On Thu, Dec 19, 2013 at 04:03:32PM +0100, Levente Kurusa wrote:
> This is required so that we give up the last reference to the device.
> Remove the kfree() as that is the job of wq_device_release which will now
> be called due to the reference count actually reaching zero.
>
> Signed-off-by: Levente Kurusa <le...@linux.com>
> ---
> kernel/workqueue.c | 2 +-
> 1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/kernel/workqueue.c b/kernel/workqueue.c
> index 987293d..f3b3398 100644
> --- a/kernel/workqueue.c
> +++ b/kernel/workqueue.c
> @@ -3361,7 +3361,7 @@ int workqueue_sysfs_register(struct workqueue_struct
> *wq)
>
> ret = device_register(&wq_dev->dev);
> if (ret) {
> - kfree(wq_dev);
> + put_device(&wq_dev->dev);
Umm... this doesn't look right. You're basically converting the code
to the following,
x = kmalloc();
if (register(x) < 0)
put(x)
They're not symmetrical anymore. register(), or any API call really,
isn't supposed to have side effects which need explicit cleanup after
a failure. The fact that x is properly initialized even after
register(x) failed is a coincidental implementation detail which
shouldn't be depended upon. Your patch is actively breaking the
convention for no good reason.
Nacked-by: Tejun Heo <t...@kernel.org>
>From the patch title, I suppose you posted a bunch of patches towards
this direction. Please consider all of them nacked if they're doing
the same thing.
Thanks.
--
tejun
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