From: "Aneesh Kumar K.V" <aneesh.kumar@linux.vnet.ibm.com>
Subject: Re: [RFC v6 01/62] powerpc: Free up four 64K PTE bits in 4K backed HPTE pages
Date: Thu, 20 Jul 2017 11:21:51 +0530
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.....

>  	/*
> @@ -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.


-aneesh

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