On 16/06/10 20:04 +0400, Pavel Emelyanov wrote:
> > +void proc_new_task(struct task_struct *task)
> > +{
> > + struct pid *pid;
> > + int i;
> > +
> > + if (!task->pid)
> > + return;
> > +
> > + pid = task_pid(task);
> > + for (i = 0; i <= pid->level; i++)
> > + mntget(pid->numbers[i].ns->proc_mnt);
>
> I feel I'm missing something significant, but this patch breaks
> the mntget/mntput balance. Doesn't it?
Why?
all mntget() here have their mntput() in proc_flush_task(). At least, this is
the intent...
Thanks,
Louis
--
Dr Louis Rilling Kerlabs
Skype: louis.rilling Batiment Germanium
Phone: (+33|0) 6 80 89 08 23 80 avenue des Buttes de Coesmes
http://www.kerlabs.com/ 35700 Rennes
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