Re: [PATCH V2 6/6] block: split .sysfs_lock into two locks

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On Wed, Aug 21, 2019 at 09:18:08AM -0700, Bart Van Assche wrote:
> On 8/21/19 2:15 AM, Ming Lei wrote:
> > diff --git a/block/blk-mq-sysfs.c b/block/blk-mq-sysfs.c
> > index 31bbf10d8149..a4cc40ddda86 100644
> > --- a/block/blk-mq-sysfs.c
> > +++ b/block/blk-mq-sysfs.c
> > @@ -247,7 +247,7 @@ void blk_mq_unregister_dev(struct device *dev, struct request_queue *q)
> >   	struct blk_mq_hw_ctx *hctx;
> >   	int i;
> > -	lockdep_assert_held(&q->sysfs_lock);
> > +	lockdep_assert_held(&q->sysfs_dir_lock);
> >   	queue_for_each_hw_ctx(q, hctx, i)
> >   		blk_mq_unregister_hctx(hctx);
> > @@ -297,7 +297,7 @@ int __blk_mq_register_dev(struct device *dev, struct request_queue *q)
> >   	int ret, i;
> >   	WARN_ON_ONCE(!q->kobj.parent);
> > -	lockdep_assert_held(&q->sysfs_lock);
> > +	lockdep_assert_held(&q->sysfs_dir_lock);
> >   	ret = kobject_add(q->mq_kobj, kobject_get(&dev->kobj), "%s", "mq");
> >   	if (ret < 0)
> 
> blk_mq_unregister_dev and __blk_mq_register_dev() are only used by
> blk_register_queue() and blk_unregister_queue(). It is the responsibility of
> the callers of these function to serialize request queue registration and
> unregistration. Is it really necessary to hold a mutex around the
> blk_mq_unregister_dev and __blk_mq_register_dev() calls? Or in other words,
> can it ever happen that multiple threads invoke one or both functions
> concurrently?

hctx kobjects can be removed and re-added via blk_mq_update_nr_hw_queues()
which may be called at the same time when queue is registering or
un-registering.

Also the change can be simpler to use a new lock to replace the old one.

> 
> > @@ -331,7 +331,7 @@ void blk_mq_sysfs_unregister(struct request_queue *q)
> >   	struct blk_mq_hw_ctx *hctx;
> >   	int i;
> > -	mutex_lock(&q->sysfs_lock);
> > +	mutex_lock(&q->sysfs_dir_lock);
> >   	if (!q->mq_sysfs_init_done)
> >   		goto unlock;
> > @@ -339,7 +339,7 @@ void blk_mq_sysfs_unregister(struct request_queue *q)
> >   		blk_mq_unregister_hctx(hctx);
> >   unlock:
> > -	mutex_unlock(&q->sysfs_lock);
> > +	mutex_unlock(&q->sysfs_dir_lock);
> >   }
> >   int blk_mq_sysfs_register(struct request_queue *q)
> > @@ -347,7 +347,7 @@ int blk_mq_sysfs_register(struct request_queue *q)
> >   	struct blk_mq_hw_ctx *hctx;
> >   	int i, ret = 0;
> > -	mutex_lock(&q->sysfs_lock);
> > +	mutex_lock(&q->sysfs_dir_lock);
> >   	if (!q->mq_sysfs_init_done)
> >   		goto unlock;
> > @@ -358,7 +358,7 @@ int blk_mq_sysfs_register(struct request_queue *q)
> >   	}
> >   unlock:
> > -	mutex_unlock(&q->sysfs_lock);
> > +	mutex_unlock(&q->sysfs_dir_lock);
> >   	return ret;
> >   }
> 
> blk_mq_sysfs_unregister() and blk_mq_sysfs_register() are only used by
> __blk_mq_update_nr_hw_queues(). Calls to that function are serialized by the
> tag_list_lock mutex. Is it really necessary to use any locking inside these
> functions?

hctx kobjects can be removed and re-added via blk_mq_update_nr_hw_queues()
which may be called at the same time when queue is registering or
un-registering.

Also the change can be simpler to use a new lock to replace the old one.

