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SubjectRe: [RFC v6 01/62] powerpc: Free up four 64K PTE bits in 4K backed HPTE pages
On Thu, Jul 20, 2017 at 11:21:51AM +0530, Aneesh Kumar K.V wrote:
>
> .....
>
> > /*
> > @@ -116,8 +104,8 @@ int __hash_page_4K(unsigned long ea, unsigned long access, unsigned long vsid,
> > * On hash insert failure we use old pte value and we don't
> > * want slot information there if we have a insert failure.
> > */
> > - old_pte &= ~(H_PAGE_HASHPTE | H_PAGE_F_GIX | H_PAGE_F_SECOND);
> > - new_pte &= ~(H_PAGE_HASHPTE | H_PAGE_F_GIX | H_PAGE_F_SECOND);
> > + old_pte &= ~(H_PAGE_HASHPTE);
> > + new_pte &= ~(H_PAGE_HASHPTE);
> > goto htab_insert_hpte;
> > }
>
> With the current path order and above hunk we will breaks the bisect I guess. With the above, when
> we convert a 64k hpte to 4khpte, since this is the first patch, we
> should clear that H_PAGE_F_GIX and H_PAGE_F_SECOND. We still use them
> for 64k. I guess you should move this hunk to second patch.

true. it should move to the next patch. Will fix it.
RP

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