* Balbir Singh <[EMAIL PROTECTED]> wrote:
> I disagree. The cost is only adding a field to cfs_rq [...]
wrong. The cost is "only" of adding a field to cfs_rq and _updating it_,
in the hottest paths of the scheduler:
@@ -256,6 +257,7 @@ static void __enqueue_entity(struct cfs_
*/
if (key < entity_key(cfs_rq, entry)) {
link = &parent->rb_left;
+ rightmost = 0;
} else {
link = &parent->rb_right;
leftmost = 0;
@@ -268,6 +270,8 @@ static void __enqueue_entity(struct cfs_
*/
if (leftmost)
cfs_rq->rb_leftmost = &se->run_node;
+ if (rightmost)
+ cfs_rq->rb_rightmost = &se->run_node;
> [...] For a large number of tasks - say 10000, we need to walk 14
> levels before we reach the node (each time). [...]
10,000 yield-ing tasks is not a common workload we care about. It's not
even a rare workload we care about. _Especially_ we dont care about it
if it slows down every other workload (a tiny bit).
> [...] Doesn't matter if the data is cached, we are still spending CPU
> time looking through pointers and walking to the right node. [...]
have you actually measured how much it takes to walk the tree that deep
on recent hardware? I have.
Ingo
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