[OpenWrt-Devel] Question about conflict between ubus_invoke and uloop_run

[Alexandru Ardelean]

Yes, you have to use ubus_invoke_sync() in this case.

You can use ubus_invoke_async() and re-design your code using a
state-machine mechanism.
This would simplify the need to add any locks & syncs.

You should not use directly:
ctx->ubus->sock.registered = true;

It could complicate things long-term.
To conclude : I think the best idea is to use a state-machine .
Whenever your callback function is called you call into the state machine.
Depending on the current state, you can execute actions, including changing
states if needed.
I'd say that if you do this, you can remove that separate thread.


On Fri, Apr 3, 2015 at 9:41 AM, XiaoFengMeng <xiaofeng.meng at pfsw.com> wrote:

>  HI Alexandru
>
>
>
> Thanks so much for your reply!
>
> In my case, the ubus_invoke will be called in a callback function which is
> in a separate thread, and uloop_run will be running in  main thread.
>
> If I have to use ubus_invoke_async, then I feel that I have to use a lock
> to sync the call.
>
>
>
> And I think you confirmed with an idea of mine:
>
> if I want to use “ubus_invoke”,  “uloop_run” can’t be used at the same
> time,
>
> or if I use uloop_run, I have to use ubus_invoke_async instread.
>
>
>
> Is that right?
>
>
>
> But I just got a tricky way of using ubus_invoke and uloop_run at the same
> time.
>
> For the ubus context which is only used to call ubus_invoke as client, I
> set the registered of usock to true,
>
> ctx->ubus->sock.registered = true;
>
>
>
> actually this socket is not added to uloop in reality, I do this so that
> following code in ubus_complete_request function will not be executed.
>
> And thus no conflict with uloop.
>
> if (!registered) {
>
>                 uloop_init();
>
>                 ubus_add_uloop(ctx);
>
> }
>
>
>
> I experimented on this, it seems working well,
>
> I like to know what’s your opinion on this?
>
>
>
> Thanks!
>
> /Kevin
>
> *From:* Alexandru Ardelean [mailto:ardeleanalex at gmail.com]
> *Sent:* 2015年4月3日 14:20
> *To:* XiaoFengMeng
> *Cc:* openwrt-devel at lists.openwrt.org
> *Subject:* Re: [OpenWrt-Devel] Question about conflict between
> ubus_invoke and uloop_run
>
>
>
> Hello Kevin,
>
> As far as my understanding goes regarding ubus and uloop, they were not
> designed with threading in mind.
>
> Also, as a general rule when using uloop, you wouldn't use threads,
> because races could occur (especially when doing ubus calls inside a
> thread) and the main uloop loop would terminate.
>
>
> One general rule would be: you should not use ubus call in threads.
>
> You could use ubus_invoke_async() instead of ubus_invoke().
>
> It is a bit more work than just using ubus_invoke() but if you want
> thread-like behaviour, it could replace all the threads you're using.
>
>
> So, is there a way to re-design your code without using threads and using
> ubus async calls ?
>
>
>
>
>
> On Fri, Apr 3, 2015 at 5:24 AM, XiaoFengMeng <xiaofeng.meng at pfsw.com>
> wrote:
>
>  HI !
>
> I am Kevin and learning the ubus code and got something that is confusing
> me very much in the code.
>
>
>
> My question is regarding code in  ubus_invoke -> ubus_complete_request,
>
>
>
> I can see that there are lots of uloop related code in
>  “ubus_complete_request”
>
> if (!registered) {
>
>                                 uloop_init();
>
>                                 ubus_add_uloop(ctx);
>
> }
>
> ….
>
> while (!req->status_msg) {
>
>                                 bool cancelled = uloop_cancelled;
>
>
>
>                                 uloop_cancelled = false;
>
>                                 if (req_timeout) {
>
>                                                 timeout = time_end -
> get_time_msec();
>
>                                                 if (timeout <= 0) {
>
>
> ubus_set_req_status(req, UBUS_STATUS_TIMEOUT);
>
>                                                                 break;
>
>                                                 }
>
>                                 }
>
>                                 ubus_poll_data(ctx, (unsigned int)
> timeout);
>
>
>
>                                 uloop_cancelled = cancelled;
>
>                 }
>
>
>
> My question is why is uloop involved here?  uloop could be working in
> another thread and also listen on the same socket.
>
> I idea that, before the ubus_invoke sends the request messge, it should
> disable the uloop for the current socket, so that code in uloop_run will
> not receive anything from the same socket.
>
>
>
> Because after calling ubus_add_uloop(ctx), the uloop_run could also be
> receiving the response data
>
> And this function could cause the uloop_run to quit the while loop.
>
> How should I understand the logic here?
>
>
>
> And How should I do if I want to call ubus_invoke as a client and use
> uloop_run as a server at  the same process?
>
>
>
> Thanks very much!
>
>
>
> Best regards
>
> Kevin
>
>
>
>
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>
>
>
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