[zeromq-dev] zactor destroy sequence

[Pieter Hintjens]

If you send and wait without a timeout, it happens that the actor
thread is gone, and its pair socket destroyed, before you get here.
Then the calling thread blocks waiting for a reply that never arrives.

The return signalling is done by the zactor wrapper, not the actor
itself. This isn't entirely symmetrical, that's true.

On Sat, Nov 1, 2014 at 10:18 PM, Arnaud Loonstra <arnaud at sphaero.org> wrote:
> Hi all,
>
> I'm going through the zactor class to understand it better. I'm slightly
> confused by the destroy sequence:
>
> At line 182 of zactor.c:
>          //  Signal the actor to end and wait for the thread exit code
>          //  If the pipe isn't connected any longer, assume child thread
>          //  has already quit due to other reasons and don't collect the
>          //  exit signal.
>          zsock_set_sndtimeo (self->pipe, 0);
>          if (zstr_send (self->pipe, "$TERM") == 0)
>              zsock_wait (self->pipe);
>          zsock_destroy (&self->pipe);
>
> So when destroy is called the actor method will receive the $TERM
> command. The destroy sequence waits for a signal from the actor command.
>
> But why the == 0. Doesn't that mean that the zstr_send send 0 bytes thus
> failed sending?
>
> Then the echo_actor example just sets: (line 268)
>              terminated = true;
> It nevers signals back.
>
> It does work that's why I'm confused about what's happening.
> Can anybody turn the light on? :)
>
> Rg,
>
> Arnaud
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