Hi Jianwen,
thread_msg does a context switch between p3 and p1 because p1 does not have
a message queue.
The second message from p3 causes p3 to go into a blocked state and give up
control otherwise the previous message will be lost.
if in p1 a msg_queue is defined:
void *thread1(void *arg)
{
(void) arg;
msg_t msg_queue[8];
msg_init_queue(msg_queue, 8);
puts("THREAD 1 start\n");
...
then thread_msg example will run as documented until the message queue is
full.
I hope this little explanation may help ...
ciao
Attilio
On Thu, Mar 31, 2016 at 10:35 AM, OuyangJianwen <jianwen.ouy...@outlook.com>
wrote:
> Hello,
>
> I am a new beginner to learn RIOT. I got some question when I run the
> "thread_msg" on board "SAMR21-xpro".
>
> For the thread1, thread2 and thread3, they have the same priority. so when
> the thread3 is running, how can scheduler switch to the thread1.
>
> I saw these in the API:
> "In case of equal priorities, the threads are scheduled in a
> semi-cooperative fashion. That means that unless an interrupt happens,
> threads with the same priority will only switch due to voluntary or
> implicit context switches."
> and
> "Some functions that unblock another thread, e.g. msg_send()
> <http://riot-os.org/api/group__core__msg.html#gac5347725c3d203ac72604c0ab8b7d6d8>
> or mutex_unlock()
> <http://riot-os.org/api/group__core__sync.html#ga05ccabe849d63b032f6317323da60187>,
> can cause a thread switch, if the target had a higher priority."
>
> but here, all the three threads have the same priority, so according the
> rule, even there is msg_send()
> <http://riot-os.org/api/group__core__msg.html#gac5347725c3d203ac72604c0ab8b7d6d8>,
> the thread3 should not be blocked and run forever, but it doesn't.
>
> Thanks for your time!
>
> Regards,
> Jianwen OUYANG
>
>
>
> _______________________________________________
> devel mailing list
> devel@riot-os.org
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>
>
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