python - 如何确保共享变量是线程安全的？

Question

我面临类似的问题:

DemoVar = 100 #### or whatever

def DemoMultiThreadMethod(mode):
    DemoRLock.acquire()

    DemoVar = 0 #### or random value
    ...
    Other resources which do not support multi-threaded access

    if mode == 0:
        do A with DemoVar

    elif mode == 1:
        do B with DemoVar
    ...

    DemoRLock.release()

...

def DecideAfterDemo(self):
    NewThread = threading.Thread(target = DemoMultiThreadMethod, args = (RandomMode, ))
    NewThread.start()

    NewThread.join()

    Result = DemoVar

    if Result == SpecificValue:
        Do something
    else:
        Do another
    ...

...

def SpawnThreads(self): 
    #### hundreds of DecideAfterDemo running
    Counter = 0

    while Counter < 1000:
        SpawnAThread = threading.Thread(target = DecideAfterDemo, args = ())
        SpawnAThread.Start()

... 

如何确保运行 Result = DemoVar 是安全的？

我的意思是，如果有大量锁定的 DemoMultiThreadMethod 等待处理，您并不真正知道 DemoVar = 0 #### 或随机值中的哪一个 和 Result = DemoVar 将首先运行，因为它们都是原子操作(如果我错了，请纠正我)，这意味着 Result = DemoVar 可能不安全如果池中有太多线程。我能想到的解决方案是在 DemoRLock.acquire() 之后添加 time.sleep(0.01) 以确保 Result = DemoVar 首先运行，但这会损失一些生产力。还有更好的主意吗？

Answer 1

您想要的是将一些值 (DemoVar) 从 DemoMultiThreadMethod 发送到 DecideAfterDemo。虽然您可以使用一些消息传递库，但最简单的方法是定义线程类......像这样(未经测试):

class DemoMultiThread(threading.Thread):
     def __init__(self, mode):
         self.mode = mode
         threading.Thread.__init__(self)
     def run(run):
         mode = self.mode
         DemoRLock.acquire()
         ...
         self.result = DemoVar
         DemoRLock.release()

def DecideAfterDemo(self):
     NewThread = DemoMultiThread(mode)
     NewThread.start()
     NewThread.join()
     Result = NewThread.result
     ....