Re: [PATCH 2/2] livepatch: fix patched module loading race

[Josh Poimboeuf <jpoimboe@redhat.com>]

On Wed, Mar 04, 2015 at 04:51:39PM +0100, Jiri Kosina wrote:
> On Wed, 4 Mar 2015, Josh Poimboeuf wrote:
> 
> > > CPU0					CPU1
> > > 
> > > delete_module()  #SYSCALL
> > > 
> > >    try_stop_module()
> > >      mod->state = MODULE_STATE_GOING;
> > > 
> > >    mutex_unlock(&module_mutex);
> > > 
> > > 					klp_register_patch()
> > > 					klp_enable_patch()
> > > 
> > > 					#save place to switch universe
> > > 
> > > 					b()     # from module that is going
> > > 					  a()   # from core (patched)
> > > 
> > > 
> > >    mod->exit();
> > > 
> > > 
> > > Note that the function b() can be called until we call mod->exit().
> > > 
> > > If we do not apply patch against b() because it is in
> > > MODULE_STATE_GOING. It will call patched a() with modified semantic
> > > and things might get wrong.
> > > 
> > > Well, the above described race is rather theoretical. It will happen
> > > only when a patched module is being removed and a module with a patch
> > > is added at the same time. Also it means that some other CPU will
> > > manage to register, enable the patch, switch the universe, and
> > > call function from the patched module during the small race window.
> > > 
> > > I am not sure if we need to handle such a type of race if the solution
> > > adds too big complexity.
> > 
> > But b() can't be called after the module is in MODULE_STATE_GOING,
> > right?  try_stop_module() makes sure it's not being used before changing
> > its state.
> 
> If b() is called from __exit() of the particular module, then you end up 
> in exactly the situation described above, don't you?

Well, maybe.  I guess it depends on the consistency model.  The current
"inconsistency" model doesn't allow for function semantic changes
anyway.

If we went to the per-task consistency model with stack checking (like
my RFC), if mod->exit() calls schedule() before calling b(), the module
task can potentially switch to the new universe, and call old b() which
calls new a().

So yeah.  Maybe this isn't the best approach.

-- 
Josh