On Sunday, 18 February 2007 17:11, Oleg Nesterov wrote:
> On 02/18, Rafael J. Wysocki wrote:
> >
> > > On Sunday, 18 February 2007 12:31, Oleg Nesterov wrote:
> > > >
> > > > A very vague idea: what if parent will do
> > > >
> > > > current->flags |= PF_PLEASE_CONSIDER_ME_AS_FROZEN_BUT_SET_TIF_FREEZE
> > > > wait_for_completion(&vfork);
> > > > try_to_freeze();
> > > >
> > > > ?
> >
> > Hm, what about the following patch instead?
> >
> > The problem is that if the child enters the refrigeratior, we can't freeze the
> > parent, because it's uninterruptible, but the child knows the parent will be
> > uninterruptible until it exits, so the child can mark the parent as frozen.
> >
> > --- linux-2.6.20-mm2.orig/kernel/power/process.c 2007-02-18 15:43:30.000000000 +0100
> > +++ linux-2.6.20-mm2/kernel/power/process.c 2007-02-18 16:09:53.000000000 +0100
> > @@ -39,6 +39,13 @@ void refrigerator(void)
> > /* Hmm, should we be allowed to suspend when there are realtime
> > processes around? */
> > long save;
> > +
> > + /* The parent is uninterruptible and will stay so until this task exits,
> > + * so we can mark it as frozen.
> > + */
> > + if (current->vfork_done)
> > + frozen_process(current->parent);
>
> This is not safe. task->flags is not atomic, we can change ->flags only
> if we know the task won't touch it itself (ptrace, thaw_process).
> The parent could be interrupted, irq may play with current->flags (slab,
> for example).
>
> Please note that ->parent may do things like ptrace_notify() before
> it actually sleeps on ->vfork_done. This means that even if we could
> set PF_FROZEN in a safe manner, this doesn't look like a good idea.
>
> > +
> > + if (current->vfork_done && frozen(current->parent))
> > + current->parent->flags &= ~PF_FROZEN;
> > }
>
> Why? If the code above works, we shouldn't take care about frozen
> ->parent?
I've added this for symmetry. thaw_tasks() should reset PF_FROZEN for it
anyway.
Okay, so I'll post the patch that implements your idea in the other thread.
Greetings,
Rafael
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