1. 两种表现形式,with和acquire,release 2. 在用一个线程中,不能连续acquire多次(进程会卡在这),必须要进行release后才能继续acquire(会报错,RuntimeError: release unlocked lock) 3. 在操作全局变量和+=,-=,*=,/=这类计算在赋值类,或lis[0]+=1,dic['key']-=1这类时,就会出现数据不一致的问题。
from multiprocessing import Process,Manager,Lock def work(data, lock): # 1.正常写法 """ # 上锁 lock.acquire() # 修改数据 data["count"] -= 1 # 解锁 lock.release() """ # 2.使用with语法可以简化上锁和解锁两步操作 with lock: data[0] += 1 if __name__ == '__main__': lst = ...
import threading num = 0 def add(): lock.acquire() global num for i in ra...
python Treading中的Lock模块提供了加锁和释放锁的方法,分别是acquire()和release(). 这两个方法可以搭配python的with语句使用. """ # 示例 fromthreadingimportLock temp_lock=Lock() withtemp_lock: print(temp_lock) # 输出是 <locked _thread.lock object at 0x10e304870> 说明temp_lock上锁了 ...
Python Threading中的Lock模块有acquire()和release()两种方法,这两种方法与with语句的搭配相当于,进入with语句块时候会先执行acquire()方法,语句块结束后会执行release方法。 举个例子: from threading impor…
#创建一个lock对象 lock = threading.Lock() #初始化共享资源 abce = 0 #本线程访问共享资源 lock.acquire() abce = abce + 1 #这个线程访问共享资源会被阻塞 lock.acquire() abce = abce + 2 lock.release() print(abce) 1. 2. 3. 4.
对于同步锁来说,一次acquire()必须对应一次release(),不能出现连续重复使用多次acquire()后再重复使用多次release()的操作,这样会引起死锁造成程序的阻塞,完全不动了,如下所示: import threadingnum = 0def add(): lock.acquire() # 上锁 lock.acquire() # 死锁 # 不执行 global num for i in range(10_000...
lock.acquire() global num for i in range(10_000_000): num -= 1 lock.release() if __name__ == "__main__": lock = threading.Lock() subThread01 = threading.Thread(target=add) subThread02 = threading.Thread(target=sub) subThread01.start() ...
): lock.acquire() withlock+=1 lock.release() def dewithlock(): global withlock for i in range(count): lock.acquire() withlock-=1 lock.release() def innolock(): global nolock for i in range(count): nolock+=1 def denolock(): global nolock for i in range(count): nolock-=...
lock.acquire()# 开始加锁 number+=1 lock.release()# 释放锁 foriinrange(2):# 用2个子线程,就可以观察到脏数据 t=threading.Thread(target=plus) t.start() time.sleep(2)# 等待2秒,确保2个子线程都已经结束运算。 print("主线程执行完毕后,number = ", number) ...