Python 线程（二）：简单锁实现线程同步

Python中有两种锁，一个锁是原始的锁（原语）， 不可重入，而另一种锁则是可重入的锁即递归锁。而是thread模块中，只提供了不可重入的锁，而在threading中则提供这两种锁。

可重入：当一个线程拥有一个锁的使用权后，再次获取锁的使用权时，不会阻塞，会立马得到使用权，则原始锁的话，则不行，会阻塞。

方法一：thead的不可重入锁

　　

import thread
import time

lock = thread.allocate_lock()

def Count(id):
    global num;

    while True:
        lock.acquire()
        if num <= 10:
            print "Thread id is : %s     The num is %s\n" % (id, str(num))
            num = num + 1
        else:
            break
        lock.release()
    else:
        thread.exit_thread()

if __name__ == "__main__":
    num = 1
    thread.start_new_thread(Count, ('A',))
    thread.start_new_thread(Count, ('B',))

    time.sleep(5)

　　

方法二：theading的Lock(不可重入锁)

import threading
import time

lock = threading.Lock()

def Count(id):
    global num;

    while True:
        lock.acquire()
        if num <= 10:

            print "Thread id is : %s     The num is %s\n" % (id, str(num))
            num = num + 1
        else:
            break
        lock.release()

if __name__ == "__main__":
    num = 1
    t1 = threading.Thread(target=Count, args=('A', ))
    t2 = threading.Thread(target=Count, args=('B', ))

    t1.start()
    t2.start()

    time.sleep(5)

　　方法三：threading的RLock(可重入)

import threading
import time

lock = threading.RLock()

def CountNum(id):
    global num
    
    lock.acquire()
    
    if num <= 10:
        print "Thread id is : %s     The num is %s\n" % (id, str(num))
        num = num + 1
        CountNum(id)

    lock.release()

if __name__ == "__main__":
    num = 1
    t1 = threading.Thread(target=CountNum, args=('A'))

    t1.start()

    time.sleep(5)