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

[Hidehiro Kawai <hidehiro.kawai.ez@hitachi.com>]

Hi,

(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)
>>  		}
>>  		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.

Sorry, you are right.  I performed same tests as for the
previous patch set, but I missed the test case for this
new logic.

> Also atomic_set doesn't imply memory barriers which
> might be a problem.

OK, I'll use atomic_xchg().

Regards,
-- 
Hidehiro Kawai
Hitachi, Ltd. Research & Development Group



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