On Tuesday 01 August 2006 19:41, Stephens, Allan wrote:
> Hi Paolo:
>
> Apparently I was unclear in my last email. I was not trying to claim
> that "uml_switch -hub" was delivering promiscuously when "ifconfig
> promisc" was not set,
This makes sense...
> but rather that it was (correctly) deliv
Hi Paolo:
Apparently I was unclear in my last email. I was not trying to claim
that "uml_switch -hub" was delivering promiscuously when "ifconfig
promisc" was not set, but rather that it was (correctly) delivering
non-promiscuously -- meaning I didn't have to trick my protocol software
into filte
"Stephens, Allan" <[EMAIL PROTECTED]> ha scritto:
> Hi Paolo:
>
> I tried configuring the Ethernet interfaces as "promiscuous" and
> this
> did indeed provide a workaround for the issue I raised. The root
> cause
> of my problems seems to have been a lack of understanding on my
> part as
> to w
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 ker
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?
If you're running UML in SKAS mode (as you should), then »untrustedProg« do
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Please enlighten me of my ignorance.
Original Belief: I can use UML as a virtual machine; jail untrusted
processes.
Problem:
Let's say I am a user U, on a machine M running Linux.
I run an instance, UML1 of User Mode Linux.
Within this instance of UML1, I create a new user "jailedUser".
"jai
