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SubjectRe: [patch] cpufreq: mark cpufreq_tsc() as core_initcall_sync
On Mon, Nov 20, 2006 at 03:01:59PM -0500, Alan Stern wrote:
> On Mon, 20 Nov 2006, Oleg Nesterov wrote:
>
> > On 11/20, Alan Stern wrote:
> > >
> > > @@ -158,6 +199,11 @@ void synchronize_srcu(struct srcu_struct
> > >
> > > [... snip ...]
> > >
> > > +#ifdef SMP__STORE_MB_LOAD_WORKS /* The fast path */
> > > + if (srcu_readers_active_idx(sp, idx) == 0)
> > > + goto done;
> > > +#endif
> >
> > I guess this is connected to another message from you,
>
> Yes.
>
> > > But of course it _is_ needed for the fastpath to work. In fact, it might
> > > not be good enough, depending on the architecture. Here's what the
> > > fastpath ends up looking like (using c[idx] is essentially the same as
> > > using hardluckref):
> > >
> > > WRITER READER
> > > ------ ------
> > > dataptr = &(new data) atomic_inc(&hardluckref)
> > > mb mb
> > > while (hardluckref > 0) ; access *dataptr
> > >
> > > Notice the pattern: Each CPU does store-mb-load. It is known that on
> > > some architectures each CPU can end up loading the old value (the value
> > > from before the other CPU's store). This would mean the writer would see
> > > hardluckref == 0 right away and the reader would see the old dataptr.
> >
> > So, if we have global A == B == 0,
> >
> > CPU_0 CPU_1
> >
> > A = 1; B = 2;
> > mb(); mb();
> > b = B; a = A;
> >
> > It could happen that a == b == 0, yes?
>
> Exactly.
>
> > Isn't this contradicts with definition
> > of mb?
>
> One might think so, at first. But if you do a careful search, you'll find
> that there _is_ no definition of mb! People state in vague terms what
> it's supposed to do, but they are almost never specific enough to tell
> whether the example above should work.

Yep -- mb() is currently defined only for specific CPUs. :-/

Some Linux kernel code has been written by considering each SMP-capable
CPU in turn, but that does not scale with increasing numbers of SMP-capable
CPUs.

> > By definition, when CPU_0 issues 'b = B', 'A = 1' should be visible to other
> > CPUs, yes?
>
> No. Memory barriers don't guarantee that any particular store will become
> visible to other CPUs at any particular time. They guarantee only that a
> certain sequence of stores will become visible in a particular order
> (provided the other CPUs also use the correct memory barriers).
>
> > Now, b == 0 means that CPU_1 did not read 'a = A' yet, otherwise
> > 'B = 2' should be visible to all CPUs (by definition again).
> >
> > Could you please clarify this?
>
> Here's an example showing how the code can fail. (Paul can correct me if
> I get this wrong.)

Looks good to me!

Thanx, Paul
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