Python中四种锁的使用示例（代码）

num = 0
def a():
    global num
    for _ in range(10000000):
        num += 1

def b():
    global num
    for _ in range(10000000):
        num += 1
if __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 = 0
def a(lock):
    global num
    for _ in range(1000000):
        with lock:
            num += 1
def b(lock):
    global num
    for _ in range(1000000):
        with lock:
            num += 1
if __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 = 0
def 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