On 2021-02-19 10:33:36 [-0800], Paul E. McKenney wrote:
> For definiteness, here is the first part of the change, posted earlier.
> The commit log needs to be updated. I will post the change that keeps
> the tick going as a reply to this email.
…
> diff --git a/kernel/softirq.c b/kernel/softirq.c
> index 9d71046..ba78e63 100644
> --- a/kernel/softirq.c
> +++ b/kernel/softirq.c
> @@ -209,7 +209,7 @@ static inline void invoke_softirq(void)
> if (ksoftirqd_running(local_softirq_pending()))
> return;
>
> - if (!force_irqthreads) {
> + if (!force_irqthreads || !__this_cpu_read(ksoftirqd)) {
> #ifdef CONFIG_HAVE_IRQ_EXIT_ON_IRQ_STACK
> /*
> * We can safely execute softirq on the current stack if
> @@ -358,8 +358,8 @@ asmlinkage __visible void __softirq_entry
> __do_softirq(void)
>
> pending = local_softirq_pending();
> if (pending) {
> - if (time_before(jiffies, end) && !need_resched() &&
> - --max_restart)
> + if (!__this_cpu_read(ksoftirqd) ||
> + (time_before(jiffies, end) && !need_resched() &&
> --max_restart))
> goto restart;
This is hunk shouldn't be needed. The reason for it is probably that the
following wakeup_softirqd() would avoid further invoke_softirq()
performing the actual softirq work. It would leave early due to
ksoftirqd_running(). Unless I'm wrong, any raise_softirq() invocation
outside of an interrupt would do the same.
I would like PeterZ / tglx to comment on this one. Basically I'm not
sure if it is okay to expect softirqs beeing served and waited on that
early in the boot.
> wakeup_softirqd();
Sebastian
