Re: Question regarding put_prev_task in preempted condition
From: Lei Wen
Date: Wed Jun 19 2013 - 01:14:36 EST
On Tue, Jun 18, 2013 at 5:55 PM, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
> On Sun, Jun 09, 2013 at 11:59:36PM +0800, Lei Wen wrote:
>> Hi Peter,
>> While I am checking the preempt related code, I find a interesting part.
>> That is when preempt_schedule is called, for its preempt_count be added
>> PREEMPT_ACTIVE, so in __schedule() it could not be dequeued from rq
>> by deactivate_task.
>> Thus in put_prev_task, which is called a little later in __schedule(), it
>> would call put_prev_task_fair, which finally calls put_prev_entity.
>> For current task is not dequeued from rq, so in this function, it would
>> enqueue it again to the rq by __enqueue_entity.
>> Is there any reason to do like this, since entity already is over rq,
>> why need to queue it again?
> Because we keep the current running task outside of the actual queue
> structure. This is because every time we update the runtime
> (__update_curr) the key on which the tree is sorted (vruntime) is
> changed and we'd need to dequeue + enqueue to keep the tree in sync.
I see... I didn't notice for this difference...
> By not having the actively running task in the tree we can avoid this;
> at the cost of having to dequeue on switching to the task and enqueue
> when switching from the task.
>> And if current rq's vruntime distribution like below, and vruntime with 8
>> is the task that would be get preempted. So in __enqueue_entity,
>> its rb_left/rb_right would be set as NULL and reinserted into this RB tree.
>> Then seems to me now, the entity with vruntime of 3 would be disappeared
>> from the RB tree.
>> / \
>> 8 19
>> / \
>> 3 11
>> I am not sure whether I understand the whole process correctly...
>> Would the example as above happen in our real life?
> No, the RB tree code will ensure we'll not loose 3. I suppose you're
> confused by rb_link_node() which does indeed clear the left and right
> node of the entity we're about to link.
Yep, since 8 is not over rq, NULL its two child would lose any info.
Thanks for detailed explanation! :)
> However, we link the previously unlinked entity as a leaf node. So your
> example is flawed; before insertion the tree would look something like:
> / \
> 11 19
> Then the lookup in __enqueue_entity would find the place to insert 8 and
> would select the right sibling of 3:
> / \
> 11 19
> We'd then link 8 as a child leaf of 3; which will indeed have NULL
> leafs. rb_insert_color() will then fix up the tree so we conform to the
> RB constraints. Please read lib/rbtree.c:__rb_insert() the code is quite
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