Python互斥锁实例（包含超时解锁）

前言

最近遇到一个小需求，大概是请求机器人&自动化设备执行某些功能，这些功能之间有共同操作资源或在业务逻辑上有冲突，如何保证多线程的请求不会同时执行造成损失。¹

这里使用的是互斥锁，当有任务正在执行时写了如下几种处理方式：

1. 直接返回
2. 等待拿到锁，继续执行
3. 等待拿到锁，继续执行或超时返回

正文

代码如下：

import threading
import timeprocess_lock = threading.Lock()def process1(id_num):name = 'process1-' + str(id_num)if process_lock.locked():print(name, ' is locked')returnif process_lock.acquire():time.sleep(0.1)print(name, ' get lock ')process_lock.release()print(name, ' release lock')def process2(id_num):name = 'process2-' + str(id_num)if process_lock.locked():print(name, ' is locked')if process_lock.acquire():time.sleep(0.2)print(name, ' get lock ')process_lock.release()print(name, ' release lock')def process3(id_num):name = 'process3-'+str(id_num)if process_lock.locked():print(name, ' is locked')if process_lock.acquire(blocking=True, timeout=0.3):try:print(name, ' get lock')time.sleep(0.2)finally:process_lock.release()print(name, ' release lock')if __name__ == '__main__':# # 如果锁被占用则直接返回# for i in range(3):#     threading.Thread(target=process1, args=(i,)).start()# # 如果拿不到锁则一直等待# for i in range(3):#     time.sleep(0.1)#     threading.Thread(target=process2, args=(i,)).start()# 如果拿不到锁则等待，直到超时for i in range(3):threading.Thread(target=process3, args=(i,)).start()

1 锁被占用直接返回

process1-1  is locked
process1-2  is locked
process1-0  get lock 
process1-0  release lock

0号线程拿到了锁，1 2线程没拿到锁直接返回

2 拿不到锁则一直等待

process2-1  is locked
process2-0  get lock 
process2-0  release lock
process2-2  is locked
process2-1  get lock 
process2-1  release lock
process2-2  get lock 
process2-2  release lock

0 1 2三个线程逐次拿到锁，执行后释放

3 拿不到锁等待直到超时

process3-0  get lock
process3-1  is locked
process3-2  is locked
process3-0  release lock
process3-1  get lock
process3-1  release lock

0 拿到锁执行完释放，2拿到锁再释放，于此同时3已经超时

总结

虽然这篇文章写的比较基础，但是实际使用中大家可能都会用的到，整理记录于此。

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1. threading — Thread-based parallelism — Python 3.12.2 documentation ↩︎