python多線程編程(4): 死結和可重新進入鎖

文章目錄

• 死結
• 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-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()

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 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()

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 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()

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