python多线程编程(3): 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：

按 Ctrl+C 复制代码

按 Ctrl+C 复制代码

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