Re: rcu alignment warning tripping on m68k
From: Mikael Pettersson
Date: Sat Jun 07 2014 - 09:17:41 EST
Paul E. McKenney writes:
> On Fri, May 30, 2014 at 11:29:41AM +1000, Greg Ungerer wrote:
> > On 29/05/14 23:11, One Thousand Gnomes wrote:
> > > On Thu, 29 May 2014 12:08:32 +1000
> > > Greg Ungerer <gerg@xxxxxxxxxxx> wrote:
> > >
> > >> Hi All,
> > >>
> > >> Inside kernel/rcy/tree.c in __call_rcu() it does an alignment check on
> > >> the head pointer passed in. This trips on m68k systems, because they only
> > >> need alignment of 32bit quantities to 16bit boundaries.
> > >
> > > __alignof perhaps ?
> > That might do. Change then becomes something like:
> > --- a/kernel/rcu/tree.c
> > +++ b/kernel/rcu/tree.c
> > @@ -2467,7 +2467,7 @@ __call_rcu(struct rcu_head *head, void (*func)(struct rcu_
> > unsigned long flags;
> > struct rcu_data *rdp;
> > - WARN_ON_ONCE((unsigned long)head & 0x3); /* Misaligned rcu_head! */
> > + WARN_ON_ONCE((unsigned long)head & (__alignof__(head) - 1)); /* Misaligned rcu_head! */
> Hmmm... The purpose of the check is to reserve the low-order bits to
> allow RCU to classify callbacks as being time-critical or not. RCU
> can probably live with a single bit, but if there is some architecture
> out there that simply refuses to do alignment, I need to know about it.
> (See "git show 0bb7b59d6e2b8" for more info.)
> So how about this instead?
> - WARN_ON_ONCE((unsigned long)head & 0x1); /* Misaligned rcu_head! */
> (Trying to remember if I have seen Linux kernel code that uses both
> the lower bits...)
As stated above, m68k-linux aligns to 16-bit boundaries by default, so you'd
get one bit but not necessarily more. If you want more free low bits, why
not attach an explicit attribute aligned to the rcu_head type declaration?
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