[uml-devel] How is Preemption implemented in User Mode Linux?

[uml-devel] How is Preemption implemented in User Mode Linux?

[Hopsing K]

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I'm running UML in Linux 3.2 (Debian) in SKAS0 mode. I compiled UML with
defconfig.
Now I try to  poked in the source to figure out how Preemption is done
in UML. I came so far to understand that there is a ptrace-ing parent and
for each UML process a userspace-slave process on the host:
arch/um/os-Linux/skas/process.c:userspace(). This routine also has the jump
to
interrupt_end() which calls schedule(). I saw that SIGALRM handler is
installed
for that ptrace-ing parent and the userspace() loop catches SIGALRM for the
traced slave. However I didnt see it being called. Only the ptrace-ing
parent
gets SIGALRM and the signal handler
arch/um/os-Linux/signal.c:hard_handler()
is called directly. With my understanding there must be some connection
between
hard_handler() and the call to interrupt_end() inside the userspace() loop,
but I
cannot find any, I cannot figure out how Preemption is initiated when the
SIGALRM arrives.
I wonder: Is all kernel code implementing cooperative threading ? Is there
no preemption when in kernel space?
If a userspace process runs a loop "while(1);" and therefore never calls out
to the kernel, how can another process been sheduled? The ptrace-ing
parent's
userspace() loop will never get control back.
I can also see a variable userspace_pid[0] that I assume is the active
slave-process and that there is only one such process running at any time,
the others are stopped?
I'm really curious how Preemption is done internally in UML, any insight
would
be welcome.
Some question at the side: can you call longjmp from within a signal
handler? My first assumption was that UML uses longjmp from within
the SIGALRM handler to implement preemption, butseems not to .
-- Thanks Konrad

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Re: [uml-devel] How is Preemption implemented in User Mode Linux?

[richard -rw- weinberger]

On Sun, May 26, 2013 at 1:01 AM, Hopsing K <hopsingk@gmail.com> wrote:
> I'm running UML in Linux 3.2 (Debian) in SKAS0 mode. I compiled UML with
> defconfig.
> Now I try to  poked in the source to figure out how Preemption is done
> in UML. I came so far to understand that there is a ptrace-ing parent and
> for each UML process a userspace-slave process on the host:
> arch/um/os-Linux/skas/process.c:userspace(). This routine also has the jump
> to
> interrupt_end() which calls schedule(). I saw that SIGALRM handler is
> installed
> for that ptrace-ing parent and the userspace() loop catches SIGALRM for the
> traced slave. However I didnt see it being called. Only the ptrace-ing
> parent
> gets SIGALRM and the signal handler arch/um/os-Linux/signal.c:hard_handler()
> is called directly. With my understanding there must be some connection
> between
> hard_handler() and the call to interrupt_end() inside the userspace() loop,
> but I
> cannot find any, I cannot figure out how Preemption is initiated when the
> SIGALRM arrives.
> I wonder: Is all kernel code implementing cooperative threading ? Is there
> no preemption when in kernel space?
> If a userspace process runs a loop "while(1);" and therefore never calls out
> to the kernel, how can another process been sheduled? The ptrace-ing
> parent's
> userspace() loop will never get control back.
> I can also see a variable userspace_pid[0] that I assume is the active
> slave-process and that there is only one such process running at any time,
> the others are stopped?
> I'm really curious how Preemption is done internally in UML, any insight
> would
> be welcome.
> Some question at the side: can you call longjmp from within a signal
> handler? My first assumption was that UML uses longjmp from within
> the SIGALRM handler to implement preemption, butseems not to .
> -- Thanks Konrad

UML installs a SIGVTALRM handler which feeds the TIMER_IRQ.
In the generic TIMER_IRQ handler Linux runs scheduler_tick() at some point.

Thanks,
//richard

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Re: [uml-devel] How is Preemption implemented in User Mode Linux?

