Re: [RFC v2 2/3] dma-buf/fence-array: add fence_array_teardown()

[Gustavo Padovan]

2016-06-28 Christian König <christian.koenig@xxxxxxx>:

> Am 28.06.2016 um 16:17 schrieb Gustavo Padovan:
> > 2016-06-28 Christian König <christian.koenig@xxxxxxx>:
> > 
> > > Am 27.06.2016 um 21:29 schrieb Gustavo Padovan:
> > > > From: Gustavo Padovan <gustavo.padovan@xxxxxxxxxxxxxxx>
> > > > 
> > > > As the array of fence callbacks held by an active struct fence_array
> > > > each has a reference to the struct fence_array, when the owner of the
> > > > fence_array is freed it must dispose of the callback references before
> > > > it can free the fence_array. This can not happen simply during
> > > > fence_release() because of the extra references and so we need a new
> > > > function to run before the final fence_put().
> > > As I said previously as well, this is completely superfluous.
> > > 
> > > The fence array keeps a reference to itself as long as not all callbacks are
> > > signaled.
> > > 
> > > So you only need to unregister your callback from the array itself and drop
> > > your reference when you don't need it any more in the sync file.
> > Exactly, this should be called from sync_file_free() because of the
> > following use case:
> > 
> > 	1. You create 2 sync_file with 1 fence each
> > 	2. Merge both fences, which creates a fence array
> > 	3. Close the sync_file fd without waiting for the fences to
> > 	signal
> > 
> > At this point you leak the fence-array because the final fence_put()
> > does not release it because of the extra references from the non
> > signalled fences so we need to clean up this somehow.
> 
> No, there won't be any leak and you don't need to cleanup anything.
> 
> The fences contained in the fence array are already enabled for signaling.
> So they will eventually signal sooner or later and drop the reference to the
> fence array freeing it when the last one finished.
> 
> This was done to avoid the need to call fence_remove_callback() all the time
> on the fences in the array because of the warning on the function.

Right. I understand it now.

	Gustavo