多线程访问共享数据不安全问题分析

例：假设两个线程t1和t2都要对num=0进行增1运算，t1和t2都各对num修改10次，num的最终的结果应该为20。
但是由于是多线程访问，有可能出现下面情况：
在num=0时，t1取得num=0。此时系统把t1调度为”sleeping”状态，把t2转换为”running”状态，t2也获得num=0。
然后t2对得到的值进行加1并赋给num，使得num=1。然后系统又把t2调度为”sleeping”，把t1转为”running”。
线程t1又把它之前得到的0加1后赋值给num。这样，明明t1和t2都完成了1次加1工作，但结果仍然是num=1

1.g_num的值大部分是在一百万到两百万之间的情况

from threading import Thread

g_num = 0
def test1():
    global g_num
    for i in range(1000000):
        # g_num += 1
        b = g_num + 1
        g_num = b

    print("---test1---g_num=%d"%g_num)

def test2():
    global g_num
    for i in range(1000000):
        a =  g_num + 1
        g_num = a

    print("---test2---g_num=%d"%g_num)


if __name__ == '__main__':
    p1 = Thread(target=test1)
    p1.start()

    p2 = Thread(target=test2)
    p2.start()
#--------------------运行结果------------------------
---test1---g_num=1283303
---test2---g_num=1257613

Process finished with exit code 0

以上结果多个线程对同一资源的访问，对数据造成破坏，使得线程运行的结果不可预期。这种现象称为“线程不安全”

2.g_num的值小于一百万的情况

from threading import Thread
import time

g_num = 0
def test1():
    global g_num
    for i in range(1000000):
        # g_num += 1
        b = g_num + 1
        time.sleep(0.1)   #此处添加阻塞状态，是为了让b结果没有赋值给g_num就被切换到另一个函数
        g_num = b

    print("---test1---g_num=%d"%g_num)

def test2():
    global g_num
    for i in range(1000000):
        a =  g_num + 1
        g_num = a

    print("---test2---g_num=%d"%g_num)


if __name__ == '__main__':
    p1 = Thread(target=test1)
    p1.start()

    p2 = Thread(target=test2)
    p2.start()
#--------------------运行结果------------------------
---test2---g_num=27712
注：此结果没有出现 Process finished这句话的原因是sleep 0.1 秒 还要运行几十万次以上，
时间太长，所以程序 test1 其实一直在运行，只能强制停止程序。

跟第一个例子一样，线程不安全

3.正常情况g_num

from threading import Thread
import time

g_num = 0

def test1():
    global g_num
    for i in range(1000000):
        # g_num += 1
        b = g_num + 1
        g_num = b

    print("---test1---g_num=%d"%g_num)

def test2():
    global g_num
    for i in range(1000000):
        a =  g_num + 1
        g_num = a

    print("---test2---g_num=%d"%g_num)

if __name__ == '__main__':
    p1 = Thread(target=test1)
    p1.start()
    time.sleep(1)   #此处阻塞1秒，给test1充分运行时间，直至该线程运行结束
    p2 = Thread(target=test2)
    p2.start()
#--------------------运行结果------------------------
---test1---g_num=1000000
---test2---g_num=2000000

Process finished with exit code 0

本应该输出的g_num正确值

4.解决办法

from threading import Thread,Lock

g_num = 0
my_lock = Lock()

def test1():
    global g_num
    if my_lock.acquire():              #同步锁防止争夺资源
        for i in range(1000000):
            g_num  = g_num+1
        my_lock.release()              #解锁

    print("---test1---g_num=%d"%g_num)

def test2():
    global g_num
    if my_lock.acquire():
        for i in range(1000000):
            g_num  = g_num+1
        my_lock.release()

    print("---test2---g_num=%d"%g_num)


if __name__ == '__main__':
    p1 = Thread(target=test1)
    p1.start()

    p2 = Thread(target=test2)
    p2.start()
#--------------------运行结果------------------------
---test1---g_num=1000000
---test2---g_num=2000000