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sched, lockdep: Annotate ->pi_lock recursion

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There's a valid ->pi_lock recursion issue where the actual PI code
tries to wake up the stop task. Make lockdep aware so it doesn't
complain about this.

Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Reviewed-by: Valentin Schneider <valentin.schneider@arm.com>
Reviewed-by: Daniel Bristot de Oliveira <bristot@redhat.com>
Link: https://lkml.kernel.org/r/20201023102347.406912197@infradead.org
This commit is contained in:
Peter Zijlstra 2020-10-01 16:13:01 +02:00
parent 95158a89dd
commit ded467dc83
1 changed files with 15 additions and 0 deletions
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15
kernel/sched/core.c
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@ -2658,6 +2658,7 @@ int select_task_rq(struct task_struct *p, int cpu, int sd_flags, int wake_flags)
void sched_set_stop_task(int cpu, struct task_struct *stop)
{
static struct lock_class_key stop_pi_lock;
struct sched_param param = { .sched_priority = MAX_RT_PRIO - 1 };
struct task_struct *old_stop = cpu_rq(cpu)->stop;
@ -2673,6 +2674,20 @@ void sched_set_stop_task(int cpu, struct task_struct *stop)
sched_setscheduler_nocheck(stop, SCHED_FIFO, &param);
stop->sched_class = &stop_sched_class;
/*
* The PI code calls rt_mutex_setprio() with ->pi_lock held to
* adjust the effective priority of a task. As a result,
* rt_mutex_setprio() can trigger (RT) balancing operations,
* which can then trigger wakeups of the stop thread to push
* around the current task.
*
* The stop task itself will never be part of the PI-chain, it
* never blocks, therefore that ->pi_lock recursion is safe.
* Tell lockdep about this by placing the stop->pi_lock in its
* own class.
*/
lockdep_set_class(&stop->pi_lock, &stop_pi_lock);
}
cpu_rq(cpu)->stop = stop;