5.1.17 死锁与死锁的解决(递归锁)

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 拿到了A锁‘ %self.name)

        mutexB.acquire()
        print(‘%s 拿到了B锁‘ %self.name)
        mutexB.release()

        mutexA.release()


    def f2(self):
        mutexB.acquire()
        print(‘%s 拿到了B锁‘ % self.name)
        time.sleep(0.1)

        mutexA.acquire()
        print(‘%s 拿到了A锁‘ % self.name)
        mutexA.release()

        mutexB.release()

if __name__ == ‘__main__‘:
    for i in range(10):
        t=MyThread()
        t.start()

# 互斥锁只能acquire一次
from threading import Thread
from threading import Lock

mutexA=Lock()

mutexA.acquire()
print(1)
mutexA.acquire()
print(2)
mutexA.release()

运行结果：
1     #输出1后卡在第二个mutexA.acquire()上

# 递归锁可以acquire多次
from threading import Thread,Lock
from threading import RLock

mutexA=RLock()

mutexA.acquire()
mutexA.acquire()
mutexA.release()

正常运行！

# 递归锁:可以连续acquire多次，每acquire一次计数器+1，只有计数为0时，才能被抢到acquire
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 拿到了A锁‘ %self.name)

        mutexB.acquire()
        print(‘%s 拿到了B锁‘ %self.name)
        mutexB.release()

        mutexA.release()


    def f2(self):
        mutexB.acquire()
        print(‘%s 拿到了B锁‘ % self.name)
        time.sleep(7)

        mutexA.acquire()
        print(‘%s 拿到了A锁‘ % self.name)
        mutexA.release()

        mutexB.release()

if __name__ == ‘__main__‘:
    for i in range(10):
        t=MyThread()
        t.start()