On Wed, Jul 13, 2022 at 09:32:33PM +0000, Joel Fernandes (Google) wrote:
> + * Furthermore, if CPU A invoked call_rcu() and CPU B invoked the
> + * resulting RCU callback function "func()", then both CPU A and CPU B are
> + * guaranteed to execute a full memory barrier during the time interval
> + * between the call to call_rcu() and the invocation of "func()" -- even
> + * if CPU A and CPU B are the same CPU (but again only if the system has
> + * more than one CPU).
> + *
> + * Implementation of these memory-ordering guarantees is described here:
> + * Documentation/RCU/Design/Memory-Ordering/Tree-RCU-Memory-Ordering.rst.
> + */
> +void call_rcu(struct rcu_head *head, rcu_callback_t func)
> +{
> + return __call_rcu_common(head, func, force_call_rcu_to_lazy);
> +
> +}
I will kill that extra new line. LOL something always has to go wrong no
matter how many hours you spend making sure eveyrthing is good :) Luckily
cosmetic here.
thanks,
- Joel
