On Fri, Jun 15, 2018 at 11:57:33AM -0400, Pavel Tatashin wrote:
> The role of zero_resv_unavail() is to make sure that every struct page that
> is allocated but is not backed by memory that is accessible by kernel is
> zeroed and not in some uninitialized state.
>
> Since struct pages are allocate
On Fri, Jun 15, 2018 at 5:48 PM Andrew Morton wrote:
>
> On Fri, 15 Jun 2018 11:57:33 -0400 Pavel Tatashin
> wrote:
>
> > The role of zero_resv_unavail() is to make sure that every struct page that
> > is allocated but is not backed by memory that is accessible by kernel is
> > zeroed and not in
On Fri, 15 Jun 2018 11:57:33 -0400 Pavel Tatashin
wrote:
> The role of zero_resv_unavail() is to make sure that every struct page that
> is allocated but is not backed by memory that is accessible by kernel is
> zeroed and not in some uninitialized state.
>
> Since struct pages are allocated in
> Hi Pavel,
>
> Thanks for the patch.
> This looks good to me.
>
> Reviewed-by: Oscar Salvador
Thank you Oscar!
Pavel
On Fri, Jun 15, 2018 at 11:57:33AM -0400, Pavel Tatashin wrote:
> The role of zero_resv_unavail() is to make sure that every struct page that
> is allocated but is not backed by memory that is accessible by kernel is
> zeroed and not in some uninitialized state.
>
> Since struct pages are allocate
The role of zero_resv_unavail() is to make sure that every struct page that
is allocated but is not backed by memory that is accessible by kernel is
zeroed and not in some uninitialized state.
Since struct pages are allocated in blocks (2M pages in x86 case), we can
skip pageblock_nr_pages at a ti
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