资源总是有限的,程序运行如果对同一个对象进行操作,则有可能造成资源的争用,甚至导致死锁
也可能导致读写混乱
锁提供如下方法:
1.Lock.acquire([blocking])
2.Lock.release()
3.threading.Lock() 加载线程的锁对象,是一个基本的锁对象,一次只能一个锁定,其余锁请求,需等待锁释放后才能获取
4.threading.RLock() 多重锁,在同一线程中可用被多次acquire。如果使用RLock,那么acquire和release必须成对出现,
调用了n次acquire锁请求,则必须调用n次的release才能在线程中释放锁对象
例如:
无锁:
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#coding=utf8 import threading import time num = 0 def sum_num(i): global num time.sleep( 1 ) num + = i print num print '%s thread start!' % (time.ctime()) try : for i in range ( 6 ): t = threading.Thread(target = sum_num,args = (i,)) t.start() except KeyboardInterrupt,e: print "you stop the threading" print '%s thread end!' % (time.ctime()) |
输出:
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Sun May 28 20 : 54 : 59 2017 thread start! Sun May 28 20 : 54 : 59 2017 thread end! 01 3 710 15 |
结果显示混乱
引入锁:
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#coding=utf8 import threading import time num = 0 def sum_num(i): lock.acquire() global num time.sleep( 1 ) num + = i print num lock.release() print '%s thread start!' % (time.ctime()) try : lock = threading.Lock() list = [] for i in range ( 6 ): t = threading.Thread(target = sum_num,args = (i,)) list .append(t) t.start() for threadinglist in list : threadinglist.join() except KeyboardInterrupt,e: print "you stop the threading" print '%s thread end!' % (time.ctime()) |
结果:
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Sun May 28 21 : 15 : 37 2017 thread start! 0 1 3 6 10 15 Sun May 28 21 : 15 : 43 2017 thread end! |
其中:
lock=threading.Lock()加载锁的方法也可以换成lock=threading.RLock()
如果将上面的sum_num修改为:
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lock.acquire() global num lock.acquire() time.sleep( 1 ) num + = i lock.release() print num lock.release() |
那么:
lock=threading.Lock() 加载的锁,则一直处于等待中,锁等待
而lock=threading.RLock() 运行正常
以上这篇对python多线程中Lock()与RLock()锁详解就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持服务器之家。
原文链接:https://blog.csdn.net/comprel/article/details/72798354