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SubjectRe: [PATCH] panic: Fix a possible deadlock in panic()
On Thu, 28 Jun 2012 16:43:05 -0700
Vikram Mulukutla <markivx@codeaurora.org> wrote:

> panic_lock is meant to ensure that panic processing takes
> place only on one cpu; if any of the other cpus encounter
> a panic, they will spin waiting to be shut down.
>
> However, this causes a regression in this scenario:
>
> 1. Cpu 0 encounters a panic and acquires the panic_lock
> and proceeds with the panic processing.
> 2. There is an interrupt on cpu 0 that also encounters
> an error condition and invokes panic.
> 3. This second invocation fails to acquire the panic_lock
> and enters the infinite while loop in panic_smp_self_stop.
>
> Thus all panic processing is stopped, and the cpu is stuck
> for eternity in the while(1) inside panic_smp_self_stop.
>
> To address this, disable local interrupts with
> local_irq_disable before acquiring the panic_lock. This will
> prevent interrupt handlers from executing during the panic
> processing, thus avoiding this particular problem.
>
> --- a/kernel/panic.c
> +++ b/kernel/panic.c
> @@ -75,6 +75,14 @@ void panic(const char *fmt, ...)
> int state = 0;
>
> /*
> + * Disable local interrupts. This will prevent panic_smp_self_stop
> + * from deadlocking the first cpu that invokes the panic, since
> + * there is nothing to prevent an interrupt handler (that runs
> + * after the panic_lock is acquired) from invoking panic again.
> + */
> + local_irq_disable();
> +
> + /*
> * It's possible to come here directly from a panic-assertion and
> * not have preempt disabled. Some functions called from here want
> * preempt to be disabled. No point enabling it later though...

Seems sane. panic() *should* work correctly when called with
interrupts disabled, so there be no bad effects from internally
disabling interrupts. If there are bad effects, we should fix them up.



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