如何将异常从一个进程传递到另一个进程？

Question

如果停止函数中的运行状态为 'stopped'，我想在上传函数中引发异常。这似乎不起作用。我正在使用 Pipe 传递异常。怎么了？

def upload(instances, u1):
    for instance in instance:
        try:
            u1.recv()
            #do_something
        except:
            #do_something_else
            continue         

def stop(instances, s1):
    for instance in instances:
        RunningStatus = instance[4]
        if RunningStatus.lower() == 'stopped'.lower():
            s1.send(Exception) # I want to raise exception in upload function 
                               # from here

if __name__ == '__main__':
    s1, u1 = multiprocessing.Pipe()
    s = multiprocessing.Process(target = stop, args = (instances, s1,))
    u = multiprocessing.Process(target = upload, args = (instances, u1))
    s.start()
    u.start()
    u.join()

Answer 1

下面是一个示例，说明如何将异常对象从一个进程发送到另一个进程。我还尝试发送完整的异常信息（由 sys.exc_info 返回），但是，很合理，这失败了。人们总是可以将回溯信息格式化为一个字符串并发送它（参见回溯模块）。

在 Ubuntu 14.04 Python 2.7、3.4（Ubuntu 提供）和 3.5（Continuum）上测试。

from __future__ import print_function

import sys
import multiprocessing
import time

def upload(u1):
    i=0
    try:
        while True:
            print('>upload>',i)
            i+=1
            if u1.poll():
                # tvt = u1.recv()
                # raise tvt[0], tvt[1], tvt[2] # Python 2.7
                e = u1.recv()
                raise e
            time.sleep(0.1)
    except Exception as e:
        print('Exception caught:',e)
        print('exiting')

def stop(s1):
    try:
        while True:
            for j in range(100,110):
                time.sleep(0.1)
                if 105==j:
                    raise RuntimeError("oh dear at j={}".format(j))
    except Exception as e:
        # tvt = sys.exc_info()
        # s1.send(tvt) # doesn't work; tracebacks are not pickle'able
        s1.send(e)

if __name__ == '__main__':
    s1, u1 = multiprocessing.Pipe()
    s = multiprocessing.Process(target = stop, args = (s1,))
    u = multiprocessing.Process(target = upload, args = (u1,))
    s.start()
    u.start()
    u.join()

输出：

>upload> 0
>upload> 1
>upload> 2
>upload> 3
>upload> 4
>upload> 5
>upload> 6
Exception caught: oh dear at j=105
exiting

Answer 2

您可能想研究使用事件对象而不是管道和异常组合。通过使用事件原语在您的进程之间共享数据，您可以让上传函数监视事件并在停止函数发现问题时触发您的逻辑。

根据您的示例，我假设 instances 中的每个实例对象都是一个数组，因此通过使用附加元素扩展该数组，您可以为每个实例创建一个唯一的事件。

def upload(instances, u1):
    for instance in instance:
        if instance[5].is_set(): # events only return TRUE if they are set
            #Exception logic would go here

    else: 
        u1.recv()
        #do_something


def stop(instances, s1):
    for instance in instances:
        RunningStatus = instance[4]
        if RunningStatus.lower() == 'stopped'.lower():
            instance[5].set() # Set the event to TRUE 

if __name__ == '__main__':
    for instance in instances:
        instance[5] = multiprocessing.Event() # create a new event semaphore 
    s = multiprocessing.Process(target = stop, args = (instances, s1,))
    u = multiprocessing.Process(target = upload, args = (instances, u1))
    s.start()
    u.start()
    u.join()

可以在此处找到多处理原语的完整列表：https://docs.python.org/2/library/multiprocessing.html#synchronization-primitives

以下是其他多处理对象的原语各种用途的一些很好的例子：https://pymotw.com/2/multiprocessing/communication.html

如何将异常从一个进程传递到另一个进程？

How to pass exception from one process to another?

python

python-multiprocessing