python多線程編程4: 死結和可重新進入鎖

死結

線上程間共用多個資源的時候，如果兩個線程分別佔有一部分資源並且同時等待對方的資源，就會造成死結。儘管死結很少發生，但一旦發生就會造成應用的停止回應。下面看一個死結的例子：

# encoding: UTF-8import threadingimport 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 = 0resB = 0  mutexA = threading.Lock()mutexB = threading.Lock()  def test():    for i in range(5):        t = MyThread()        t.start()if __name__ == '__main__':    test()

執行結果：

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

此時進程已經死掉。

可重新進入鎖

更簡單的死結情況是一個線程“迭代”請求同一個資源，直接就會造成死結：

import threadingimport 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 = 0mutex = threading.Lock()def test():    for i in range(5):        t = MyThread()        t.start()if __name__ == '__main__':    test()

為了支援在同一線程中多次請求同一資源，python提供了“可重新進入鎖”：threading.RLock。RLock內部維護著一個Lock和一個counter變數，counter記錄了acquire的次數，從而使得資源可以被多次require。直到一個線程所有的acquire都被release，其他的線程才能獲得資源。上面的例子如果使用RLock代替Lock，則不會發生死結：

import threadingimport 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 = 0mutex = threading.RLock()def test():    for i in range(5):        t = MyThread()        t.start()if __name__ == '__main__':    test()

執行結果：

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

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