递归锁
一个线程拿到锁，counter加1,该线程内又碰到加锁的情况，则counter继续加1，这期间所有其他线程都只能等待，等待该线程释放所有锁，即counter递减到0为止
from threading import Thread, RLock import time mutexB = mutexA = RLock() class Mythread(Thread): def run(self): self.f1() self.f2() def f1(self): mutexA.acquire() print(‘%s 拿到A锁‘ % self.name) mutexB.acquire() print(‘%s 拿到B锁‘ % self.name) mutexB.release() mutexA.release() def f2(self): mutexB.acquire() print(‘%s 拿到B锁‘ % self.name) time.sleep(1) mutexA.acquire() print(‘%s 拿到A锁‘ % self.name) mutexA.release() mutexB.release() if __name__ == ‘__main__‘: for i in range(3): t = Mythread() t.start()
死锁
from threading import Thread, Lock import time mutexA = Lock() mutexB = Lock() class Mythread(Thread): def run(self): self.f1() self.f2() def f1(self): mutexA.acquire() print(‘%s 拿到A锁‘ % self.name) mutexB.acquire() print(‘%s 拿到B锁‘ % self.name) mutexB.release() mutexA.release() def f2(self): mutexB.acquire() print(‘%s 拿到B锁‘ % self.name) time.sleep(1) mutexA.acquire() print(‘%s 拿到A锁‘ % self.name) mutexA.release() mutexB.release() if __name__ == ‘__main__‘: for i in range(3): t = Mythread() t.start()
原文：https://www.cnblogs.com/zhengyiqun1992/p/10409059.html