On Wed, Nov 02, 2022 at 03:45:27PM -0600, Alex Williamson wrote: > > open_count was from before we changed the core code to call > > open_device only once. If we are only a call chain from > > open_device/close_device then we know that the open_count must be 1. > > That accounts for the first test, we don't need to test open_count on > the calling device in this path, but the idea here is that we want to > test whether we're the last close_device among the set. Not sure how > we'd do that w/o some visibility to open_count. Maybe we need a > vfio_device_set_open_count() that when 1 we know we're the first open > or last close? Thanks, Right, we are checking the open count on all the devices in the set, so I think that just this hunk is fine: > > > > - if (cur->vdev.open_count) > > > > + if (cur != vdev && cur->vdev.open_count) > > > > return false; Because vfio_pci_dev_set_needs_reset() is always called from open/close (which is how it picks up the devset lock), so we never need to consider the current device by definition, it is always "just being closed" A little comment to explain this and that should be it? Jason
