线程死锁问题以及递归锁解法

from threading import Thread, Lock, RLock
import time

mutexA = Lock()
muteXB = Lock()

'''
# 将上述的
mutexA = Lock()
mutexB = Lock()
# 换成
mutexA = mutexB = RLock() # 递归锁 可解决本例的死锁问题
'''

class MyThead(Thread):
    def run(self) -> None:
        self.func1()
        self.func2()

    def func1(self):
        mutexA.acquire()
        print('%s 抢到A锁'%self.name) # 获取当前线程名
        muteXB.acquire()
        print('%s 抢到B锁'%self.name)
        muteXB.release()
        mutexA.release()

    def func2(self):
        muteXB.acquire()
        print('%s 抢到B锁' % self.name)
        time.sleep(2)
        mutexA.acquire()
        print('%s 抢到A锁' % self.name)
        mutexA.release()
        muteXB.release()

'''
Thread-1 抢到A锁
Thread-1 抢到B锁
Thread-1 抢到B锁
Thread-2 抢到A锁
线程1接下来要抢A锁，但A锁在线程2那里
线程2接下来要抢B锁，但B锁在线程1那里
形成死锁
'''

if __name__ == '__main__':
    for i in range(10):
        t = MyThead()
        t.start()