Python中四种锁的使用示例（代码）

num = 0def a():    global num    for _ in range(10000000):        num += 1def b():    global num    for _ in range(10000000):        num += 1if __name__ == '__main__':    t1=Thread(target=a)    t1.start()    t2=Thread(target=b)    t2.start()    t1.join()    t2.join()    print(num)    #基本永远会小于20000000

num = 0def a(lock):    global num    for _ in range(1000000):        with lock:            num += 1def b(lock):    global num    for _ in range(1000000):        with lock:            num += 1if __name__ == '__main__':    lock = threading.Lock()    t1=Thread(target=a, args=(lock,))    t1.start()    t2=Thread(target=b, args=(lock,))    t2.start()    t1.join()    t2.join()    print(num)    #永远会输出20000000

#在之前的代码中永远不可能出现锁在没释放之前重新获得锁，但rlock可以做到，但只能发生在一个线程中，如：num = 0def a(lock):    with lock:        print("我是A")        b(lock)def b(lock):    with lock:        print("我是b")if __name__ == '__main__':    lock = threading.Lock()    t1 = Thread(target=a, args=(lock,))    t1.start()    #会发生死锁，因为在第一次还没释放锁后，b就准备上锁，并阻止a释放锁

if __name__ == '__main__':    lock = threading.RLock()    #只需要改变锁为RLock程序马上恢复    t1 = Thread(target=a, args=(lock,))    t1.start()

#这个程序我们模拟甲乙对话Jlist = ["在吗", "干啥呢", "去玩儿不", "好吧"]Ylist = ["在呀", "玩儿手机", "不去"]def J(list):    for i in list:        print(i)        time.sleep(0.1)def Y(list):    for i in list:        print(i)        time.sleep(0.1)if __name__ == '__main__':    t1 = Thread(target=J, args=(Jlist,))    t1.start()    t1.join()    t2 = Thread(target=Y, args=(Ylist,))    t2.start()    t2.join()    #上面的程序输出后发现效果就是咱们想要的，但是我们每次输出后都要等待0.1秒，也无法正好确定可以拿到时间片的最短时间值，并且不能保证每次正好都是另一个线程执行。因此，我们用以下方式，完美解决这些问题。

Jlist = ["在吗", "干啥呢", "去玩儿不", "好吧"]Ylist = ["在呀", "玩儿手机", "不去","哦"]def J(cond, list):    for i in list:        with cond:            print(i)            cond.notify()            cond.wait()def Y(cond, list):    for i in list:        with cond:            cond.wait()            print(i)            cond.notify()if __name__ == '__main__':    cond = threading.Condition()    t1 = Thread(target=J, args=(cond, Jlist))    t2 = Thread(target=Y, args=(cond, Ylist))    t2.start()    t1.start()    #一定保证t1启动在t2之后,因为notify发送的信号要被t2接受到，如果t1先启动，会发生阻塞。

class B(threading.Thread):    def __init__(self, name):        super().__init__()        self.name = name    def run(self):        time.sleep(1)        print(self.name)class A(threading.Thread):    def __init__(self):        super().__init__()    def run(self):        for i in range(100):            b = B(i)            b.start()if __name__ == '__main__':    a = A()    a.start()    #执行后发现不断在输出

class B(threading.Thread):    def __init__(self, name, sem):        super().__init__()        self.name = name        self.sem = sem    def run(self):        time.sleep(1)        print(self.name)        sem.release()class A(threading.Thread):    def __init__(self, sem):        super().__init__()        self.sem = sem    def run(self):        for i in range(100):            self.sem.acquire()            b = B(i, self.sem)            b.start()if __name__ == '__main__':    sem = threading.Semaphore(value=3)    a = A(sem)    a.start()    #通过执行上面的代码，我们发现一次只能输出三个数字，sem控制访问并发量