On Wed, Mar 23, 2005 at 10:46:45PM +0100, Ingo Molnar wrote:
>
> * Ingo Molnar <[email protected]> wrote:
>
> > i think the 'migrate read-count' method is not adequate either,
> > because all callbacks queued within an RCU read section must be called
> > after the lock has been dropped - while with the migration method
> > CPU#1 would be free to process callbacks queued in the RCU read
> > section still active on CPU#2.
> >
> > i'm wondering how much of a problem this is though. Can there be stale
> > pointers at that point? Yes in theory, because code like:
> >
> > rcu_read_lock();
> > call_rcu(&dentry->d_rcu, d_callback);
> > func(dentry->whatever);
> > rcu_read_unlock();
>
> but, this cannot happen, because call_rcu() is used by RCU-write code.
The code would not look exactly like this, but acquiring the update-side
lock inside an RCU read-side critical section can be thought of as
a reader-to-writer lock upgrade. RCU can do this unconditionally,
which was one of the walls I was banging my head against when trying
to think up a realtime-safe RCU implementation.
So something like the following would be legitimate RCU code:
rcu_read_lock();
spin_lock(&dcache_lock);
call_rcu(&dentry->d_rcu, d_callback);
spin_unlock(&dcache_lock);
rcu_read_unlock();
The spin_lock() call upgrades from a read-side to a write-side critical
section.
> so the important property seems to be that any active RCU-read section
> should keep at least one CPU's active_readers count elevated
> permanently, for the duration of the RCU-read section.
Yep!
> It doesnt matter
> that the reader migrates between CPUs - because the RCU code itself
> guarantees that no callbacks will be executed until _all_ CPUs have been
> in quiescent state. I.e. all CPUs have to go through a zero
> active_readers value before the callback is done.
Yep again!
Thanx, Paul
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