ZeroMQ 订阅者未通过 inproc 从发布者接收消息：传输 class

Question

我是 pyzmq 的新手。我试图理解 inproc: 传输 class 并创建了这个示例示例来玩。

看起来 Publisher 实例正在发布消息，但 Subscriber 实例是没有收到任何。

如果我将 Subscriber 个实例移动到单独的 process 并将 inproc: 更改为 tcp: 传输 class，示例有效。

代码如下：

import threading
import time

import zmq

context = zmq.Context.instance()

address = 'inproc://test' 


class Publisher(threading.Thread):
    def __init__(self):
        self.socket = context.socket(zmq.PUB)

        self.socket.bind(address)

    def run(self):
        while True:
            message = 'snapshot,current_time_%s' % str(time.time())
            print 'sending message %s' % message
            self.socket.send(message)
            time.sleep(1)


class Subscriber(object):
    def __init__(self, sub_name):
        self.name = sub_name
        self.socket = context.socket(zmq.SUB)
        self.socket.connect(address)

    def listen(self):
        while True:
            try:
                msg = self.socket.recv()
                a, b = msg.split(' ', 1)
                print 'Received message -> %s-%s-%s' % (self.name, a, b)
            except zmq.ZMQError as e:
                logger.exception(e)


if __name__ == '__main__':
    thread_a = []
    for i in range(0, 1):
        subs = Subscriber('subscriber_%s' % str(i))
        th = threading.Thread(target=subs.listen)
        thread_a.append(th)
        th.start()

    pub = Publisher()
    pub_th = threading.Thread(target=pub.run)

    pub_th.start()

Answer 1

没有错，但是

ZeroMQ 是一个很棒的工具箱。

它在引擎盖下充满了智能、明亮和自适应的服务，在许多方面确实拯救了我们可怜的生活。

仍然值得阅读并遵守文档中的一些规则。

inproc 运输 class 有一个这样的。 .bind()先，.connect()-s

之前

^{[ Page 38, Code Connected, Volume I ]}

... inproc is an inter-thread signalling transport ... it is faster than tcp or ipc. This transport has a specific limitation compared to tpc and icp: the server must issue a bind before any client issues a connect. This is something future versions of ØMQ may fix, but at present this defines how you use inproc sockets.

所以，举个例子：

if __name__ == '__main__':

    pub    = Publisher()
    pub_th = threading.Thread( target = pub.run )
    pub_th.start()
    # give it a place to start before .connect()-s may take place
    # give it a time  to start before .connect()-s may take place
    sleep(0.5)

    thread_a = []
    for i in range( 0, 1 ):
        subs = Subscriber( 'subscriber_%s' % str( i ) )
        th   = threading.Thread( target = subs.listen )
        thread_a.append( th )
        th.start()

ZeroMQ 订阅者未通过 inproc 从发布者接收消息：传输 class

ZeroMQ Subscribers not receiving message from Publisher over an inproc: transport class

distributed-computing

publish-subscribe

zeromq

low-latency

pyzmq

没有错，但是