python多线程编程(4): 死锁和可重入锁

死锁

在线程间共享多个资源的时候,如果两个线程分别占有一部分资源并且同时等待对方的资源,就会造成死锁。尽管死锁很少发生,但一旦发生就会造成应用的停止响应。下面看一个死锁的例子:

# encoding: UTF-8
import threading
import time

class MyThread(threading.Thread):
def do1(self):
global resA, resB
if mutexA.acquire():
msg = self.name+' got resA'
print msg

if mutexB.acquire(1):
msg = self.name+' got resB'
print msg
mutexB.release()
mutexA.release()
def do2(self):
global resA, resB
if mutexB.acquire():
msg = self.name+' got resB'
print msg

if mutexA.acquire(1):
msg = self.name+' got resA'
print msg
mutexA.release()
mutexB.release()


def run(self):
self.do1()
self.do2()
resA = 0
resB = 0

mutexA = threading.Lock()
mutexB = threading.Lock()

def test():
for i in range(5):
t = MyThread()
t.start()
if __name__ == '__main__':
test()

执行结果:

Thread-1 got resA
Thread-1 got resB
Thread-1 got resB
Thread-1 got resA
Thread-2 got resA
Thread-2 got resB
Thread-2 got resB
Thread-2 got resA
Thread-3 got resA
Thread-3 got resB
Thread-3 got resB
Thread-3 got resA
Thread-5 got resA
Thread-5 got resB
Thread-5 got resB
Thread-4 got resA

此时进程已经死掉。

可重入锁

更简单的死锁情况是一个线程“迭代”请求同一个资源,直接就会造成死锁:

import threading
import time

class MyThread(threading.Thread):
def run(self):
global num
time.sleep(1)

if mutex.acquire(1):
num = num+1
msg = self.name+' set num to '+str(num)
print msg
mutex.acquire()
mutex.release()
mutex.release()
num = 0
mutex = threading.Lock()
def test():
for i in range(5):
t = MyThread()
t.start()
if __name__ == '__main__':
test()

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”:threading.RLock。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:

import threading
import time

class MyThread(threading.Thread):
def run(self):
global num
time.sleep(1)

if mutex.acquire(1):
num = num+1
msg = self.name+' set num to '+str(num)
print msg
mutex.acquire()
mutex.release()
mutex.release()
num = 0
mutex = threading.RLock()
def test():
for i in range(5):
t = MyThread()
t.start()
if __name__ == '__main__':
test()

执行结果:

Thread-1 set num to 1
Thread-3 set num to 2
Thread-2 set num to 3
Thread-5 set num to 4
Thread-4 set num to 5







posted @ 2012-03-08 16:31 心内求法 阅读(...) 评论(...) 编辑 收藏