Re: [PATCH v5 04/45] percpu_rwlock: Implement the core design ofPer-CPU Reader-Writer Locks

From: Paul E. McKenney
Date: Fri Feb 08 2013 - 17:48:18 EST


On Tue, Jan 29, 2013 at 08:12:37PM +0900, Namhyung Kim wrote:
> On Thu, 24 Jan 2013 10:00:04 +0530, Srivatsa S. Bhat wrote:
> > On 01/24/2013 01:27 AM, Tejun Heo wrote:
> >> On Thu, Jan 24, 2013 at 01:03:52AM +0530, Srivatsa S. Bhat wrote:
> >>> CPU 0 CPU 1
> >>>
> >>> read_lock(&rwlock)
> >>>
> >>> write_lock(&rwlock) //spins, because CPU 0
> >>> //has acquired the lock for read
> >>>
> >>> read_lock(&rwlock)
> >>> ^^^^^
> >>> What happens here? Does CPU 0 start spinning (and hence deadlock) or will
> >>> it continue realizing that it already holds the rwlock for read?
> >>
> >> I don't think rwlock allows nesting write lock inside read lock.
> >> read_lock(); write_lock() will always deadlock.
> >>
> >
> > Sure, I understand that :-) My question was, what happens when *two* CPUs
> > are involved, as in, the read_lock() is invoked only on CPU 0 whereas the
> > write_lock() is invoked on CPU 1.
> >
> > For example, the same scenario shown above, but with slightly different
> > timing, will NOT result in a deadlock:
> >
> > Scenario 2:
> > CPU 0 CPU 1
> >
> > read_lock(&rwlock)
> >
> >
> > read_lock(&rwlock) //doesn't spin
> >
> > write_lock(&rwlock) //spins, because CPU 0
> > //has acquired the lock for read
> >
> >
> > So I was wondering whether the "fairness" logic of rwlocks would cause
> > the second read_lock() to spin (in the first scenario shown above) because
> > a writer is already waiting (and hence new readers should spin) and thus
> > cause a deadlock.
>
> In my understanding, current x86 rwlock does basically this (of course,
> in an atomic fashion):
>
>
> #define RW_LOCK_BIAS 0x10000
>
> rwlock_init(rwlock)
> {
> rwlock->lock = RW_LOCK_BIAS;
> }
>
> arch_read_lock(rwlock)
> {
> retry:
> if (--rwlock->lock >= 0)
> return;
>
> rwlock->lock++;
> while (rwlock->lock < 1)
> continue;
>
> goto retry;
> }
>
> arch_write_lock(rwlock)
> {
> retry:
> if ((rwlock->lock -= RW_LOCK_BIAS) == 0)
> return;
>
> rwlock->lock += RW_LOCK_BIAS;
> while (rwlock->lock != RW_LOCK_BIAS)
> continue;
>
> goto retry;
> }
>
>
> So I can't find where the 'fairness' logic comes from..

I looked through several of the rwlock implementations, and in all of
them the writer backs off if it sees readers -- or refrains from asserting
ownership of the lock to begin with.

So it should be OK to use rwlock as shown in the underlying patch.

Thanx, Paul

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