How can the linux CFS scheduler prevent a task with a very small vruntime from starving the processor?

Question

So after a task eats up its time slice, it would be re-inserted into the red-black tree. If the task has slept for a long time previously , leading to a very small vruntime compared to other tasks in the runqueue, then it will be repetitively re-inserted as the left-most node in the red-black tree, right? Consequently it will always be picked up as the next task to run? I have checked the source code in core.c and fair.c, I didn't see any place where this task should yield to other tasks. Though in the funciton pick_next_entity(), I do see some tasks such cfs_rq->next,cfs_rq->last or etc.. which might have higher running priority,I do not think this is the correct place to prevent a task with very small vruntime from taking a processor for a too long time,right? Does anyone have a clue? Thanks,

Answer 1

I found the answer. When task is dequeue from the runqueue, this will be called： se->vruntime -= cfs_rq->min_vruntime When task is again enqueued to the runqueue, this will be called: se->vruntime += cfs_rq->min_vruntime So actually only the offset of the vruntime will be stored when the task is sleeping and the offset will be added again when it wakes up.