Re: [PATCH v4 2/2] mm: Init page count in reserve_bootmem_region when MEMINIT_EARLY

[David Hildenbrand <david@redhat.com>]

On 16.10.23 10:10, Yajun Deng wrote:
> 
> On 2023/10/16 14:33, Mike Rapoport wrote:
>> On Fri, Oct 13, 2023 at 05:29:19PM +0800, Yajun Deng wrote:
>>> On 2023/10/13 16:48, Mike Rapoport wrote:
>>>> On Thu, Oct 12, 2023 at 05:53:22PM +0800, Yajun Deng wrote:
>>>>> On 2023/10/12 17:23, David Hildenbrand wrote:
>>>>>> On 10.10.23 04:31, Yajun Deng wrote:
>>>>>>> On 2023/10/8 16:57, Yajun Deng wrote:
>>>>>>>>> That looks wrong. if the page count would by pure luck be 0
>>>>>>>>> already for hotplugged memory, you wouldn't clear the reserved
>>>>>>>>> flag.
>>>>>>>>>
>>>>>>>>> These changes make me a bit nervous.
>>>>>>>> Is 'if (page_count(page) || PageReserved(page))' be safer? Or do I
>>>>>>>> need to do something else?
>>>>>>>>
>>>>>>> How about the following if statement? But it needs to add more patch
>>>>>>> like v1 ([PATCH 2/4] mm: Introduce MEMINIT_LATE context).
>>>>>>>
>>>>>>> It'll be safer, but more complex. Please comment...
>>>>>>>
>>>>>>>        if (context != MEMINIT_EARLY || (page_count(page) ||
>>>>>>> PageReserved(page)) {
>>>>>>>
>>>>>> Ideally we could make initialization only depend on the context, and not
>>>>>> check for count or the reserved flag.
>>>>>>
>>>>> This link is v1,
>>>>> https://lore.kernel.org/all/20230922070923.355656-1-yajun.deng@linux.dev/
>>>>>
>>>>> If we could make initialization only depend on the context, I'll modify it
>>>>> based on v1.
>>>> Although ~20% improvement looks impressive, this is only optimization of a
>>>> fraction of the boot time, and realistically, how much 56 msec saves from
>>>> the total boot time when you boot a machine with 190G of RAM?
>>> There are a lot of factors that can affect the total boot time. 56 msec
>>> saves may be insignificant.
>>>
>>> But if we look at the boot log, we'll see there's a significant time jump.
>>>
>>> before:
>>>
>>> [    0.250334] ACPI: PM-Timer IO Port: 0x508
>>> [    0.618994] Memory: 173413056K/199884452K available (18440K kernel code,
>>>
>>> after:
>>>
>>> [    0.260229] software IO TLB: area num 32.
>>> [    0.563497] Memory: 173413056K/199884452K available (18440K kernel code,
>>> Memory:
>>> Memory initialization is time consuming in the boot log.
>> You just confirmed that 56 msec is insignificant and then you send again
>> the improvement of ~60 msec in memory initialization.
>>
>> What does this improvement gain in percentage of total boot time?
> 
> 
> before:
> 
> [   10.692708] Run /init as init process
> 
> 
> after:
> 
> [   10.666290] Run /init as init process
> 
> 
> About 0.25%. The total boot time is variable, depending on how many
> drivers need to be initialized.
> 
> 
>>    
>>>> I still think the improvement does not justify the churn, added complexity
>>>> and special casing of different code paths of initialization of struct pages.
>>>
>>> Because there is a loop, if the order is MAX_ORDER, the loop will run 1024
>>> times. The following 'if' would be safer:
>>>
>>> 'if (context != MEMINIT_EARLY || (page_count(page) || >> PageReserved(page))
>>> {'
>> No, it will not.
>>
>> As the matter of fact any condition here won't be 'safer' because it makes
>> the code more complex and less maintainable.
>> Any future change in __free_pages_core() or one of it's callers will have
>> to reason what will happen with that condition after the change.
> 
> 
> To avoid introducing MEMINIT_LATE context and make code simpler. This
> might be a better option.
> 
> if (page_count(page) || PageReserved(page))

I'll have to side with Mike here; this change might not be worth it.

-- 
Cheers,

David / dhildenb