On Mon, Feb 01, 2016 at 12:34:32PM +0530, Anshuman Khandual wrote: > 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 > > --- > > 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, Well, that don't really know, it's just copied from the existing code. > why minimum (1UL << 11 = 2048) number of PTE groups required, Ok. > why > (1U << 7 = 128) entries per PTE group Um.. what? Because that's how big a PTEG is, I don't think re-explaining the HPT structure here is useful. > and also remove the existing > confusing comments above ? Just a suggestion. Not sure which comment you mean. > > > + 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. Oops, yes. That would have broken all non-LPAR platforms. -- David Gibson | I'll have my music baroque, and my code david AT gibson.dropbear.id.au | minimalist, thank you. NOT _the_ _other_ | _way_ _around_! http://www.ozlabs.org/~dgibson