Re: [PATCH v4 1/7] mm/zone_device: Add order argument to folio_free callback

Zi Yan
2026-01-12T17:47:53Z

On 12 Jan 2026, at 11:50, Jason Gunthorpe wrote:

> On Mon, Jan 12, 2026 at 11:31:04AM -0500, Zi Yan wrote:
>>> folio_free()
>>>
>>> 1) Allocator finds free memory
>>> 2) zone_device_page_init() allocates the memory and makes refcount=1
>>> 3) __folio_put() knows the recount 0.
>>> 4) free_zone_device_folio() calls folio_free(), but it doesn't
>>> actually need to undo prep_compound_page() because *NOTHING* can
>>> use the page pointer at this point.
>>> 5) Driver puts the memory back into the allocator and now #1 can
>>> happen. It knows how much memory to put back because folio->order
>>> is valid from #2
>>> 6) #1 happens again, then #2 happens again and the folio is in the
>>> right state for use. The successor #2 fully undoes the work of the
>>> predecessor #2.
>>
>> But how can a successor #2 undo the work if the second #1 only allocates
>> half of the original folio? For example, an order-9 at PFN 0 is
>> allocated and freed, then an order-8 at PFN 0 is allocated and another
>> order-8 at PFN 256 is allocated. How can two #2s undo the same order-9
>> without corrupting each other’s data?
>
> What do you mean? The fundamental rule is you can't read the folio or
> the order outside folio_free once it's refcount reaches 0.

There is no such a rule. In core MM, folio_split(), which splits a high
order folio to low order ones, freezes the folio (turning refcount to 0)
and manipulates the folio order and all tail pages compound_head to
restructure the folio. Your fundamental rule breaks this.
Allowing compound information to stay after a folio is freed means
you cannot tell whether a folio is under split or freed.

>
> So the successor #2 will write updated heads and order to the order 8
> pages at PFN 0 and the ones starting at PFN 256 will remain with
> garbage.
>
> This is OK because nothing is allowed to read them as their refcount
> is 0.
>
> If later PFN256 is allocated then it will get updated head and order
> at the same time it's refcount becomes 1.
>
> There is corruption and they don't corrupt each other's data.
>
>>> If the allocator is using the struct page memory then step #5 should
>>> also clean up the struct page with the allocator data before returning
>>> it to the allocator.
>>
>> Do you mean ->folio_free() callback should undo prep_compound_page()
>> instead?
>
> I wouldn't say undo, I was very careful to say it needs to get the
> struct page memory into a state that the allocator algorithm expects,
> whatever that means.
>
> Jason

Best Regards,
Yan, Zi