On Mon, Feb 29, 2016 at 01:58:17PM +0100, Michal Hocko wrote:
> @@ -215,16 +216,34 @@ void __sched __down_write(struct rw_semaphore *sem)
> */
> if (sem->count == 0)
> break;
> - set_task_state(tsk, TASK_UNINTERRUPTIBLE);
> + set_task_state(tsk, state);
> raw_spin_unlock_irqrestore(&sem->wait_lock, flags);
> schedule();
> + if (signal_pending_state(state, current)) {
> + ret = -EINTR;
> + raw_spin_lock_irqsave(&sem->wait_lock, flags);
> + goto out;
> + }
> raw_spin_lock_irqsave(&sem->wait_lock, flags);
> }
> /* got the lock */
> sem->count = -1;
> @@ -487,20 +488,38 @@ struct rw_semaphore __sched
> *rwsem_down_write_failed(struct rw_semaphore *sem)
> /* Block until there are no active lockers. */
> do {
> schedule();
> - set_current_state(TASK_UNINTERRUPTIBLE);
> + if (signal_pending_state(state, current)) {
> + raw_spin_lock_irq(&sem->wait_lock);
> + ret = ERR_PTR(-EINTR);
> + goto out;
> + }
> + set_current_state(state);
> } while ((count = sem->count) & RWSEM_ACTIVE_MASK);
>
> raw_spin_lock_irq(&sem->wait_lock);
> }
> __set_current_state(TASK_RUNNING);
Why is the signal_pending_state() test _after_ the call to schedule()
and before the 'trylock'.
__mutex_lock_common() has it before the call to schedule and after the
'trylock'.
The difference is that rwsem will now respond to the KILL and return
-EINTR even if the lock is available, whereas mutex will acquire it and
ignore the signal (for a little while longer).
Neither is wrong per se, but I feel all the locking primitives should
behave in a consistent manner in this regard.
