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  • 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
    
    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
    ……
    

      

     

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  • 原文地址:https://www.cnblogs.com/xiao-apple36/p/9501869.html
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