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Message-ID: <8a42d52c-6275-4798-19c0-dfc530c04b0e@linux.ibm.com>
Date:   Wed, 8 Jun 2022 10:00:41 +0530
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Subject: Re: [PATCH v5 1/9] mm/demotion: Add support for explicit memory tiers
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To:     Tim Chen <tim.c.chen@linux.intel.com>, linux-mm@kvack.org,
        akpm@linux-foundation.org
Cc:     Wei Xu <weixugc@google.com>, Huang Ying <ying.huang@intel.com>,
        Greg Thelen <gthelen@google.com>,
        Yang Shi <shy828301@gmail.com>,
        Davidlohr Bueso <dave@stgolabs.net>,
        Tim C Chen <tim.c.chen@intel.com>,
        Brice Goglin <brice.goglin@gmail.com>,
        Michal Hocko <mhocko@kernel.org>,
        Linux Kernel Mailing List <linux-kernel@vger.kernel.org>,
        Hesham Almatary <hesham.almatary@huawei.com>,
        Dave Hansen <dave.hansen@intel.com>,
        Jonathan Cameron <Jonathan.Cameron@huawei.com>,
        Alistair Popple <apopple@nvidia.com>,
        Dan Williams <dan.j.williams@intel.com>,
        Feng Tang <feng.tang@intel.com>,
        Jagdish Gediya <jvgediya@linux.ibm.com>,
        Baolin Wang <baolin.wang@linux.alibaba.com>,
        David Rientjes <rientjes@google.com>
References: <20220603134237.131362-1-aneesh.kumar@linux.ibm.com>
 <20220603134237.131362-2-aneesh.kumar@linux.ibm.com>
 <92649c9a6e0b6931b34aeaaf22c0a1e874484b7f.camel@linux.intel.com>
From:   Aneesh Kumar K V <aneesh.kumar@linux.ibm.com>
In-Reply-To: <92649c9a6e0b6931b34aeaaf22c0a1e874484b7f.camel@linux.intel.com>
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On 6/8/22 12:13 AM, Tim Chen wrote:
> On Fri, 2022-06-03 at 19:12 +0530, Aneesh Kumar K.V wrote:
>>
>>
>> The nodes which are part of a specific memory tier can be listed
>> via
>> /sys/devices/system/memtier/memtierN/nodelist
>>
>> "Rank" is an opaque value. Its absolute value doesn't have any
>> special meaning. But the rank values of different memtiers can be
>> compared with each other to determine the memory tier order.
>>
>> For example, if we have 3 memtiers: memtier0, memtier1, memiter2, and
>> their rank values are 300, 200, 100, then the memory tier order is:
>> memtier0 -> memtier2 -> memtier1,
> 
> Why is memtier2 (rank 100) higher than memtier1 (rank 200)?  Seems like
> the order should be memtier0 -> memtier1 -> memtier2?
>                      (rank 300)  (rank 200)  (rank 100)
> 
>> where memtier0 is the highest tier
>> and memtier1 is the lowest tier.
> 
> I think memtier2 is the lowest as it has the lowest rank value.


typo error. Will fix that in the next update

>>
>> The rank value of each memtier should be unique.
>>
>>
>> +
>> +static void memory_tier_device_release(struct device *dev)
>> +{
>> +	struct memory_tier *tier = to_memory_tier(dev);
>> +
> 
> Do we need some ref counts on memory_tier?
> If there is another device still using the same memtier,
> free below could cause problem.
> 
>> +	kfree(tier);
>> +}
>> +
>>
> ...
>> +static struct memory_tier *register_memory_tier(unsigned int tier)
>> +{
>> +	int error;
>> +	struct memory_tier *memtier;
>> +
>> +	if (tier >= MAX_MEMORY_TIERS)
>> +		return NULL;
>> +
>> +	memtier = kzalloc(sizeof(struct memory_tier), GFP_KERNEL);
>> +	if (!memtier)
>> +		return NULL;
>> +
>> +	memtier->dev.id = tier;
>> +	memtier->rank = get_rank_from_tier(tier);
>> +	memtier->dev.bus = &memory_tier_subsys;
>> +	memtier->dev.release = memory_tier_device_release;
>> +	memtier->dev.groups = memory_tier_dev_groups;
>> +
> 
> Should you take the mem_tier_lock before you insert to
> memtier-list?


Both register_memory_tier and unregister_memory_tier get called with 
memory_tier_lock held.

> 
>> +	insert_memory_tier(memtier);
>> +
>> +	error = device_register(&memtier->dev);
>> +	if (error) {
>> +		list_del(&memtier->list);
>> +		put_device(&memtier->dev);
>> +		return NULL;
>> +	}
>> +	return memtier;
>> +}
>> +
>> +__maybe_unused // temporay to prevent warnings during bisects
>> +static void unregister_memory_tier(struct memory_tier *memtier)
>> +{
> 
> I think we should take mem_tier_lock before modifying memtier->list.
> 

unregister_memory_tier get called with memory_tier_lock held.

>> +	list_del(&memtier->list);
>> +	device_unregister(&memtier->dev);
>> +}
>> +
>>

-aneesh