[zeromq-dev] Concurrent send/receive with a ROUTER socket - efficient and idiomatic solution

Sean Robertson sprobertson at gmail.com
Sat Aug 30 22:03:54 CEST 2014

I ran into this as well and came up with a potentially interesting
solution (though probably not idiomatic).

What I have is a single "dispatch" goroutine to keep all socket reads
and writes in one place. The dispatch interacts with two channels,
Requests (in) and Responses (out). On every loop it tries to read the
socket (with DONTWAIT), adds any incoming messages to the Requests
channel, and then looks in the Responses channel for outgoing messages
to send. No mutex necessary as the operations always happen one after

The rest of the code then only has to worry about reading and writing
these channels, which are of course thread-safe.


On Sat, Aug 30, 2014 at 8:35 AM, Eyal Arubas <eyalarubas at gmail.com> wrote:
> I am implementing a server in Go with a ROUTER socket.
> In one goroutine, messages are received from the socket, and distributed to
> workers through work queues (buffered channels).
> Workers produce some result and queue it to another buffered channel.
> In another goroutine, results are dequeued and sent back to clients.
> The same ROUTER socket needs to be accessed from those two goroutines (one
> to receive and one to send), which creates concurrency problems; as sockets
> are not threads-safe.
> As a demonstration, the following Go code panics randomly with:
>> fatal error: unexpected signal during runtime execution
> Link to demo:
> https://gist.github.com/EyalAr/117125b0e72a69584dee#file-error_demo-go
> My current solution is to use a poller with a short timeout, and a mutex to
> lock access to the socket. But this seems inefficient and not idiomatic. Why
> should I poll and lock the socket, with some arbitrary timeout, if nothing
> is there?
> Link to my current solution:
> https://gist.github.com/EyalAr/117125b0e72a69584dee#file-solution-go
> Is there a better way?
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