RLock（递归锁）

import threading, time
def run1():
    print("grab the first part data")
    lock.acquire()#进入大门后的一间屋子
    global num
    num += 1
    lock.release()#退出这间屋子
    return num

def run2():
    print("grab the second part data")
    lock.acquire()#进入大门后的另一间屋子，和上个屋子平行
    global num2
    num2 += 1
    lock.release()#退出这间屋子
    return num2

def run3():
    lock.acquire()#第一道门，大门
    res = run1()
    print('--------between run1 and run2-----')
    res2 = run2()
    lock.release()
    print(res, res2)

if __name__ == '__main__':

    num, num2 = 0, 0
    lock = threading.Lock()
    for i in range(10):
        t = threading.Thread(target=run3)
        t.start()
#一共11个线程
while threading.active_count() != 1:
    print(threading.active_count())
else:#只剩一个线程了，表示所有子线程执行完毕
    print('----all threads done---')
    print(num, num2)

　　结果：

11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11
11

一直执行，不停的打印11......，锁死了，怎么办？其实是各个门之间的钥匙混了，所以打不开锁了，锁死了，需要以下操作才能解决。

RLock（递归锁）

说白了就是在一个大锁中还要再包含子锁

import threading, time
def run1():
    print("grab the first part data")
    lock.acquire()
    global num
    num += 1
    lock.release()
    return num

def run2():
    print("grab the second part data")
    lock.acquire()
    global num2
    num2 += 1
    lock.release()
    return num2

def run3():
    lock.acquire()
    res = run1()
    print('--------between run1 and run2-----')
    res2 = run2()
    lock.release()
    print(res, res2)

if __name__ == '__main__':

    num, num2 = 0, 0
    lock = threading.RLock()#改为递归锁
    for i in range(10):
        t = threading.Thread(target=run3)
        t.start()

while threading.active_count() != 1:
    print(threading.active_count())
else:
    print('----all threads done---')
    print(num, num2)

　　结果：

D:\Anaconda3\python.exe C:/Users/Administrator/Desktop/py_work/lib/aa.py
grab the first part data
--------between run1 and run2-----
grab the second part data
1 1
grab the first part data
--------between run1 and run2-----
grab the second part data
2 2
grab the first part data
--------between run1 and run2-----
grab the second part data
3 3
grab the first part data
--------between run1 and run2-----
grab the second part data
4 4
grab the first part data
--------between run1 and run2-----
grab the second part data
5 5
grab the first part data
--------between run1 and run2-----
grab the second part data
6 6
grab the first part data
--------between run1 and run2-----
grab the second part data
7 7
grab the first part data
--------between run1 and run2-----
grab the second part data
8 8
grab the first part data
--------between run1 and run2-----
grab the second part data
9 9
grab the first part data
--------between run1 and run2-----
grab the second part data
10 10
----all threads done---
10 10

Process finished with exit code 0

概念总结：

死锁： 是指两个或两个以上的进程或线程在执行过程中，因争夺资源而造成的一种互相等待的现象，若无外力作用，它们都将无法推进下去。此时称系统处于死锁状态或系统产生了死锁，这些永远在互相等待的进程称为死锁进程

那么怎么解决死锁现象呢？

解决方法，递归锁：在Python中为了支持在同一线程中多次请求同一资源，python提供了可重入锁RLock。这个RLock内部维护着一个Lock和一个counter变量，counter记录了acquire的次数，从而使得资源可以被多次require。直到一个线程所有的acquire都被release，其他的线程才能获得资源。上面的例子如果使用RLock代替Lock，则不会发生死锁。

threading.Lock和threading.RLock区别：RLock允许在同一线程中被多次acquire。而Lock却不允许这种情况。注意：如果使用RLock，那么acquire和release必须成对出现，即调用了n次acquire，必须调用n次的release才能真正释放所占用的琐。