On 02/01/21 at 04:34pm, Mike Rapoport wrote: > On Mon, Feb 01, 2021 at 07:26:05PM +0800, Baoquan He wrote: > > On 02/01/21 at 10:32am, David Hildenbrand wrote: > > > > > > 2) In init_zone_unavailable_mem(), similar to round_up(max_pfn, > > > PAGES_PER_SECTION) handling, consider range > > > [round_down(min_pfn, PAGES_PER_SECTION), min_pfn - 1] > > > which would handle in the x86-64 case [0..0] and, therefore, initialize PFN > > > 0. > > > > Sounds reasonable. Maybe we can change to get the real expected lowest > > pfn from find_min_pfn_for_node() by iterating memblock.memory and > > memblock.reserved and comparing. > > As I've found out the hard way [1], reserved memory is not necessary present. > > There could be a system that instead of reserving memory at 0xfe000000 like > in Guillaume's report, could have it reserved at 0x0 and populated only > from the first gigabyte... OK. I thought that we can even compare memblock.memory.regions[0].base with memblock.reserved.regions[0].base and take the smaller one as the lowest pfn and assign it to arch_zone_lowest_possible_pfn[0]. When we try to get the present pages, we still check memblock.memory with for_each_mem_pfn_range(). Since we will consider and take reserved memory into zone anyway, arch_zone_lowest_possible_pfn[] only impact the boundary of zone. Just rough thought, please ignore it if something is missed. Thanks Baoquan
