[nanomsg] Re: accessing control IDs
- From: Achille Roussel <achille.roussel@xxxxxxxxx>
- To: nanomsg@xxxxxxxxxxxxx
- Date: Mon, 5 May 2014 14:32:00 -0700
You have to use AF_SP_RAW sockets to get access to these info in the control
header when receiving a message with nn_recvmsg.
On May 5, 2014, at 2:27 PM, Drew Crawford <drew@xxxxxxxxxxxxxxxxxx> wrote:
> I have a REP socket. I’m trying to identify the channel (sender or
> forwarder) on which some message has arrived to the socket. A
> transport-layer understanding of the sender is not required; any identifying
> value, such as an integer, is sufficient. Consulting the REQREP spec
> suggests that the topmost “channel ID”, one of the records in the
> “backtrace”, is the identifier I’m looking for.
>
> Clearly this identifier is not exposed over the nn_recv interface. I had
> some hopes that it would be accessible in the nn_recvmsg interface, possibly
> as control information, but it seems not to be the case:
>
>> int client = nn_socket(AF_SP,NN_REQ);
>> int server = nn_socket(AF_SP,NN_REP);
>> nn_connect(client,"inproc://test");
>> nn_bind(server,"inproc://test");
>> nn_send(client,"A",1,0);
>>
>> int rc;
>> void *body;
>> void *control;
>> struct nn_iovec iov;
>> struct nn_msghdr hdr;
>>
>> iov.iov_base = &body;
>> iov.iov_len = NN_MSG;
>> memset (&hdr, 0, sizeof (hdr));
>> hdr.msg_iov = &iov;
>> hdr.msg_iovlen = 1;
>> hdr.msg_control = &control;
>> hdr.msg_controllen = NN_MSG;
>> rc = nn_recvmsg (server, &hdr, 0);
>> print_array(body,rc,"body”); //contains only A
>>
>> printf("msg_iovlen %d\n",hdr.msg_iovlen); //1
>> printf("msg_controllen %d\n",hdr.msg_controllen); //-1
>
>
> I have consulted a previous mailing thread on this topic which suggests
> channel IDs are manipulated in rep.c. Indeed, the information I’m looking
> for seems to be moved around between nn_sockbase, nn_msg, nn_rep, and similar
> structures. However I cannot work out a sane way to get those structures
> from application code.
>
> Any suggestions on identifying the sender of a remote message?
>
> Drew
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