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SubjectRe: [RFC GIT PULL] softirq: Consolidation and stack overrun fix

* Linus Torvalds <torvalds@linux-foundation.org> wrote:

> On Fri, Sep 20, 2013 at 9:26 AM, Frederic Weisbecker <fweisbec@gmail.com> wrote:
> >
> > Now just for clarity, what do we then do with inline sofirq
> > executions: on local_bh_enable() for example, or explicit calls to
> > do_softirq() other than irq exit?
>
> If we do a softirq because it was pending and we did a
> "local_bh_enable()" in normal code, we need a new stack. The
> "local_bh_enable()" may be pretty deep in the callchain on a normal
> process stack, so I think it would be safest to switch to a separate
> stack for softirq handling.
>
> So you have a few different cases:
>
> - irq_exit(). The irq stack is by definition empty (assuming itq_exit()
> is done on the irq stack), so doing softirq in that context should be
> fine. However, that assumes that if we get *another* interrupt, then
> we'll switch stacks again, so this does mean that we need two irq
> stacks. No, irq's don't nest, but if we run softirq on the first irq
> stack, the other irq *can* nest that softirq.
>
> - process context doing local_bh_enable, and a bh became pending while
> it was disabled. See above: this needs a stack switch. Which stack to
> use is open, again assuming that a hardirq coming in will switch to yet
> another stack.
>
> Hmm?

I'd definitely argue in favor of never letting unknown-size stacks nest
(i.e. to always switch if we start a new context on top of a non-trivial
stack).

Known (small) size stack nesting is not real stack nesting, it's just a
somewhat unusual (and faster) way of stack switching.

Thanks,

Ingo


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