On 09/19/2010 02:55 PM, Vasiliy Kulikov wrote:
> If device_register() fails then call put_device().
> See comment to device_register.
>
> Signed-off-by: Vasiliy Kulikov <segooon@xxxxxxxxx>
> ---
> compile tested.
>
> drivers/scsi/osd/osd_uld.c | 4 +++-
> 1 files changed, 3 insertions(+), 1 deletions(-)
>
> diff --git a/drivers/scsi/osd/osd_uld.c b/drivers/scsi/osd/osd_uld.c
> index cefb2c0..3e0edc2 100644
> --- a/drivers/scsi/osd/osd_uld.c
> +++ b/drivers/scsi/osd/osd_uld.c
> @@ -474,7 +474,7 @@ static int osd_probe(struct device *dev)
> error = device_register(&oud->class_dev);
> if (error) {
> OSD_ERR("device_register failed => %d\n", error);
> - goto err_put_cdev;
> + goto err_put_device;
> }
>
> get_device(&oud->class_dev);
> @@ -482,6 +482,8 @@ static int osd_probe(struct device *dev)
> OSD_INFO("osd_probe %s\n", disk->disk_name);
> return 0;
>
> +err_put_device:
> + put_device(&oud->class_dev);
I'm not sure we can do this here. We might need to disregard the
comment at device_register. Because this put_ will try to call the
registered __release which will try to free the oud structure which
has the ->class_dev embedded, and now we have a double free.
But I will add a fat comment if all agree.
I'm assuming that if the device_register has failed then we are not
yet on any exposed system lists. (proof of we don't need to call
device_unregister). Since we don't yet let anyone see this device
we can go head and free it regardless of it's initialized ref-count
== 1. The motivation here is to tear down the device without any
possible users. Is that guaranteed? From my code audit it is.
> err_put_cdev:
> cdev_del(&oud->cdev);
> err_put_disk:
And I think device_register has a very bad API side effect with this put.
If you are going to monitor all places that do not call put_device. Why
not go to all places that do, and remove them and fix device_register.
Do a majority vote. What is done more? put_device called or not called.
Thanks
Boaz
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