On Thu, Dec 03, 2015 at 11:29:21AM +0000, ???? / KAWAI?HIDEHIRO wrote:
> I was arguing about the case of oops_end --> crash_kexec
> --> return from crash_kexec because of !kexec_crash_image -->
> panic.
Aha.
> In the case of panic --> __crash_kexec, __crash_kexec is called
> only once, so we don't need to check the return value of __crash_kexec
> as you suggested. So I thought you stated about crash_kexec --> panic
> case.
No, I meant the other way around.
> I also mentioned !kexec_crash_image case...
I must've missed it.
> No. The first CPU calls panic, and then it calls __crash_kexec.
> Because of !kexec_crash_image, it returns from __crash_kexec and
> continues to the panic procedure. At the same time, another CPU
> tries to call panic(), but it doesn't run the panic procedure;
> panic_cpu prevents the second CPU from running it.
>
> This means __crash_kexec is called only once even if we don't
> check the return value of __crash_kexec.
I think we're on the same page, even if we express it differently - the
other CPUs entering panic() will loop in panic_smp_self_stop() so they
won't reach __crash_kexec().
> (Please note that crash_kexec can be called multiple times in the
> case of oops_end() --> crash_kexec().)
Right, and that was the case that was bugging me - calling into
crash_kexec() on multiple CPUs but it is a trylock and a pointer test -
I guess that's diminishingly small overhead to care.
Thanks.
--
Regards/Gruss,
Boris.
ECO tip #101: Trim your mails when you reply.
