python递归锁

Posted 2020-10-11

一把大锁在加一把小锁。

import threading
import time


def run1():
    print("grab the first part data")
    lock.acquire()
    global num
    num += 1
    lock.release()
    return num


def run2():
    print("grab the second part data")
    lock.acquire()
    global num2
    num2 += 1
    lock.release()
    return num2


def run3():
    lock.acquire()
    res = run1()
    print(‘--------between run1 and run2-----‘)
    res2 = run2()
    lock.release()
    print(res, res2)


if __name__ == ‘__main__‘:

    num, num2 = 0, 0
    lock = threading.Lock()
    for i in range(10):
        t = threading.Thread(target=run3)
        t.start()

while threading.active_count() != 1:
    print(threading.active_count())
else:
    print(‘----all threads done---‘)
    print(num, num2)

 过程分享:
1、启动10个线程，每个线程都执行run3。

2、run3中先加了一个锁，然后执行run1。

3、run1中也加了一个锁，计算完成后解锁返回数据。执行run2。

4、run2中同样加了一个锁，计算完成后解锁打印run1,run2返回值。

5、最后打印num1、num2。

需要注意的是在执行过程中是有两把锁的，一个run3，一个run1或者run2。

运行结果：

11
11
11
11
11
11
11
11
# 程序进入死循环

 死循环的原因是解锁的时候 钥匙拿错了导致进程退不出来，解决死锁就是要使用递归锁。

threading.Lock() 改为 threading.RLock()

import threading
import time


def run1():
    print("grab the first part data")
    lock.acquire()
    global num
    num += 1
    lock.release()
    return num


def run2():
    print("grab the second part data")
    lock.acquire()
    global num2
    num2 += 1
    lock.release()
    return num2


def run3():
    lock.acquire()
    res = run1()
    print(‘--------between run1 and run2-----‘)
    res2 = run2()
    lock.release()
    print(res, res2)


if __name__ == ‘__main__‘:

    num, num2 = 0, 0
    lock = threading.RLock()
    for i in range(1):
        t = threading.Thread(target=run3)
        t.start()

while threading.active_count() != 1:
    print(threading.active_count())
else:
    print(‘----all threads done---‘)
    print(num, num2)

运行结果

grab the first part data
--------between run1 and run2-----
grab the second part data
1 1
----all threads done---
1 1

 实现原理：

解锁的时候找对应的钥匙。