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Ensure things are dequeued before > freeing. > > NOTE: switched_from_fair() causes spurious wakeups due to clearing > sched_delayed after enqueueing a task in another class that should've > been dequeued. This *should* be harmless. > It might bring some expected behavior in some corner cases reported here: https://lore.kernel.org/lkml/202408161619.9ed8b83e-lkp@intel.com/ As the block task might return from schedule() with TASK_INTERRUPTIBLE. We cooked a patch to workaround it(as below). thanks, Chenyu >From 9251b25073d43aeac04a6ee69b590fbfa1b8e1a5 Mon Sep 17 00:00:00 2001 From: Chen Yu Date: Mon, 26 Aug 2024 22:16:38 +0800 Subject: [PATCH] sched/eevdf: Dequeue the delayed task when changing its schedule policy [Problem Statement] The following warning was reported: do not call blocking ops when !TASK_RUNNING; state=1 set at kthread_worker_fn (kernel/kthread.c:?) WARNING: CPU: 1 PID: 674 at kernel/sched/core.c:8469 __might_sleep handle_bug exc_invalid_op asm_exc_invalid_op __might_sleep __might_sleep kthread_worker_fn kthread_worker_fn kthread __cfi_kthread_worker_fn ret_from_fork __cfi_kthread ret_from_fork_asm [Symptom] kthread_worker_fn() ... repeat: set_current_state(TASK_INTERRUPTIBLE); ... if (work) { // false __set_current_state(TASK_RUNNING); ... } else if (!freezing(current)) { schedule(); // after schedule, the state is still *TASK_INTERRUPTIBLE* } try_to_freeze() might_sleep() <--- trigger the warning [Analysis] The question is after schedule(), the state remains TASK_INTERRUPTIBLE rather than TASK_RUNNING. The short answer is, someone has incorrectly picked the TASK_INTERRUPTIBLE task from the tree. The scenario is described below, and all steps happen on 1 CPU: time | | | v kthread_worker_fn() <--- t1 set_current_state(TASK_INTERRUPTIBLE) schedule() block_task(t1) dequeue_entity(t1) t1->sched_delayed = 1 t2 = pick_next_task() put_prev_task(t1) enqueue_entity(t1) <--- TASK_INTERRUPTIBLE in the tree t1 switches to t2 erofs_init_percpu_worker() <--- t2 sched_set_fifo_low(t1) sched_setscheduler_nocheck(t1) __sched_setscheduler(t1) t1->sched_class = &rt_sched_class check_class_changed(t1) switched_from_fair(t1) t1->sched_delayed = 0 <--- gotcha ** from now on, t1 in the tree is TASK_INTERRUPTIBLE ** ** and sched_delayed = 0 ** preempt_enable() preempt_schedule() t1 = pick_next_task() <--- because sched_delayed = 0, eligible t2 switches back to t1, now t1 is TASK_INTERRUPTIBLE. The cause is, switched_from_fair() incorrectly clear the sched_delayed flag and confuse the pick_next_task() that it thinks a delayed task is a eligible task(without dequeue it). [Proposal] In the __sched_setscheduler() when trying to change the policy of that delayed task, do not re-enqueue the delayed task thus to avoid being picked again. The side effect that, the delayed task can not wait for its 0-vlag time to be dequeued, but its effect should be neglect. Reported-by: kernel test robot Closes: https://lore.kernel.org/oe-lkp/202408161619.9ed8b83e-lkp@intel.com Signed-off-by: Chen Yu --- kernel/sched/syscalls.c | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/kernel/sched/syscalls.c b/kernel/sched/syscalls.c index 4fae3cf25a3a..10859536e509 100644 --- a/kernel/sched/syscalls.c +++ b/kernel/sched/syscalls.c @@ -818,7 +818,8 @@ int __sched_setscheduler(struct task_struct *p, if (oldprio < p->prio) queue_flags |= ENQUEUE_HEAD; - enqueue_task(rq, p, queue_flags); + if (!p->se.sched_delayed) + enqueue_task(rq, p, queue_flags); } if (running) set_next_task(rq, p); -- 2.25.1