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Date: Tue, 16 Sep 2014 13:48:24 -0300
From: Marcelo Tosatti <mtosatti@redhat.com>
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Subject: Re: [Qemu-devel] [PATCH] kvmclock: clarify usage of
	cpu_clean_all_dirty
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To: Paolo Bonzini <pbonzini@redhat.com>
Cc: qemu-devel <qemu-devel@nongnu.org>, Alexander Graf <agraf@suse.de>

On Tue, Sep 16, 2014 at 06:22:15PM +0200, Paolo Bonzini wrote:
> Il 16/09/2014 18:07, Marcelo Tosatti ha scritto:
> >> > The cpu_synchronize_all_states() call in kvmclock_vm_state_change() is
> >> > needed to make env->tsc up to date with the value on the source, right?
> > Its there to make sure the pair
> > 
> > env->tsc, s->clock = data.clock
> > 
> > are relative to point close in time.
> 
> Ok.  But why are they not close in time?

Scenario 1

A. s->clock = get_clock()
B. env->tsc = rdtsc()

Scenario 2

A. s->clock = get_clock()
C. VM callbacks, bdrv_flush, ...
B. env->tsc = rdtsc()

They are not "close in time" because of C.

> Could we have the opposite situation where env->tsc is loaded a long
> time _after_ s->clock, and something breaks?

This particular read avoids an overflow. See 

+    assert(time.tsc_timestamp <= migration_tsc);


About your question, perhaps, would have to make up an
env->tsc, s->clock pair which breaks the code or a guest
application.

> Also, there is no reason to do kvmclock_current_nsec() during normal
> execution of the VM.  Is the s->clock sent by the source ever good
> across migration, and could the source send kvmclock_current_nsec()
> instead of whatever KVM_GET_CLOCK returns?

Yes it could. What difference does it make?

> I don't understand this code very well, but it seems to me that the
> migration handling and VM state change handler are mixed up...

I don't see what the problem is. I am sure you can understand the code.