Messages in this thread | | | Date | Fri, 29 Jun 2012 13:29:35 -0700 | From | Andrew Morton <> | Subject | Re: [PATCH] panic: Fix a possible deadlock in panic() |
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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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