And it's not just 0MQ...it's the default behavior with TCP sockets. Take
a look at the code for a slow server emulation:
http://pastebin.com/raw.php?i=jDYFYNz3
If you compile/run it and then pipe a bunch of data at it (e.g. 'man cat
| telnet <servers-listen-port>') , you'll see that the client ends long
before the server and yet the server continues to get data on its recv
calls. A netstat will show the server's socket in a CLOSE_WAIT state
with a RECVQ that's decreasing ever so slowly, and the server itself
won't detect socket close until the RECVQ empties.
On 11/30/12 12:57 PM, Andrew Hume wrote:
chuck answered this more fully, but here is at least one good reason:
one thing i love about 0mq is that i can start sender and receiver in
any order.
and this is precisely the case where just one end is up.
but as chuck said, read that part of the guide.
On Nov 30, 2012, at 9:38 AM, Marco Trapanese wrote:
Il 30/11/2012 00:36, Michel Pelletier ha scritto:
Yes. 0mq pushes messages as far to the receiver as possible as soon as
possible. Linger only effects messages that have not yet been
transmitted from the senders queue. If your message got sent to the
router and it's in the routers queue, then the router will receive it,
regardless of the state of the sender at that point.
What is the purpose of this behavior?
I can't imagine such an application where I want to either transmit or
receive something if *both* ends aren't running.
I have to do several workarounds in my code to get all the stuff to work.
Of course, if the zmq's guys decided to do this there is good reason.
But I can't see which one.
Marco
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