On Tue, Aug 01, 2006 at 12:51:44AM -0700, TongKe Xue wrote:
> Now, we know that M is running in ring 0.
> UML1 is running as a process, so it's in ring 3.
>
> If this is the case, what protects "untrustedProg" from playing around
> with the kernel memory of UML1?
The process can't form a UML kernel address. The UML kernel is in an
entirely different host address space.
Jeff
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