[Hopsing K]

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Hi, Ok, thanks. I got a little better picture. Is the following true:
- The ptrace-ing parent waits in waitpid() called from within the
 userspace() loops.
- The SIGVTALRM comes in and the signal handler is executed, it'll end up
  in scheduler_tick() and set need_resched() true.
- Because of the signal the waitpid() system call is interrupted and the
  ptrace-ing parent returnes from waitpid().
- At the same time the ptraced userspace slave is stopped (is this the
  case?) therefore the condition WIFSTOPPED(status) is true and
  therefore you end up calling interrupt_end() ?
-- Thanks Konrad




2013/5/26 richard -rw- weinberger <richard.weinberger@gmail.com>

> On Sun, May 26, 2013 at 1:01 AM, Hopsing K <hopsingk@gmail.com> wrote:
> > I'm running UML in Linux 3.2 (Debian) in SKAS0 mode. I compiled UML with
> > defconfig.
> > Now I try to  poked in the source to figure out how Preemption is done
> > in UML. I came so far to understand that there is a ptrace-ing parent and
> > for each UML process a userspace-slave process on the host:
> > arch/um/os-Linux/skas/process.c:userspace(). This routine also has the
> jump
> > to
> > interrupt_end() which calls schedule(). I saw that SIGALRM handler is
> > installed
> > for that ptrace-ing parent and the userspace() loop catches SIGALRM for
> the
> > traced slave. However I didnt see it being called. Only the ptrace-ing
> > parent
> > gets SIGALRM and the signal handler
> arch/um/os-Linux/signal.c:hard_handler()
> > is called directly. With my understanding there must be some connection
> > between
> > hard_handler() and the call to interrupt_end() inside the userspace()
> loop,
> > but I
> > cannot find any, I cannot figure out how Preemption is initiated when the
> > SIGALRM arrives.
> > I wonder: Is all kernel code implementing cooperative threading ? Is
> there
> > no preemption when in kernel space?
> > If a userspace process runs a loop "while(1);" and therefore never calls
> out
> > to the kernel, how can another process been sheduled? The ptrace-ing
> > parent's
> > userspace() loop will never get control back.
> > I can also see a variable userspace_pid[0] that I assume is the active
> > slave-process and that there is only one such process running at any
> time,
> > the others are stopped?
> > I'm really curious how Preemption is done internally in UML, any insight
> > would
> > be welcome.
> > Some question at the side: can you call longjmp from within a signal
> > handler? My first assumption was that UML uses longjmp from within
> > the SIGALRM handler to implement preemption, butseems not to .
> > -- Thanks Konrad
>
> UML installs a SIGVTALRM handler which feeds the TIMER_IRQ.
> In the generic TIMER_IRQ handler Linux runs scheduler_tick() at some point.
>
> Thanks,
> //richard
>

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Re: [uml-devel] How is Preemption implemented in User Mode Linux?

[richard -rw- weinberger]

On Sun, May 26, 2013 at 4:48 PM, Hopsing K <hopsingk@gmail.com> wrote:
> Hi, Ok, thanks. I got a little better picture. Is the following true:
> - The ptrace-ing parent waits in waitpid() called from within the
> userspace() loops.

Correct.

> - The SIGVTALRM comes in and the signal handler is executed, it'll end up
>   in scheduler_tick() and set need_resched() true.

Yeah. We use SIGVTALRM timer tick.

> - Because of the signal the waitpid() system call is interrupted and the
>   ptrace-ing parent returnes from waitpid().

Correct.

> - At the same time the ptraced userspace slave is stopped (is this the
>   case?) therefore the condition WIFSTOPPED(status) is true and
>   therefore you end up calling interrupt_end() ?

Correct.

--
Thanks,
//richard

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Re: [uml-devel] How is Preemption implemented in User Mode Linux?

[Rob Landley]

On 05/26/2013 03:08:17 AM, richard -rw- weinberger wrote:
> On Sun, May 26, 2013 at 1:01 AM, Hopsing K <hopsingk@gmail.com> wrote:
> > Some question at the side: can you call longjmp from within a signal
> > handler? My first assumption was that UML uses longjmp from within
> > the SIGALRM handler to implement preemption, butseems not to .

Yes, "man longjmp" shows you sigsetjmp() and siglongjmp(), which save  
and restore the signal mask. This is an important detail because the  
signal you're handling is disabled in the signal handler (to avoid an  
infinite loop). You can just setjmp() and longjmp() and fiddle with the  
signal mask yourself, but signals can nest (a different signal can  
interrupt a signal handler) so re-enabling just one may not be enough.

On return, the kernel re-enables the signal you were handling, even  
from nested signal handlers. But if you longjmp() you don't call the  
kernel return code...

Rob
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