Re: [PATCH 2/7] mm: Free device private pages have zero refcount

[Alistair Popple <apopple@nvidia.com>]

Jason Gunthorpe <jgg@nvidia.com> writes:

> On Mon, Sep 26, 2022 at 04:03:06PM +1000, Alistair Popple wrote:
>> Since 27674ef6c73f ("mm: remove the extra ZONE_DEVICE struct page
>> refcount") device private pages have no longer had an extra reference
>> count when the page is in use. However before handing them back to the
>> owning device driver we add an extra reference count such that free
>> pages have a reference count of one.
>>
>> This makes it difficult to tell if a page is free or not because both
>> free and in use pages will have a non-zero refcount. Instead we should
>> return pages to the drivers page allocator with a zero reference count.
>> Kernel code can then safely use kernel functions such as
>> get_page_unless_zero().
>>
>> Signed-off-by: Alistair Popple <apopple@nvidia.com>
>> ---
>>  arch/powerpc/kvm/book3s_hv_uvmem.c       | 1 +
>>  drivers/gpu/drm/amd/amdkfd/kfd_migrate.c | 1 +
>>  drivers/gpu/drm/nouveau/nouveau_dmem.c   | 1 +
>>  lib/test_hmm.c                           | 1 +
>>  mm/memremap.c                            | 5 -----
>>  mm/page_alloc.c                          | 6 ++++++
>>  6 files changed, 10 insertions(+), 5 deletions(-)
>
> I think this is a great idea, but I'm surprised no dax stuff is
> touched here?

free_zone_device_page() shouldn't be called for pgmap->type ==
MEMORY_DEVICE_FS_DAX so I don't think we should have to worry about DAX
there. Except that the folio code looks like it might have introduced a
bug. AFAICT put_page() always calls
put_devmap_managed_page(&folio->page) but folio_put() does not (although
folios_put() does!). So it seems folio_put() won't end up calling
__put_devmap_managed_page_refs() as I think it should.

I think you're right about the change to __init_zone_device_page() - I
should limit it to DEVICE_PRIVATE/COHERENT pages only. But I need to
look at Dan's patch series more closely as I suspect it might be better
to rebase this patch on top of that.

> Jason