enqueue_task_rt function in ./kernel/sched/rt.c is responsible for queuing the task to the run queue. enqueue_task_rt contains call to enqueue_rt_entity which calls dequeue_rt_stack. Most part of the code seems logical but I am a bit lost because of the function dequeue_rt_stack unable to understand what it does. Can somebody tell what is the logic that I am missing or suggest some good read.
Edit: The following is the code for dequeue_rt_stack function
struct sched_rt_entity *back = NULL;
/* macro for_each_sched_rt_entity defined as
for(; rt_se; rt_se = rt_se->parent)*/
for_each_sched_rt_entity(rt_se) {
rt_se->back = back;
back = rt_se;
}
for (rt_se = back; rt_se; rt_se = rt_se->back) {
if (on_rt_rq(rt_se))
__dequeue_rt_entity(rt_se);
}
More specifically, I do not understand why there is a need for this code:
for_each_sched_rt_entity(rt_se) {
rt_se->back = back;
back = rt_se;
}
What is its relevance.
When a task is to be added to some queue, it must first be removed from the queue that it currently is on, if any.
With the group scheduler, a task is always at the lowest level of the tree, and might have multiple ancestors:
To remove the task from the tree, it must be removed from all groups’ queues, and this must be done first at the top-level group (otherwise, the scheduler might try to run an already partially-removed task). Therefore,
dequeue_rt_stackuses thebackpointers to constructs a list in the opposite direction:That
backlist can then be used to walk down the tree to remove the entities in the correct order.