Re: rcu alignment warning tripping on m68k

[Mikael Pettersson]

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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