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Date:	Sat, 21 Sep 2013 09:47:26 +0200
From:	Ingo Molnar <mingo@...nel.org>
To:	Linus Torvalds <torvalds@...ux-foundation.org>
Cc:	Frederic Weisbecker <fweisbec@...il.com>,
	Thomas Gleixner <tglx@...utronix.de>,
	LKML <linux-kernel@...r.kernel.org>,
	Benjamin Herrenschmidt <benh@...nel.crashing.org>,
	Paul Mackerras <paulus@....ibm.com>,
	Peter Zijlstra <peterz@...radead.org>,
	"H. Peter Anvin" <hpa@...or.com>,
	James Hogan <james.hogan@...tec.com>,
	"James E.J. Bottomley" <jejb@...isc-linux.org>,
	Helge Deller <deller@....de>,
	Martin Schwidefsky <schwidefsky@...ibm.com>,
	Heiko Carstens <heiko.carstens@...ibm.com>,
	"David S. Miller" <davem@...emloft.net>,
	Andrew Morton <akpm@...ux-foundation.org>
Subject: Re: [RFC GIT PULL] softirq: Consolidation and stack overrun fix


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

> On Fri, Sep 20, 2013 at 9:26 AM, Frederic Weisbecker <fweisbec@...il.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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