Re: [RFC PATCH 1/3 v2] futex: introduce FUTEX_SWAP operation

[Andrei Vagin <avagin@gmail.com>]

On Tue, Jun 23, 2020 at 11:30:30AM -0700, Peter Oskolkov wrote:
...
> > >  /**
> > > +static int futex_swap(u32 __user *uaddr, unsigned int flags, u32
> > > val,
> > > +		      ktime_t *abs_time, u32 __user *uaddr2)
> > > +{
> > > +	u32 bitset = FUTEX_BITSET_MATCH_ANY;
> > > +	struct task_struct *next = NULL;
> > > +	DEFINE_WAKE_Q(wake_q);
> > > +	int ret;
> > > +
> > > +	ret = prepare_wake_q(uaddr2, flags, 1, bitset, &wake_q);
> > > +	if (!wake_q_empty(&wake_q)) {
> > > +		/* Pull the first wakee out of the queue to swap into.
> > > */
> > > +		next = container_of(wake_q.first, struct task_struct,
> > > wake_q);
> > > +		wake_q.first = wake_q.first->next;
> > > +		next->wake_q.next = NULL;
> > > +		/*
> > > +		 * Note that wake_up_q does not touch wake_q.last, so
> > > we
> > > +		 * do not bother with it here.
> > > +		 */
> > > +		wake_up_q(&wake_q);
> > 
> > wake_up_q() doesn't seem to serve any purpose in that the above
> > assignment of wake_q.first shall make it an empty queue now?
> > Also, I don't see a need to touch wake_q.first either so I think we
> > can
> > get rid of wake_q altogether here.
> 
> The futex at uaddr2 may have more than one waiter, so we cannot assume
> that wake_q will be empty when we remove the first element.

The third argument of prepare_wake_q is nr_wake which is one in this
case, so we can be sure that wake_q will be empty, can't we?

> 
> > 
> > > +	}
> > > +	if (ret < 0)
> > > +		return ret;
> > > +
> > > +	return futex_wait(uaddr, flags, val, abs_time, bitset, next);
> > > +}
> >