barriers: was: [RFC PATCH v2 17/18] livepatch: change to a per-task consistency model
Petr Mladek
pmladek at suse.com
Thu May 5 21:21:32 AEST 2016
On Wed 2016-05-04 12:25:17, Josh Poimboeuf wrote:
> On Wed, May 04, 2016 at 04:12:05PM +0200, Petr Mladek wrote:
> > On Wed 2016-05-04 14:39:40, Petr Mladek wrote:
> > > *
> > > * Note that the task must never be migrated to the target
> > > * state when being inside this ftrace handler.
> > > */
> > >
> > > We might want to move the second paragraph on top of the function.
> > > It is a basic and important fact. It actually explains why the first
> > > read barrier is not needed when the patch is being disabled.
> >
> > I wrote the statement partly intuitively. I think that it is really
> > somehow important. And I am slightly in doubts if we are on the safe side.
> >
> > First, why is it important that the task->patch_state is not switched
> > when being inside the ftrace handler?
> >
> > If we are inside the handler, we are kind-of inside the called
> > function. And the basic idea of this consistency model is that
> > we must not switch a task when it is inside a patched function.
> > This is normally decided by the stack.
> >
> > The handler is a bit special because it is called right before the
> > function. If it was the only patched function on the stack, it would
> > not matter if we choose the new or old code. Both decisions would
> > be safe for the moment.
> >
> > The fun starts when the function calls another patched function.
> > The other patched function must be called consistently with
> > the first one. If the first function was from the patch,
> > the other must be from the patch as well and vice versa.
> >
> > This is why we must not switch task->patch_state dangerously
> > when being inside the ftrace handler.
> >
> > Now I am not sure if this condition is fulfilled. The ftrace handler
> > is called as the very first instruction of the function. Does not
> > it break the stack validity? Could we sleep inside the ftrace
> > handler? Will the patched function be detected on the stack?
> >
> > Or is my brain already too far in the fantasy world?
>
> I think this isn't a possibility.
>
> In today's code base, this can't happen because task patch states are
> only switched when sleeping or when exiting the kernel. The ftrace
> handler doesn't sleep directly.
>
> If it were preempted, it couldn't be switched there either because we
> consider preempted stacks to be unreliable.
This was the missing piece.
> In theory, a DWARF stack trace of a preempted task *could* be reliable.
> But then the DWARF unwinder should be smart enough to see that the
> original function called the ftrace handler. Right? So the stack would
> be reliable, but then livepatch would see the original function on the
> stack and wouldn't switch the task.
>
> Does that make sense?
Yup. I think that we are on the safe side. Thanks for explanation.
Best Regards,
Petr
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