[zeromq-dev] Watermarks & socket types
Pavel Gushcha
pavimus at gmail.com
Tue May 25 11:11:03 CEST 2010
2010/5/25 Martin Sustrik <sustrik at 250bpm.com>:
> Pavel,
>
>> I'm began developing of my client-server apps when no xrep/xreq
>> sockets was available and ZMQ_HWM/ZMQ_LWM was present, but not
>> implemented. My clients are asyncronous (it can send multiple
>> request(s) at one time, and then wait for all or some from
>> response(s)). When i started developing, the only plausible solution
>> was:
>> 1) using ZMQ_DOWNSTREAM socket to send request to one from servers,
>> and include client ID in request.
>> 2) using ZMQ_PUB on server to send responses to topic with client ID
>> 3) using ZMQ_SUB socket, subscribed to client ID for receive responses
>>
>> As i understand, now zmq has more better solution:
>> 1) using ZMQ_XREQ socket on client for sending multiple requests
>> without waiting a reply.
>> 2) using ZMQ_XREP socket on server.
>
> If you don't mind requester working in lock-step fashion you can use
> REQ/REP. If you have to have multiple request in the air at the same
> time use XREQ/XREP.
Yes, i want to have multiple requests in the air
>
>> Currently, now my old solution is not working properly, when i use
>> watermarks. When i use low values for watermarks, some messages that
>> sent to ZMQ_PUB are dropped and not received by ZMQ_SUB. In
>> documentation i didn't found detailed information about how watermarks
>> works for different socket types (for some types messages will be
>> dropped, for other types sending thread will be blocked). Can you get
>> detailed info about this?
>
> PUB/SUB = drop the message
> otherwise = block the sender
Thanks for the info. I think, this info must be added to manpages...
>
>> Is there some workaround to prevent message dropping in my old
>> solution when using watermarks, or i must rewrite all this logic to
>> ZMQ_XREQ/ZMQ_XREP sockets?
>
> No, sorry. The behaviour in case of full queue is defined by messaging
> pattern, i.e. PUB/SUB => drop messages if queue is full.
Thanks for fast reply!
>
> Martin
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