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Python学习day39-并发编程(各种锁)

同步锁

前文的进程中我们已经了解到,锁通常被用来实现对共享资源的同步访问.为每一个共享资源创建一个Lock对象,当我们需要访问该共享资源的时候,就需要先用acquire来获取锁对象,然后执行中间内容,执行完毕之后再调用release来释放锁.

线程中的同步锁其实就类似于我们之前用的进程锁,还有一种叫法是线程的互斥锁.

另外线程中的锁还可以保证数据的安全,具体体现如下例:

xxxxxxxxxx

from threading import Thread, Lock

x = 0

mutex = Lock()

def task():

    global x

    for i in range(1000000):

        x += 1

if __name__ == '__main__':

    t1 = Thread(target=task)

    t2 = Thread(target=task)

    t3 = Thread(target=task)

    t1.start()

    t2.start()

    t3.start()

    t1.join()

    t2.join()

    t3.join()

    print(x)

# 这个例程的执行结果应该是大多数人没想到的,因为结果并不是所预料的3000000,而是一个小于3000000的随机值,为什么呢?

# 我们可以看到,这个例子开启了三个子线程,之前我们说过,线程之间的切换一般会有两种情况,一种是线程遇到了IO,第二种就是一个线程运行的时间过长的时候,就会切换至下一个要执行的线程.这个例子就是第二种情况,因为task()里面的循环太长,所以一个线程短时间内并不能执行完,所以就会在执行期间被CPU切换至别的进程,这就出现了问题!?!?

# 问题就是,假设当时Thread-1正在执行x+=1这个语句,刚执行完x+1,还没有赋值给原x,此时CPU切换走了,那么这次这个循环就相当于执行了,但是没有结果,x也没有增加,这就是为什么最后的值会小于我们的预定值.

# 当然这种概率会很小,但是随着循环量的变大,这种概率会变大大,尤其是在循环次数大于100000以上时,造成最后的数据不正确,也不安全

# 所以此时我们就要考虑加入同步锁,或者说互斥锁,来保证数据的安全,添加的方法也很简单,就和之前的进程锁一样,在要执行的任务的前后两端加上acquire和release就可以了

加入同步锁以后就是这样的:

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from threading import Thread, Lock

x = 0

mutex = Lock()

def task():

    # with mutex:# 我们也可以这么添加同步锁,可以减少代码量

    mutex.acquire()# 保证数据安全,取到同步锁acquire

    global x

    for i in range(1000000):

        x += 1

    mutex.release()# 释放同步锁release

if __name__ == '__main__':

    t1 = Thread(target=task)

    t2 = Thread(target=task)

    t3 = Thread(target=task)

    t1.start()

    t2.start()

    t3.start()

    t1.join()

    t2.join()

    t3.join()

    print(x)

# 加入同步锁以后再执行程序,我们会看到结果就不会出错了,而且是丝毫不差

死锁与递归锁

死锁问题的根源在于过度使用同步锁,其原理类似于我们之前学习模块的时候的循环导入问题,具体描述下来就是: 两个线程: 线程1拿到了锁2,想要往下执行,需要锁1 线程2拿到了锁1,想要往下执行,需要锁2 互相都拿到了对方想要往下继续运行的锁,且彼此都不会释放自己的锁,就会造成死锁

死锁实例如下:

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# 死锁问题的演示:

from threading import Thread, Lock

import time

mutex1 = Lock()   # 互斥锁,同一时间仅有一个任务可以抢到这个锁

mutex2 = Lock()

class MyThread(Thread):

    def run(self):

        self.task1()

        self.task2()

    def task1(self):

        mutex1.acquire()

        print(f'{self.name}抢到了 锁1')

        mutex2.acquire()

        print(f'{self.name}抢到了 锁2')

        mutex1.release()

        print(f'{self.name}释放了 锁1')

        mutex2.release()

        print(f'{self.name}释放了 锁2')

    def task2(self):

        mutex2.acquire()

        print(f'{self.name}抢到了 锁2')

        time.sleep(2)

        mutex1.acquire()

        print(f'{self.name}抢到了 锁1')

        mutex1.release()

        print(f'{self.name}释放了 锁1')

        mutex2.release()

        print(f'{self.name}释放了 锁2')

if __name__ == '__main__':

    for i in range(3):

        t = MyThread()

        t.start()

# 结果为:

'''

Thread-1抢到了 锁1

Thread-1抢到了 锁2

Thread-1释放了 锁1

Thread-1释放了 锁2

Thread-1抢到了 锁2

Thread-2抢到了 锁1

'''

# 到这里程序并没有结束,而是卡住了,Thread-2想抢锁2,但是锁2在Thread-1里面,而Thread-1想抢锁1,锁1在Thread-2里面,就造成了死锁

解决死锁问题比较好的一个方法就是递归锁: 递归锁: 在同一个线程内可以被多次acquire 如何释放: 递归锁内部相当于维护了一个计数器同一个线程acquire几次,就要release几次,计数为零的时候就可以结束

加入递归锁之后的代码如下:

