Re: [PATCH 1/1] timer: Warn about schedule_timeout() called for tasks in TASK_RUNNING state

[Steven Rostedt]

On Thu, 16 Jan 2020 17:02:18 +0300
Alexander Popov <alex.popov@xxxxxxxxx> wrote:

> When we were preparing the patch 6dcd5d7a7a29c1e, we made a mistake noticed
> by Linus: schedule_timeout() was called without setting the task state to
> anything particular. It calls the scheduler, but doesn't delay anything,
> because the task stays runnable. That happens because sched_submit_work()
> does nothing for tasks in TASK_RUNNING state.
> 
> Let's add a WARN_ONCE() under CONFIG_SCHED_DEBUG to detect such kernel
> API misuse.
> 
> Signed-off-by: Alexander Popov <alex.popov@xxxxxxxxx>
> ---
>  kernel/time/timer.c | 5 +++++
>  1 file changed, 5 insertions(+)
> 
> diff --git a/kernel/time/timer.c b/kernel/time/timer.c
> index 4820823515e9..52ad2d6ce352 100644
> --- a/kernel/time/timer.c
> +++ b/kernel/time/timer.c
> @@ -1887,6 +1887,11 @@ signed long __sched schedule_timeout(signed long timeout)
>  		}
>  	}
>  
> +#ifdef CONFIG_SCHED_DEBUG
> +	WARN_ONCE(current->state == TASK_RUNNING,
> +			"schedule_timeout for TASK_RUNNING\n");
> +#endif
> +

But this can trigger false warnings. For example, if we are waiting on
an event with a timeout:


	DEFINE_WAIT(wait);

	for (;;) {
		prepare_to_wait(&waitq, &wait, TASK_UNINTERRUPTIBLE);
		if (event)
			break;
		timeout = schedule_timeout(timeout);
		if (!timeout)
			break;
	}
	finish_wait(&waitq, &wait);


If the event happens between "prepare_to_wait" and just before
schedule_timeout(), the wait queue will set this task's state to
TASK_RUNNING, which in turn triggers your warning.

-- Steve


>  	expire = timeout + jiffies;
>  
>  	timer.task = current;