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Message-ID: <20120524153409.GM1663@somewhere>
Date:	Thu, 24 May 2012 17:34:13 +0200
From:	Frederic Weisbecker <fweisbec@...il.com>
To:	Russ Anderson <rja@....com>
Cc:	linux-kernel@...r.kernel.org, x86@...nel.org,
	Thomas Gleixner <tglx@...utronix.de>,
	Ingo Molnar <mingo@...hat.com>,
	"H. Peter Anvin" <hpa@...or.com>,
	Andrew Morton <akpm@...ux-foundation.org>,
	Greg Kroah-Hartman <gregkh@...uxfoundation.org>
Subject: Re: [PATCH] x86: Avoid intermixing cpu dump_stack output on
 multi-processor systems

On Thu, May 24, 2012 at 09:42:29AM -0500, Russ Anderson wrote:
> When multiple cpus on a multi-processor system call dump_stack()
> at the same time, the backtrace lines get intermixed, making 
> the output worthless.  Add a lock so each cpu stack dump comes
> out as a coherent set.
> 
> For example, when a multi-processor system is NMIed, all of the
> cpus call dump_stack() at the same time, resulting in output for
> all of cpus getting intermixed, making it impossible to tell what
> any individual cpu was doing.  With this patch each cpu prints
> its stack lines as a coherent set, so one can see what each cpu
> was doing.
> 
> It has been tested on a 4069 cpu system.
> 
> Signed-off-by: Russ Anderson <rja@....com>

I don't think this is a good idea. What if an interrupt comes
and calls this at the same time? Sure you can mask irqs but NMIs
can call that too. In this case I prefer to have a messy report
rather than a deadlock on the debug path.

May be something like that:

static atomic_t dump_lock = ATOMIC_INIT(-1);

static void dump_stack(void)
{
	int was_locked;
	int old;
	int cpu;

	preempt_disable();
retry:
	cpu = smp_processor_id();
	old = atomic_cmpxchg(&dump_lock, -1, cpu);
	if (old == -1) {
		was_locked = 0;
	} else if (old == cpu) {
		was_locked = 1;
	} else {
		cpu_relax();
		goto retry;
	}

	__dump_trace();

	if (!was_locked)
		atomic_set(&dump_lock, -1);

	preempt_enable();
}

You could also use a spinlock with irq disabled and test in_nmi()
but we could have a dump_trace() in an NMI before the nmi count is
incremented. So the above is perhaps more robust.
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