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# 死锁问题的解决,递归锁

from threading import Thread, Lock, RLock# RLock就是递归锁

import time

mutex1 = mutex2 = RLock()# 我们用RLock来替代Lock

class MyThread(Thread):

    def run(self):

        self.task1()

        self.task2()

    def task1(self):

        mutex1.acquire()

        print(f'{self.name}抢到了 锁1')

        mutex2.acquire()

        print(f'{self.name}抢到了 锁2')

        mutex1.release()

        print(f'{self.name}释放了 锁1')

        mutex2.release()

        print(f'{self.name}释放了 锁2')

    def task2(self):

        mutex2.acquire()

        print(f'{self.name}抢到了 锁2')

        time.sleep(1)

        mutex1.acquire()

        print(f'{self.name}抢到了 锁1')

        mutex1.release()

        print(f'{self.name}释放了 锁1')

        mutex2.release()

        print(f'{self.name}释放了 锁2')

if __name__ == '__main__':

    for i in range(3):

        t = MyThread()

        t.start()

# 运行结果为:

'''

Thread-1抢到了 锁1

Thread-1抢到了 锁2

Thread-1释放了 锁1

Thread-1释放了 锁2

Thread-1抢到了 锁2

Thread-1抢到了 锁1

Thread-1释放了 锁1

Thread-1释放了 锁2

Thread-2抢到了 锁1

Thread-2抢到了 锁2

Thread-2释放了 锁1

Thread-2释放了 锁2

Thread-2抢到了 锁2

Thread-2抢到了 锁1

Thread-2释放了 锁1

Thread-2释放了 锁2

Thread-3抢到了 锁1

Thread-3抢到了 锁2

Thread-3释放了 锁1

Thread-3释放了 锁2

Thread-3抢到了 锁2

Thread-3抢到了 锁1

Thread-3释放了 锁1

Thread-3释放了 锁2

Process finished with exit code 0

'''

# 从上面结果中我们可以看到,程序的执行没有卡主,而是每个进程都取到了锁,并且释放了锁,最后都运行完毕,没有报错

信号量Semaphore(带上限的锁)

Semaphore的用法和同步锁基本一致,不同的是他可以限制同时取到这个锁的线程的数量,类比到生活也许就是,给一个锁开了几个钥匙孔?

实例如下:

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from threading import Thread, Semaphore

import threading

import time

def task():

    s.acquire()

    print(f'{threading.currentThread().name} 在执行')

    time.sleep(2)

    s.release()

if __name__ == '__main__':

    s = Semaphore(5)# Semaphore括号里的数字就是能同时取到锁的线程的数量,也就是说这个实例中最多允许五个进程同时取到锁,进行下面的运算

    for i in range(20):

        t = Thread(target=task)

        t.start()

# 运行结果:

'''

Thread-1 在执行

Thread-2 在执行

Thread-3 在执行

Thread-4 在执行

Thread-5 在执行

Thread-6 在执行

Thread-7 在执行

Thread-8 在执行

Thread-9 在执行

Thread-10 在执行

Thread-11 在执行

Thread-12 在执行

Thread-13 在执行

Thread-14 在执行

Thread-15 在执行

Thread-16 在执行

Thread-17 在执行

Thread-18 在执行

Thread-20 在执行

Thread-19 在执行

Process finished with exit code 0

'''

# 可能你的运行结果并不太能看的出来五个线程一组,我们可以在程序运行的时候在run窗口里面一直按回车键,虽然这很蠢....但是这种方法是最直观能看得出来五个线程同时执行的方法

GIL锁

Cpython里面有一把GIL锁(全局解释器锁),GIL锁本质上是一把互斥锁因为GIL锁,同一时刻一个进程下只有一个线程在执行,无法利用多核优势同一个进程下, 多个线程只能并发不能并行为什么要有GIL: 因为Cpython自带的垃圾回收机制不是线程安全的,所以要有GIL锁

分析: 假设有四个任务要处理,处理方式要体现出并发的效果方案一:开启四个进程方案二:开启一个进程,四个线程

下面我们用实例来证明这两个方案的不同之处:

# 计算密集型任务

def work1():

    res = 0

    for i in range(109090000):

        res *= i

# IO密集型任务

# def work1():

#     res = 0

#     time.sleep(5)

#     for i in range(2):

#         res *= i

if __name__ == '__main__':

    t_list = list()

    start = time.time()

    for i in range(4):

        t = Thread(target=work1)

        t_list.append(t)

        t.start()

    for j in t_list:

        j.join()

    end = time.time()

    print(f'线程总共的运行时间为{end - start:6.2f}')  # 计算密集型19.13s,    IO密集型,5.00s

    # t_list = list()

    # start = time.time()

    # for i in range(4):

    #     t = Process(target=work1)

    #     t_list.append(t)

    #     t.start()

    # for j in t_list:

    #     j.join()

    # end = time.time()

    # print(f'进程总共的运行时间为{end - start:6.2f}')  # 计算密集型6.49s,   IO密集型5.31s

# 上例中把计算密集型任务和IO密集型任务放在了一起,读者在尝试的时候可以自己手输分开,或者是分别注释其中的一半,否则会报错

# 从结果中我们可以看出:

# 计算密集型任务,线程的时间19.13s>>进程的时间6.49s

# 而IO密集型任务,线程的时间5.00s<进程的时间5.31s

# 再总和考虑线程和进程的成本以及切换速度,我们就可以得出以下结论:

'''

如果是计算密集型任务:(推荐使用多进程的方案)(比如挖矿)

如果是IO密集型任务:(推荐使用多线程)(比如web,后端,爬虫)

'''