Question

根据ZeroMQ文档，一旦排队的消息数量达到最高水位，发布套接字应该丢弃消息。

在以下示例中，这似乎不起作用（是的，我确实在绑定/连接之前设置了hwm）：

import time
import pickle
from threading import Thread
import zmq

ctx = zmq.Context()

def pub_thread():
    pub = ctx.socket(zmq.PUB)
    pub.set_hwm(2)
    pub.bind('tcp://*:5555')

    i = 0
    while True:
        # Send message every 100ms
        time.sleep(0.1)
        pub.send_string("test", zmq.SNDMORE)
        pub.send_pyobj(i)
        i += 1

def sub_thread():
    sub = ctx.socket(zmq.SUB)
    sub.subscribe("test")
    sub.connect('tcp://localhost:5555')
    while True:
        # Receive messages only every second
        time.sleep(1)
        msg = sub.recv_multipart()
        print("Sub: %d" % pickle.loads(msg[1]))

t_pub = Thread(target=pub_thread)
t_sub = Thread(target=sub_thread)
t_pub.start()
t_sub.start()

while True:
    pass

我在pub上发送消息的速度比在子套接字上读取消息快10倍，hwm设置为2。我希望只能收到大约每10条消息。相反，我看到以下输出：

Sub: 0
Sub: 1
Sub: 2
Sub: 3
Sub: 4
Sub: 5
Sub: 6
Sub: 7
Sub: 8
Sub: 9
Sub: 10
Sub: 11
Sub: 12
Sub: 13
Sub: 14
...

所以我看到所有消息都到达了，因此它们被保留在队列中，直到我阅读它们为止。在连接之前在子插座上添加hwm = 2时也是如此。

我在做什么错或者我误解了hwm？

我使用的是 pyzmq版本17.1.2

Answer 1

借用了issue which I opened in Github的答案，我将答案更新如下：

消息保存在操作系统的网络缓冲区中。我找到因此，HWM没那么有用。这是修改后的代码订户错过消息的地方：

import time
import pickle
import zmq
from threading import Thread
import os

ctx = zmq.Context()

def pub_thread():
    pub = ctx.socket(zmq.PUB)
    pub.setsockopt(zmq.SNDHWM, 2)
    pub.setsockopt(zmq.SNDBUF, 2*1024)  # See: http://api.zeromq.org/4-2:zmq-setsockopt
    pub.bind('tcp://*:5555')
    i = 0
    while True:
        time.sleep(0.001)
        pub.send_string(str(i), zmq.SNDMORE)
        pub.send(os.urandom(1024))
        i += 1

def sub_thread():
    sub = ctx.socket(zmq.SUB)
    sub.setsockopt(zmq.SUBSCRIBE, b'')
    sub.setsockopt(zmq.RCVHWM, 2)
    sub.setsockopt(zmq.RCVBUF, 2*1024)
    sub.connect('tcp://localhost:5555')
    while True:
        time.sleep(0.1)
        msg, _ = sub.recv_multipart()
        print("Received:", msg.decode())

t_pub = Thread(target=pub_thread)
t_pub.start()
sub_thread()

输出看起来像这样：

Received: 0
Received: 1
Received: 2
Received: 3
Received: 4
Received: 5
Received: 6
Received: 47
Received: 48
Received: 64
Received: 65
Received: 84
Received: 85
Received: 159
Received: 160
Received: 270

由于所有队列/缓冲区已满并且发布者未收到消息开始丢弃消息（请参阅ZMQ_PUB的文档： http://api.zeromq.org/4-2:zmq-socket。

[注意]：

您应该在侦听器/订阅者和广告商/发布者中使用高水印选项。
这些帖子也相关（Post1-Post2）
sock.setsockopt(zmq.CONFLATE, 1)是仅获取在订户方定义的最后一条消息的另一种选择。

Pyzmq高水位标记在酒吧插座上不起作用

1 个答案: