Re: [PATCH v4 1/2] x86/setup: always add the beginning of RAM as memblock.memory

[Baoquan He]

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