Python 多线程编程

import threading
import time
g_nums = [11, 22]
g_num = 0
# 创建一个互斥锁，默认是没有上锁的
mutex = threading.Lock()


def test1():
    for i in range(5):
        print("test1------%d------" % i)
        time.sleep(1)


def test2():
    for i in range(5):
        print("test2------%d------" % i)
        time.sleep(1)


def test3(temp):
    temp.append(33)
    print(temp)


def test4():
    global g_num
    # 上锁，如果之前没有被上锁，那么此时 上锁成功
    # 如果上锁之前 已经被上锁了，那么此时会堵塞在这里，直到 这个锁被解开为止
    mutex.acquire()
    for i in range(1000000):
        g_num += 1
    # 解锁
    mutex.release()
    print(g_num)


def test5():
    global g_num
    mutex.acquire()
    for i in range(1000000):
        g_num += 1
    mutex.release()
    print(g_num)


def main():
    t1 = threading.Thread(target=test1)
    t2 = threading.Thread(target=test2)
    # args指定将来调用 函数的时候 传递什么数据过去
    t3 = threading.Thread(target=test3, args=(g_nums,))
    t4 = threading.Thread(target=test4)
    t5 = threading.Thread(target=test5)
    t1.start()
    t2.start()
    t3.start()
    t4.start()
    t5.start()
    # 可以打印当前程序有多少个线程（下面调用返回一个列表）
    print(threading.enumerate())
    time.sleep(5)
    print(g_num)


if __name__ == '__main__':
    main()

 

转载于:https://www.cnblogs.com/duxie/p/11336554.html