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SubjectRe: [PATCH -tip v5 03/10] kprobes: Introduce kprobes jump optimization
On Tue, Nov 24, 2009 at 10:34:16AM -0500, Masami Hiramatsu wrote:
> Frederic Weisbecker wrote:
> > I _might_ have understood.
> > You have set up the optimized flags, then you wait for
> > any old-style int 3 kprobes to complete and route
> > to detour buffer so that you can patch the jump
> > safely in the dead code? (and finish with first byte
> > by patching the int 3 itself)
> >
>
> Yeah, you might get almost correct answer.
> The reason why we have to wait scheduling on all processors
> is that this code may modify N instructions (not a single
> instruction). This means, there is a chance that 2nd to nth
> instructions are interrupted on other cpus when we start
> code modifying.


Aaah ok!

In this case, you probably just need the synchronize_sched()
thing. The delayed work looks unnecessary.


> Please imagine that 2nd instruction is interrupted and
> stop_machine() replaces the 2nd instruction with jump
> *address* while running interrupt handler. When the interrupt
> returns to original address, there is no valid instructions
> and it causes unexpected result.


Yeah.


>
> To avoid this situation, we have to wait a scheduler quiescent
> state on all cpus, because it also ensure that all current
> interruption are done.


Ok.


> This also excuses why we don't need to wait when unoptimizing
> and why it has not supported preemptive kernel yet.


I see...so the non-preemptible kernel requirement looks
hard to workaround :-s


> In unoptimizing case, since there is just a single instruction
> (jump), there is no nth instruction which can be interrupted.
> Thus we can just use a stop_machine(). :-)


Ok.


>
> On the preemptive kernel, waiting scheduling is not work as we
> see on non-preemptive kernel. Since processes can be preempted
> in interruption, we can't ensure that the current running
> interruption is done. (I assume that a pair of freeze_processes
> and thaw_processes may possibly ensure that, or maybe we can
> share some stack rewinding code with ksplice.)
> So it depends on !PREEMPT.



Right.
However using freeze_processes() and thaw_processes() would be
probably too costly and it's not a guarantee that every processes
go to the refrigerator() :-), because some tasks are not freezable,
like the kernel threads by default if I remember well, unless they
call set_freezable(). That's a pity, we would just have needed
to set __kprobe in refrigerator().


PS: hmm btw I remember about a patch that
tagged refrigerator() as __cold but it looks like it hasn't been
applied....

Thanks.



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