一、递归锁
# Lock ：互斥锁 效率高 # RLock ：递归(recursion)锁 效率相对低 在同一个线程中可以被acquire多次，如果想要释放锁，acquire多少次就要release多少次 from threading import RLock def func(i,rlock): rlock .acquire() rlock .acquire() print(i,': start') rlock .release() rlock .release() print(i, ': end') rlock = RLock () for i in range(5): Thread(target=func,args=(i,rlock)).start()
二、死锁
# import time # from threading import Thread,Lock,RLock # fork_lock = noodle_lock = RLock() # # fork_lock = RLock() # # def eat(name): # noodle_lock.acquire() # print(name,'抢到面了') # fork_lock.acquire() # print(name, '抢到叉子了') # print(name,'吃面') # time.sleep(0.1) # fork_lock.release() # print(name, '放下叉子了') # noodle_lock.release() # print(name, '放下面了') # # def eat2(name): # fork_lock.acquire() # print(name, '抢到叉子了') # noodle_lock.acquire() # print(name,'抢到面了') # print(name,'吃面') # noodle_lock.release() # print(name, '放下面了') # fork_lock.release() # print(name, '放下叉子了') # # Thread(target=eat,args=('alex',)).start() # Thread(target=eat2,args=('wusir',)).start() # Thread(target=eat,args=('taibai',)).start() # Thread(target=eat2,args=('大壮',)).start() import time from threading import Thread,Lock,RLock fork_noodle_lock = Lock() # fork_lock = RLock() def eat(name): fork_noodle_lock.acquire() print(name,'抢到面了') print(name, '抢到叉子了') print(name,'吃面') time.sleep(0.1) fork_noodle_lock.release() print(name, '放下叉子了') print(name, '放下面了') def eat2(name): fork_noodle_lock.acquire() print(name, '抢到叉子了') print(name,'抢到面了') print(name,'吃面') fork_noodle_lock.release() print(name, '放下面了') print(name, '放下叉子了') Thread(target=eat,args=('a',)).start() Thread(target=eat2,args=('b',)).start() Thread(target=eat,args=('c',)).start() Thread(target=eat2,args=('d',)).start() # # 1:死锁现象是怎么产生的? # 多把(互斥/递归)锁 并且在多个线程中 交叉使用 # fork_lock.acquire() # noodle_lock.acquire() # # fork_lock.release() # noodle_lock.release() # 2:如果是互斥锁,出现了死锁现象,最快速的解决方案把所有的互斥锁都改成一把递归锁,但是程序的效率会降低的 # 3:递归锁 效率低 但是解决死锁现象有奇效 # 4:互斥锁 效率高 但是多把锁容易出现死锁现象 # 5:一把互斥锁就够了
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