Re: vm86.c audit_syscall_exit() call trashes registers

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The corruption originally looked like a race condition.

Sometimes the EDID buffer would be all zeros.
Sometimes it would contain partial data, and then the rest of the buffer filled with zeros. The amount of data transferred into the buffer before going to all zeros is non-deterministic.

When we put a known value in each byte of the buffer before making the vm86 call, the known data would always be overwritten either with EDID data or zeros.

-Bill

----

William Cattey
Linux Platform Coordinator
MIT Information Services & Technology

W92-176, 617-253-0140, [email protected]
http://web.mit.edu/wdc/www/


On Aug 14, 2007, at 4:42 PM, Andi Kleen wrote:

Chuck Anderson <[email protected]> writes:

If I'm reading correctly, it appears that the code above trashes the
%fs and %gs registers, or otherwise doesn't leave them at zero before
returning from the system call as the old code did. Is this a correct
analysis?

The kernel runs with defined fs -- saved and set at system call entry/exit --
and shouldn't touch gs (except on a context switch, but then it should
be set back when you get scheduled again)

It's in theory possible that something went wrong with the gs saving
for the vm86 path, but this changed long 2.6.16. But I assume
when you just remove the call in 2.6.16 it already works? If yes
it cannot be that (2.6.16 didn't use either fs or gs in the kernel)

How should this be fixed?

The problem first needs to be fully understood. Do you have more
details on the corruption?

One suspicious thing is that the audit code does mutex_lock (&tty_mutex)
and could sleep there. It's a long shot, but does the problem go
away when you comment that out? [such a patch is incorrect in theory,
but should be unlikely enough to crash for a quick test]

But actually sleeping should be ok here and a preemptible kernel could do
it anyways.

-Andi

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