[V2 PATCH 2/3] kexec: Fix race between panic() and crash_kexec() called directly
Michal Hocko
mhocko at kernel.org
Mon Jul 27 07:55:06 PDT 2015
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
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