RE: [uml-devel] Question about the kernel thread in UML TT mode

[Alex L. <ale...@st...>]

On May 18, 2005 9:31 PM, Jeff wrote:

>> But P and Q are different threads and they will be scheduled=20
>> differently,right?

> No, they're two different ways of looking at the same thread.  They're =

> scheduled somewhat differently, but that's just reflecting the two =
ways of
looking at it.  If it's runnable on the host, it's runnable inside UML, =
but
not the other way around.

Then how does the UML thread interact with its corresponding host thread =
in
TT mode? Also use the switch_pipe?

>> And in the function outer_tramp, the thread A see the thread B was=20
>> stopped with a SIGSTOP signal. Why? Since the thread B is blocked by=20
>> the signal SIGUSR1...

> The handler calls suspend_new_thread, which stops itself.

Exactly! I'm wrong...

Then I have another question:

In UML function new_thread_handler, the current sigcontext is defined as
following:

	UPT_SC(&current->thread.regs.regs) =3D (void *) (&sig + 1);

Why? What does "&sig + 1" mean?

Thanks a lot!

Alex