Re: [RFCv2 5/9] arch/powerpc: Split hash page table sizing heuristic into a helper

[Anshuman Khandual <khandual@linux.vnet.ibm.com>]

On 01/29/2016 10:53 AM, David Gibson wrote:
> htab_get_table_size() either retrieve the size of the hash page table (HPT)
> from the device tree - if the HPT size is determined by firmware - or
> uses a heuristic to determine a good size based on RAM size if the kernel
> is responsible for allocating the HPT.
> 
> To support a PAPR extension allowing resizing of the HPT, we're going to
> want the memory size -> HPT size logic elsewhere, so split it out into a
> helper function.
> 
> Signed-off-by: David Gibson <david@gibson.dropbear.id.au>
> ---
>  arch/powerpc/include/asm/mmu-hash64.h |  3 +++
>  arch/powerpc/mm/hash_utils_64.c       | 30 +++++++++++++++++-------------
>  2 files changed, 20 insertions(+), 13 deletions(-)
> 
> diff --git a/arch/powerpc/include/asm/mmu-hash64.h b/arch/powerpc/include/asm/mmu-hash64.h
> index 7352d3f..cf070fd 100644
> --- a/arch/powerpc/include/asm/mmu-hash64.h
> +++ b/arch/powerpc/include/asm/mmu-hash64.h
> @@ -607,6 +607,9 @@ static inline unsigned long get_kernel_vsid(unsigned long ea, int ssize)
>  	context = (MAX_USER_CONTEXT) + ((ea >> 60) - 0xc) + 1;
>  	return get_vsid(context, ea, ssize);
>  }
> +
> +unsigned htab_shift_for_mem_size(unsigned long mem_size);
> +
>  #endif /* __ASSEMBLY__ */
>  
>  #endif /* _ASM_POWERPC_MMU_HASH64_H_ */
> diff --git a/arch/powerpc/mm/hash_utils_64.c b/arch/powerpc/mm/hash_utils_64.c
> index e88a86e..d63f7dc 100644
> --- a/arch/powerpc/mm/hash_utils_64.c
> +++ b/arch/powerpc/mm/hash_utils_64.c
> @@ -606,10 +606,24 @@ static int __init htab_dt_scan_pftsize(unsigned long node,
>  	return 0;
>  }
>  
> -static unsigned long __init htab_get_table_size(void)
> +unsigned htab_shift_for_mem_size(unsigned long mem_size)
>  {
> -	unsigned long mem_size, rnd_mem_size, pteg_count, psize;
> +	unsigned memshift = __ilog2(mem_size);
> +	unsigned pshift = mmu_psize_defs[mmu_virtual_psize].shift;
> +	unsigned pteg_shift;
> +
> +	/* round mem_size up to next power of 2 */
> +	if ((1UL << memshift) < mem_size)
> +		memshift += 1;
> +
> +	/* aim for 2 pages / pteg */

While here I guess its a good opportunity to write couple of lines
about why one PTE group for every two physical pages on the system,
why minimum (1UL << 11 = 2048) number of PTE groups required, why
(1U << 7 = 128) entries per PTE group and also remove the existing
confusing comments above ? Just a suggestion.

> +	pteg_shift = memshift - (pshift + 1);
> +
> +	return max(pteg_shift + 7, 18U);
> +}
>  
> +static unsigned long __init htab_get_table_size(void)
> +{
>  	/* If hash size isn't already provided by the platform, we try to
>  	 * retrieve it from the device-tree. If it's not there neither, we
>  	 * calculate it now based on the total RAM size
> @@ -619,17 +633,7 @@ static unsigned long __init htab_get_table_size(void)
>  	if (ppc64_pft_size)
>  		return 1UL << ppc64_pft_size;
>  
> -	/* round mem_size up to next power of 2 */
> -	mem_size = memblock_phys_mem_size();
> -	rnd_mem_size = 1UL << __ilog2(mem_size);
> -	if (rnd_mem_size < mem_size)
> -		rnd_mem_size <<= 1;
> -
> -	/* # pages / 2 */
> -	psize = mmu_psize_defs[mmu_virtual_psize].shift;
> -	pteg_count = max(rnd_mem_size >> (psize + 1), 1UL << 11);
> -
> -	return pteg_count << 7;
> +	return htab_shift_for_mem_size(memblock_phys_mem_size());

Would it be 1UL << htab_shift_for_mem_size(memblock_phys_mem_size())
instead ? It was returning the size of the HPT not the shift of HPT
originally or I am missing something here.