锁——死锁——单个锁锁死

锁：

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()

 

死锁：

 

from threading import Thread,Lock
import time,random

mutex1 = 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

 

单个锁：（递归锁/重入锁）

 单个锁能不能锁死____答案是肯定的：

from threading import  Thread,Lock,RLock,current_thread

l = 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()

 

信号量：（也是一种锁）

 

from threading import Thread,Semaphore,current_thread,active_count

import 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())