Re: [PATCH] panic: Fix a possible deadlock in panic()

[Andrew Morton]

On Thu, 28 Jun 2012 16:43:05 -0700
Vikram Mulukutla <markivx@xxxxxxxxxxxxxx> 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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