from threading import Thread,Lock
# 创建一个互斥锁
mutex = Lock()def task1():# 锁定
    mutex.acquire()for i in range(100):print("===================")# 打开
    mutex.release()
def task2():mutex.acquire()for i in range(100):print("!!!!!!!!!!!!!!!!!!")mutex.release()def task3():mutex.acquire()for i in range(100):print("********************")mutex.release()t1 = Thread(target=task1)
t2 = Thread(target=task2)
t3 = Thread(target=task3)t1.start()
t2.start()
t3.start()

from threading import Thread,Lock
import time,randommutex1 = Lock()   #创建一个锁1
mutex2 = Lock()   #创建一个锁2
def fun1():mutex1.acquire()print("抢到了锁1")time.sleep(1)mutex2.acquire()print("抢到了锁2")mutex2.release()print("释放了锁2")mutex1.release()print("释放了锁1")def fun2():mutex2.acquire()print("抢到了锁2")time.sleep(1)mutex1.acquire()print("抢到了锁1")mutex1.release()print("释放了锁2")mutex2.release()print("释放了锁1")# def fun3():
#     fun1()
#     fun2(

t1 = Thread(target=fun1)
t1.start()
t2 = Thread(target=fun2)
t2.start()

from threading import Thread,Lock,current_thread,RLock
import time
# 叉子
locka = RLock()
# 盘子
lockb = RLock()def task1():print(current_thread())locka.acquire()print("抢到叉子 需要盘子")time.sleep(0.1)lockb.acquire()print("吃饭")lockb.release()locka.release()def task2():print(current_thread())lockb.acquire()print("抢到盘子 需要叉子")time.sleep(0.1)locka.acquire()print("吃饭")locka.release()lockb.release()t1 = Thread(target=task1)
t1.start()
t2 = Thread(target=task2)
t2.start()# 死锁发生的条件  有多个线程 多个锁   如果只有一个锁 无论是LOCK RLOK 卡不死(前提是逻辑上没有错误)
# RLock 就算你的代码逻辑不对 同一个线程多次对一个锁执行acquire 也不会卡死

from threading import  Thread,Lock,RLock,current_threadl = Lock()  # 互斥锁# l.acquire()
# print("zxxzxxxzxzx")
# l.acquire()
# print("aaaaaaaaa")# RLock   递归锁 重入锁  可以多次执行acquire# lock = RLock()
#
# lock.acquire()
# print("aaaaaaaaaaa")
# lock.acquire()
# print("bbbbbbbbbbb")import time
lock = RLock()
# 对于同一个线程而言 可以多次acquire  其他线程会被阻塞
def task():lock.acquire()for i in range(5):time.sleep(1)print(current_thread())lock.release()Thread(target=task).start()
Thread(target=task).start()

from threading import Thread,Semaphore,current_thread,active_countimport time
# 用于控制 同时执行被锁定代码的线程数量   也就是线程的并发数量
# 也是一种锁
sm = Semaphore(1)def task():sm.acquire()for i in range(10):print(current_thread())time.sleep(0.5)sm.release()def task2():for i in range(10):print(current_thread())time.sleep(0.5)for i in range(5):Thread(target=task).start()Thread(target=task2).start()
print(active_count())