[zeromq-dev] What prevents reading a message self sent ?

[Goswin von Brederlow]

On Tue, Jan 21, 2014 at 11:09:54AM +0100, Laurent Alebarde wrote:
> Hi Devs,
> 
> This is my thuesday morning stupid question: say I have a socket on
> which I send a message, and just after I put a receive on this same
> socket in the code. What prevents in libzmq to receive the message I
> have just sent ?
> 
> Let's raise the question differently, what prevents zmq_proxy to
> loop indefinitely since what it receives on socket A, it sends it on
> socket B, and vice versa ?
> 
> socket A                                        socket B
>    | ______________________\______________________ |
>    |/                      /                      \|
>    |                                               |
>    |\______________________/______________________/|
>    |                       \                       |
>     zmq_proxy(A,B)does not loop on itself like this
> 
> Why does it manage to work like this:
> 
> socket A                                        socket B
>    |                                               |
> -->|---------------------------------------------->|------------>
>    |                                               |
>    |                                               |
> <--|<----------------------------------------------|<------------
> 
> zmq_proxy(A,B)does what it is expected to do, yeah !

I imagine it is more like:

socket A           socket B
   |                  |
-->| recv() -> send() |------------>
   |                  |
   |                  |      
<--| send() <- recv() |<------------

Each socket has an incoming and outgoing queue. Recv() looks in the
incoming queue and send() puts messages in the ougoing queue and never
the two shall meet.

> Can I chain: zmq_proxy(A,B); zmq_proxy(B,C), zmq_proxy(C,D),
> assuming I start them in different threads ?

I don't see why not.
 
> Cheers,
> 
> 
> Laurent

MfG
	Goswin