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，假如两个线程别离占有一部门资源而且同时期待对方的资源，就会造成死锁。尽量死锁很少产生，但一旦产生就会造成应用的遏制响应。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()

执行功效：

Thread-1 got resA

Thread-1 got resB

Thread-1 got resA

Thread-2 got resA

Thread-2 got resB

Thread-2 got resA

Thread-3 got resA

Thread-3 got resB

Thread-3 got resA

Thread-5 got resA

Thread-5 got resB

Thread-4 got resA

此时历程已经死掉。

可重入锁

更简朴的死锁环境是一个线程“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，其他的线程才气得到资源。上面的例子假如利用RLock取代Lock，则不会发存亡锁：

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()

执行功效：

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

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