Re: Possible kernel lock in semaphore's __down()

Peter Zijlstra
2007-08-30T07:16:20Z

On Wed, 2007-08-29 at 23:52 +0200, Aleksandar Dezelin wrote:
> Hi!
>
> I'm a newbie here on the list and also as a "kernel hacker". There's a
> bug reported in bugzilla (Bug 7927), cite:
>
>
> > In the function __down
> >
> > fastcall void __sched __down(struct semaphore * sem)
> > {
> > struct task_struct *tsk = current;
> > DECLARE_WAITQUEUE(wait, tsk);
> > unsigned long flags;
> >
> > tsk->state = TASK_UNINTERRUPTIBLE;
> > spin_lock_irqsave(&sem->wait.lock, flags);
> > add_wait_queue_exclusive_locked(&sem->wait, &wait);
> > ...
> > }
> >
> >
> > From this code fragment, it sets the tsk->state to TASK_UNINTERRUPTIBLE before
> > gets the spinlock. Assume at that moment, a interrupt ocuur and and after the
> > interrupt handle ends, an other process is scheduled to run (assume the kernel
> > is preemptalbe). In this case, the previous process ( its state has set to
> > TASK_UNINTERRUPTIBLE) has been picked off the run queue, and it has not yet add
> > to the wait queue( sem->wait ), so it may be never waited up forever.
> >
>
> I have marked it as rejected as as I can see at the time this function is called,
> it is guaranteed that ret_from_intr() will not call schedule() on return from an
> interrupt handler to either kernel space or user space because of the call
> to macro might_sleep() in semaphore's down(). Am I wrong?

I think the reported meant interrupt driven involuntary preemption. So
ret_from_intr() is not the right place to look. But afaict you're still
right, see how preempt_schedule*() adds PREEMPT_ACTIVE to the
preempt_count, and how that makes the scheduler ignore the task state.