Re: [PATCH 02/17] mm: Add folio_pincount_available()

[John Hubbard <jhubbard@nvidia.com>]

On 1/2/22 13:57, Matthew Wilcox (Oracle) wrote:
> Convert hpage_pincount_available() into folio_pincount_available() and
> turn hpage_pincount_available() into a wrapper.  We don't need to
> check folio_test_large() before checking folio_order() as
> folio_order() includes a check of folio_test_large().
> 
> Signed-off-by: Matthew Wilcox (Oracle) <willy@infradead.org>
> ---
>   include/linux/mm.h | 12 ++++++++----
>   1 file changed, 8 insertions(+), 4 deletions(-)
> 
> diff --git a/include/linux/mm.h b/include/linux/mm.h
> index 98a10412d581..269b5484d66e 100644
> --- a/include/linux/mm.h
> +++ b/include/linux/mm.h
> @@ -927,15 +927,19 @@ static inline void destroy_compound_page(struct page *page)
>   	compound_page_dtors[page[1].compound_dtor](page);
>   }
>   
> -static inline bool hpage_pincount_available(struct page *page)
> +static inline bool folio_pincount_available(struct folio *folio)
>   {
>   	/*
> -	 * Can the page->hpage_pinned_refcount field be used? That field is in
> +	 * Can the folio->hpage_pinned_refcount field be used? That field is in
>   	 * the 3rd page of the compound page, so the smallest (2-page) compound
>   	 * pages cannot support it.
>   	 */
> -	page = compound_head(page);
> -	return PageCompound(page) && compound_order(page) > 1;
> +	return folio_order(folio) > 1;

I see, no need to look at the compound page head, because folio_order() returns
zero for tail pages, and neatly avoids all of that.


> +}
> +
> +static inline bool hpage_pincount_available(struct page *page)
> +{
> +	return folio_pincount_available(page_folio(page));
>   }
>   
>   static inline int head_compound_pincount(struct page *head)


Reviewed-by: John Hubbard <jhubbard@nvidia.com>

thanks,
-- 
John Hubbard
NVIDIA