gevent
例2:
import gevent
import time
from gevent import monkey
monkey.patch_all()
def eat(name):
print('%s eat 1' % name)
gevent.sleep(2)
# time.sleep(2)
print('%s eat 2' % name)
def play(name):
print('%s play 1'% name)
# gevent.sleep(1)
time.sleep(2)
print('%s play 2' % name)
g1 = gevent.spawn(eat, 'lqz')
g2 = gevent.spawn(play, name='lqz')
g1.join()
g2.join()
运行结果:
lqz eat 1
lqz play 1
lqz eat 2
lqz play 2