Queuing in the timeout_q of timeouts expiring on the same tick queue
them in reverse order: as soon as the new timeout finds a timeout
expiring on the same tick or later, it get prepended to that timeout:
this allows exiting the traversal of the timeout as soon as possible,
which is done with interrupts locked, thus reducing interrupt latency.
However, this has the side-effect of handling the timeouts expiring on
the same tick in the reverse order that they are queued.
For example:
thread_c, prio 4:
uint32_t uptime = k_uptime_get_32();
while(uptime == k_uptime_get_32()); /* align on tick */
k_timer_start(&timer_a, 5, 0);
k_timer_start(&timer_b, 5, 0);
thread_a, prio 5:
k_timer_status_sync(&timer_a);
printk("thread_a got timer_a\n");
thread_b, prio 5:
k_timer_status_sync(&timer_b);
printk("thread_b got timer_b\n");
One could "reasonably" expect thread_a to run first, since both threads
have the same prio, and timer_a was started before timer_b, thus
inserted first in the timeout_q first (time-wise). However, thread_b
will run before thread_a, since timer_b's timeout is prepended to
timer_a's.
This patch keeps the reversing of the order when adding timeouts in the
timeout_q, thus preserving the same interrupt latency; however, when
dequeuing them and adding them to the expired queue, we now reverse that
order _again_, causing the timeouts to be handled in the expected order.
Change-Id: Id83045f63e2be88809d6089b8ae62034e4e3facb
Signed-off-by: Benjamin Walsh <walsh.benj@gmail.com>
6f4bc80901