[V2 PATCH 2/3] kexec: Fix race between panic() and crash_kexec() called directly
hidehiro.kawai.ez at hitachi.com
Mon Jul 27 19:15:07 PDT 2015
(2015/07/27 23:55), Michal Hocko wrote:
> 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)
>> + /*
>> + * 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.
Sorry, you are right. I performed same tests as for the
previous patch set, but I missed the test case for this
> Also atomic_set doesn't imply memory barriers which
> might be a problem.
OK, I'll use atomic_xchg().
Hitachi, Ltd. Research & Development Group
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