[PATCH 1/2] kvm: Fix mmu_notifier release race
Radim Krčmář
rkrcmar at redhat.com
Tue Apr 25 14:49:05 EDT 2017
2017-04-24 11:10+0100, Suzuki K Poulose:
> The KVM uses mmu_notifier (wherever available) to keep track
> of the changes to the mm of the guest. The guest shadow page
> tables are released when the VM exits via mmu_notifier->ops.release().
> There is a rare chance that the mmu_notifier->release could be
> called more than once via two different paths, which could end
> up in use-after-free of kvm instance (such as [0]).
>
> e.g:
>
> thread A thread B
> ------- --------------
>
> get_signal-> kvm_destroy_vm()->
> do_exit-> mmu_notifier_unregister->
> exit_mm-> kvm_arch_flush_shadow_all()->
> exit_mmap-> spin_lock(&kvm->mmu_lock)
> mmu_notifier_release-> ....
> kvm_arch_flush_shadow_all()-> .....
> ... spin_lock(&kvm->mmu_lock) .....
> spin_unlock(&kvm->mmu_lock)
> kvm_arch_free_kvm()
> *** use after free of kvm ***
I don't understand this race ...
a piece of code in mmu_notifier_unregister() says:
/*
* Wait for any running method to finish, of course including
* ->release if it was run by mmu_notifier_release instead of us.
*/
synchronize_srcu(&srcu);
and code before that removes the notifier from the list, so it cannot be
called after we pass this point. mmu_notifier_release() does roughly
the same and explains it as:
/*
* synchronize_srcu here prevents mmu_notifier_release from returning to
* exit_mmap (which would proceed with freeing all pages in the mm)
* until the ->release method returns, if it was invoked by
* mmu_notifier_unregister.
*
* The mmu_notifier_mm can't go away from under us because one mm_count
* is held by exit_mmap.
*/
synchronize_srcu(&srcu);
The call of mmu_notifier->release is protected by srcu in both cases and
while it seems possible that mmu_notifier->release would be called
twice, I don't see a combination that could result in use-after-free
from mmu_notifier_release after mmu_notifier_unregister() has returned.
Doesn't [2/2] solve the exact same issue (that the release method cannot
be called twice in parallel)?
Thanks.
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