Re: [PATCH 11/18] seqcount: Introduce raw_write_seqcount_barrier()

[Oleg Nesterov]

On 06/11, Paul E. McKenney wrote:
>
> > > + *      seqcount_t seq;
> > > + *      bool X = true, Y = false;
> > > + *
> > > + *      void read(void)
> > > + *      {
> > > + *              bool x, y;
> > > + *
> > > + *              do {
> > > + *                      int s = read_seqcount_begin(&seq);
> > > + *
> > > + *                      x = X; y = Y;
> > > + *
> > > + *              } while (read_seqcount_retry(&seq, s));
> > > + *
> > > + *              BUG_ON(!x && !y);
> > > + *      }
> > > + *
> > > + *      void write(void)
> > > + *      {
> > > + *              Y = true;
> > > + *
> > > + *              write_seqcount_begin(seq);
> > > + *              write_seqcount_end(seq);
> > > + *
> > > + *              X = false;
> > > + *      }
> >
> > > +static inline void raw_write_seqcount_barrier(seqcount_t *s)
> > > +{
> > > +	s->sequence++;
> > > +	smp_wmb();
> > > +	s->sequence++;
> > > +}
> > > +
> > >  /*
> > >   * raw_write_seqcount_latch - redirect readers to even/odd copy
> > >   * @s: pointer to seqcount_t
> >
> > Looks good otherwise.
> >
> > Reviewed-by: Paul E. McKenney <paulmck@xxxxxxxxxxxxxxxxxx>
>
> Color me slow and stupid.  Maybe due to reviewing a patch too early in
> the morning, who knows?
>
> There is nothing above that prevents the compiler and the CPU from
> reordering the assignments to X and Y with the increment of s->sequence++.

Yes, but this doesn't matter, I think. The writer does

	Y = true;
	1st_increment;

	wmb();

	2nd_increment;
	X = false;

and we do not care about reordering before or after wmnb() at all. But we
rely on the fact that 1st_increment can not be reordered with "X = false",
and that "Y = true" can not be reordered with the 2nd_increment.


And another simple "proof" is that  seqcount_barrier() is equivalent to
write_seqcount_begin() + + write_seqcount_end() and thus the code above
is correct, or the ACQUIRE/RELEASE semantics of seqcount_t is broken ;)

Oleg.

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