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?
--
To unsubscribe from this list: send the line "unsubscribe linux-m68k" in
the body of a message to majordomo@xxxxxxxxxxxxxxx
More majordomo info at http://vger.kernel.org/majordomo-info.html
