import threading, time
a = 5
alock = threading.Lock()
b = 5
block = threading.Lock()
def thread1calc():
print "Thread1 acquiring lock a"
alock.acquire()
time.sleep(5)
print "Thread1 acquiring lock b"
block.acquire()
time.sleep(5)
a += 5
b += 5
print "Thread1 releasing both locks"
block.release()
alock.release()
def thread2calc():
print "Thread2 acquiring lock b"
block.acquire()
time.sleep(5)
print "Thread2 acquiring lock a"
alock.acquire()
time.sleep(5)
a += 10
b += 10
print "Thread2 releasing both locks"
block.release()
alock.release()
t = threading.Thread(target = thread1calc)
#t.setDaemon(1)
t.start()
t1 = threading.Thread(target = thread2calc)
#t1.setDaemon(2)
t1.start()
t.join()
t1.join()
print a
print b
上述程序的输出如下:
python thread_deadlock.py
Thread1 acquiring lock a
Thread2 acquiring lock b
Thread2 acquiring lock a Thread1 acquiring lock b
我试图模拟一个死锁场景,并对此行为感到困惑。在线程1获得“ a”锁之后,如果不释放线程2怎么能获得已经获得的相同锁?