> 
> > diff --git a/block/blk-sysfs.c b/block/blk-sysfs.c
> > index 5b0b5224cfd4..5941a0176f87 100644
> > --- a/block/blk-sysfs.c
> > +++ b/block/blk-sysfs.c
> > @@ -938,6 +938,7 @@ int blk_register_queue(struct gendisk *disk)
> >   	int ret;
> >   	struct device *dev = disk_to_dev(disk);
> >   	struct request_queue *q = disk->queue;
> > +	bool has_elevator = false;
> >   	if (WARN_ON(!q))
> >   		return -ENXIO;
> > @@ -945,7 +946,6 @@ int blk_register_queue(struct gendisk *disk)
> >   	WARN_ONCE(blk_queue_registered(q),
> >   		  "%s is registering an already registered queue\n",
> >   		  kobject_name(&dev->kobj));
> > -	blk_queue_flag_set(QUEUE_FLAG_REGISTERED, q);
> >   	/*
> >   	 * SCSI probing may synchronously create and destroy a lot of
> > @@ -966,7 +966,7 @@ int blk_register_queue(struct gendisk *disk)
> >   		return ret;
> >   	/* Prevent changes through sysfs until registration is completed. */
> > -	mutex_lock(&q->sysfs_lock);
> > +	mutex_lock(&q->sysfs_dir_lock);
> >   	ret = kobject_add(&q->kobj, kobject_get(&dev->kobj), "%s", "queue");
> >   	if (ret < 0) {
> > @@ -987,26 +987,37 @@ int blk_register_queue(struct gendisk *disk)
> >   		blk_mq_debugfs_register(q);
> >   	}
> > -	kobject_uevent(&q->kobj, KOBJ_ADD);
> > -
> > -	wbt_enable_default(q);
> > -
> > -	blk_throtl_register_queue(q);
> > -
> > +	/*
> > +	 * The queue's kobject ADD uevent isn't sent out, also the
> > +	 * flag of QUEUE_FLAG_REGISTERED isn't set yet, so elevator
> > +	 * switch won't happen at all.
> > +	 */
> >   	if (q->elevator) {
> > -		ret = elv_register_queue(q);
> > +		ret = elv_register_queue(q, false);
> >   		if (ret) {
> > -			mutex_unlock(&q->sysfs_lock);
> > -			kobject_uevent(&q->kobj, KOBJ_REMOVE);
> > +			mutex_unlock(&q->sysfs_dir_lock);
> >   			kobject_del(&q->kobj);
> >   			blk_trace_remove_sysfs(dev);
> >   			kobject_put(&dev->kobj);
> >   			return ret;
> >   		}
> > +		has_elevator = true;
> >   	}
> > +
> > +	mutex_lock(&q->sysfs_lock);
> > +	blk_queue_flag_set(QUEUE_FLAG_REGISTERED, q);
> > +	wbt_enable_default(q);
> > +	blk_throtl_register_queue(q);
> > +	mutex_unlock(&q->sysfs_lock);
> > +
> > +	/* Now everything is ready and send out KOBJ_ADD uevent */
> > +	kobject_uevent(&q->kobj, KOBJ_ADD);
> > +	if (has_elevator)
> > +		kobject_uevent(&q->elevator->kobj, KOBJ_ADD);
> > +
> >   	ret = 0;
> >   unlock:
> > -	mutex_unlock(&q->sysfs_lock);
> > +	mutex_unlock(&q->sysfs_dir_lock);
> >   	return ret;
> >   }
> 
> My understanding is that the mutex_lock() / mutex_unlock() calls in this
> function are necessary today to prevent concurrent changes of the scheduler
> from this function and from sysfs. If the kobject_uevent(KOBJ_ADD) call is
> moved, does that mean that all mutex_lock() / mutex_unlock() calls can be
> left out from this function?


hctx kobjects can be removed and re-added via blk_mq_update_nr_hw_queues()
which may be called at the same time when queue is registering or
un-registering.

Also the change can be simpler to use a new lock to replace the old one.

> 
> >   EXPORT_SYMBOL_GPL(blk_register_queue);
> > @@ -1021,6 +1032,7 @@ EXPORT_SYMBOL_GPL(blk_register_queue);
> >   void blk_unregister_queue(struct gendisk *disk)
> >   {
> >   	struct request_queue *q = disk->queue;
> > +	bool has_elevator;
> >   	if (WARN_ON(!q))
> >   		return;
> > @@ -1035,25 +1047,25 @@ void blk_unregister_queue(struct gendisk *disk)
> >   	 * concurrent elv_iosched_store() calls.
> >   	 */
> >   	mutex_lock(&q->sysfs_lock);
> > -
> >   	blk_queue_flag_clear(QUEUE_FLAG_REGISTERED, q);
> > +	has_elevator = !!q->elevator;
> > +	mutex_unlock(&q->sysfs_lock);
> > +	mutex_lock(&q->sysfs_dir_lock);
> >   	/*
> >   	 * Remove the sysfs attributes before unregistering the queue data
> >   	 * structures that can be modified through sysfs.
> >   	 */
> >   	if (queue_is_mq(q))
> >   		blk_mq_unregister_dev(disk_to_dev(disk), q);
> > -	mutex_unlock(&q->sysfs_lock);
> >   	kobject_uevent(&q->kobj, KOBJ_REMOVE);
> >   	kobject_del(&q->kobj);
> >   	blk_trace_remove_sysfs(disk_to_dev(disk));
> > -	mutex_lock(&q->sysfs_lock);
> > -	if (q->elevator)
> > +	if (has_elevator)
> >   		elv_unregister_queue(q);
> > -	mutex_unlock(&q->sysfs_lock);
> > +	mutex_unlock(&q->sysfs_dir_lock);
> >   	kobject_put(&disk_to_dev(disk)->kobj);
> >   }
> 
> If this function would call kobject_del(&q->kobj) before doing anything
> else, does that mean that all mutex_lock() / mutex_unlock() calls can be
> left out from this function?

As I mentioned above, we need to sync between registering/un-registering
queue and updating nr_hw_queues, so the lock of sysfs_dir_lock is needed.

Thanks, 
Ming



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