Question regarding put_prev_task in preempted condition

Question regarding put_prev_task in preempted condition

[Lei Wen]

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?

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.
            13
           /  \
         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?

Thanks,
Lei

Re: Question regarding put_prev_task in preempted condition

[Peter Zijlstra]

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.

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.
>             13
>            /  \
>          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.

However, we link the previously unlinked entity as a leaf node. So your
example is flawed; before insertion the tree would look something like:


         13
	/  \
       11  19
      /
     3

Then the lookup in __enqueue_entity would find the place to insert 8 and
would select the right sibling of 3:

         13
	/  \
       11  19
      / 
     3  
      \
       (8)

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
readable.
 

Re: Question regarding put_prev_task in preempted condition

[Lei Wen]

Hi Peter,

On Tue, Jun 18, 2013 at 5:55 PM, Peter Zijlstra <peterz@infradead.org> 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.
>>             13
>>            /  \
>>          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! :)

Thanks,
Lei

>
> However, we link the previously unlinked entity as a leaf node. So your
> example is flawed; before insertion the tree would look something like:
>
>
>          13
>         /  \
>        11  19
>       /
>      3
>
> Then the lookup in __enqueue_entity would find the place to insert 8 and
> would select the right sibling of 3:
>
>          13
>         /  \
>        11  19
>       /
>      3
>       \
>        (8)
>
> 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
> readable.
>