Re: [V2 PATCH 2/3] kexec: Fix race between panic() and crash_kexec() called directly

[Michal Hocko <mhocko@kernel.org>]

On Mon 27-07-15 10:58:50, Hidehiro Kawai wrote:
[...]
> @@ -1472,6 +1472,18 @@ void __weak crash_unmap_reserved_pages(void)
>  
>  void crash_kexec(struct pt_regs *regs)
>  {
> +	int old_cpu, this_cpu;
> +
> +	/*
> +	 * `old_cpu == -1' means we are the first comer and crash_kexec()
> +	 * was called without entering panic().
> +	 * `old_cpu == this_cpu' means crash_kexec() was called from panic().
> +	 */
> +	this_cpu = raw_smp_processor_id();
> +	old_cpu = atomic_cmpxchg(&panicking_cpu, -1, this_cpu);
> +	if (old_cpu != -1 && old_cpu != this_cpu)
> +		return;
> +
>  	/* Take the kexec_mutex here to prevent sys_kexec_load
>  	 * running on one cpu from replacing the crash kernel
>  	 * we are using after a panic on a different cpu.
> @@ -1491,6 +1503,14 @@ void crash_kexec(struct pt_regs *regs)
>  		}
>  		mutex_unlock(&kexec_mutex);
>  	}
> +
> +	/*
> +	 * If we came here from panic(), we have to keep panicking_cpu
> +	 * to prevent other cpus from entering panic().  Otherwise,
> +	 * resetting it so that other cpus can enter panic()/crash_kexec().
> +	 */
> +	if (old_cpu == this_cpu)
> +		atomic_set(&panicking_cpu, -1);

This do the opposite what the comment says, wouldn't it? You should
check old_cpu == -1. Also atomic_set doesn't imply memory barriers which
might be a problem.
-- 
Michal Hocko
SUSE Labs