excel表格vba编程的方法步骤(excel vba编程基础知识)
573
2022-05-30
线程 --> 锁
GIL 是python 解释器全局锁
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
import threading R=threading.Lock() R.acquire() ''' 对公共数据的操作 '''
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
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-1 is running Thread-2 is running ...... Thread-100 is running 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 '''
进程也有死锁与递归锁,在进程那里忘记说了,放到这里一切说了额
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
死锁:
from threading import Lock as Lock import time mutexA=Lock() mutexA.acquire() mutexA.acquire() print(123) mutexA.release() mutexA.release()
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:
递归锁:
from threading import RLock as Lock import time mutexA=Lock() mutexA.acquire() mutexA.acquire() print(123) mutexA.release() mutexA.release()
典型问题:科学家吃面
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','egon','yuan']: t1 = Thread(target=eat1,args=(name,)) t2 = Thread(target=eat2,args=(name,)) t1.start() t2.start()
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','egon','yuan']: t1 = Thread(target=eat1,args=(name,)) t2 = Thread(target=eat2,args=(name,)) t1.start() t2.start()
软件开发 人工智能 云计算 机器学习
版权声明:本文内容由网络用户投稿,版权归原作者所有,本站不拥有其著作权,亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容,请联系我们jiasou666@gmail.com 处理,核实后本网站将在24小时内删除侵权内容。