python之线程锁

摘要：信号量使用前执行后发现不断在输出使用后通过执行上面的代码，我们发现一次只能输出三个数字，控制访问并发量事件等待后，才能往下执行定义五个线程重置，使得起到阻塞作用启动所有线程等待唤醒所有线程等待唤醒所有线程

本文讲python中的四种锁
Lock互斥锁
使用前

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

使用后

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

RLock重用锁

#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

使用后

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()

Condition同步锁

#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。

使用后

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。

Seamplore信号量
使用前

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

使用后

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

Event事件

import time
import threading


class MyThread(threading.Thread):
    def __init__(self, name, event):
        super().__init__()
        self.name = name
        self.event = event

    def run(self):
        print("Thread: {} start at {}".format(self.name, time.ctime(time.time())))
        # 等待event.set()后，才能往下执行
        self.event.wait()
        print("Thread: {} finish at {}".format(self.name, time.ctime(time.time())))


event = threading.Event()

# 定义五个线程
threads = [MyThread(str(i), event) for i in range(1,5)]

# 重置event，使得event.wait()起到阻塞作用
event.clear()

# 启动所有线程
[t.start() for t in threads]

print("等待5s...")
time.sleep(5)

print("唤醒所有线程...")
event.set()

# output:
"""
Thread: 1 start at Sun May 13 20:38:08 2018
Thread: 2 start at Sun May 13 20:38:08 2018
Thread: 3 start at Sun May 13 20:38:08 2018
Thread: 4 start at Sun May 13 20:38:08 2018

等待5s...

唤醒所有线程...
Thread: 1 finish at Sun May 13 20:38:13 2018
Thread: 4 finish at Sun May 13 20:38:13 2018
Thread: 2 finish at Sun May 13 20:38:13 2018
Thread: 3 finish at Sun May 13 20:38:13 2018
"""