python线程死锁与递归锁

死锁现象

所谓死锁： 是指两个或两个以上的进程或线程在执行过程中，因争夺资源而造成的一种互相等待的现象，若无外力作用，它们都将无法推进下去。

此时称系统处于死锁状态或系统产生了死锁，这些永远在互相等待的进程称为死锁进程，如下就是死锁

from threading import Thread,Lock
import time

mutexA = Lock()
mutexB = Lock()

class MyThread(Thread):
    def run(self):
        self.task1()
        self.task2()


    def task1(self):
        mutexA.acquire()
        print('%s task1 get A' %self.name)

        mutexB.acquire()
        print('%s task1 get B' % self.name)
        mutexB.release()

        mutexA.release()

    def task2(self):
        mutexB.acquire()
        print('%s task2 get B' % self.name)
        time.sleep(1)  # Thread-2 拿到执行权，执行get A出现死锁，此时thread2需要B锁，而thread1占用，与此同时，thread1需要A锁，thread2占用
        mutexA.acquire()
        print('%s task2 get A' % self.name)

        mutexA.release()
        mutexB.release()


if __name__ == '__main__':
    for i in range(10):
        t = MyThread()
        t.start()

-------------------输出
Thread-1 task1 get A
Thread-1 task1 get B
Thread-1 task2 get B
Thread-2 task1 get A # 出现死锁，整个程序被阻塞

　

递归锁

解决方法，递归锁，在Python中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RLock。

这个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，二者的区别是：递归锁可以连续acquire多次，而互斥锁只能acquire一次

from threading import Thread,RLock
import time

mutexA = mutexB = RLock()

class MyThread(Thread):
    def run(self):
        self.task1()
        self.task2()


    def task1(self):
        mutexA.acquire()
        print('%s task1 get A' %self.name)

        mutexB.acquire()
        print('%s task1 get B' % self.name)
        mutexB.release()

        mutexA.release()
        time.sleep(1) # Thread-2 拿到执行权，,此时counter=0，thread2执行task1

    def task2(self):
        mutexB.acquire()
        print('%s task2 get B' % self.name)

        mutexA.acquire()
        print('%s task2 get A' % self.name)

        mutexA.release()
        mutexB.release()


if __name__ == '__main__':
    for i in range(10):
        t = MyThread()
        t.start()


------------输出
from threading import Thread,RLock
import time

mutexA = mutexB = RLock()

class MyThread(Thread):
    def run(self):
        self.task1()
        self.task2()


    def task1(self):
        mutexA.acquire()
        print('%s task1 get A' %self.name)

        mutexB.acquire()
        print('%s task1 get B' % self.name)
        mutexB.release()

        mutexA.release()
        time.sleep(1) # Thread-2 拿到执行权，,此时counter=0，thread2执行task1

    def task2(self):
        mutexB.acquire()
        print('%s task2 get B' % self.name)

        mutexA.acquire()
        print('%s task2 get A' % self.name)

        mutexA.release()
        mutexB.release()


if __name__ == '__main__':
    for i in range(10):
        t = MyThread()
        t.start()

------------------------输出

Thread-1 task1 get A
Thread-1 task1 get B
Thread-2 task1 get A
Thread-2 task1 get B
Thread-3 task1 get A
Thread-3 task1 get B
Thread-4 task1 get A
Thread-4 task1 get B
Thread-5 task1 get A
Thread-5 task1 get B
Thread-6 task1 get A
Thread-6 task1 get B
Thread-7 task1 get A
Thread-7 task1 get B
Thread-8 task1 get A
Thread-8 task1 get B
Thread-9 task1 get A
Thread-9 task1 get B
Thread-10 task1 get A
Thread-10 task1 get B
Thread-1 task2 get B
Thread-1 task2 get A
Thread-2 task2 get B
Thread-2 task2 get A
Thread-4 task2 get B
Thread-4 task2 get A
Thread-3 task2 get B
Thread-3 task2 get A
Thread-5 task2 get B
Thread-5 task2 get A
Thread-6 task2 get B
Thread-6 task2 get A
……