On Thu, 24 Jan 2013 10:00:04 +0530, Srivatsa S. Bhat wrote:
On 01/24/2013 01:27 AM, Tejun Heo wrote:
quoted
On Thu, Jan 24, 2013 at 01:03:52AM +0530, Srivatsa S. Bhat wrote:
quoted
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..
Thanks,
Namhyung
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:
quoted
On 01/24/2013 01:27 AM, Tejun Heo wrote:
quoted
On Thu, Jan 24, 2013 at 01:03:52AM +0530, Srivatsa S. Bhat wrote:
quoted
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
On 02/09/2013 04:17 AM, Paul E. McKenney wrote:
On Tue, Jan 29, 2013 at 08:12:37PM +0900, Namhyung Kim wrote:
quoted
On Thu, 24 Jan 2013 10:00:04 +0530, Srivatsa S. Bhat wrote:
quoted
On 01/24/2013 01:27 AM, Tejun Heo wrote:
quoted
On Thu, Jan 24, 2013 at 01:03:52AM +0530, Srivatsa S. Bhat wrote:
quoted
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.
Thanks a lot for confirming that Paul! So I guess we can use rwlocks
as it is, since its behaviour suits our needs perfectly. So I won't tinker
with atomic counters for a while, atleast not until someone starts making
rwlocks fair.. ;-)
Regards,
Srivatsa S. Bhat