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SubjectRe: [PATCH V2 2/2] mm/memblock.c: check return value of memblock_reserve() in memblock_virt_alloc_internal()
On Wed, Dec 21, 2016 at 02:22:01PM +0100, Michal Hocko wrote:
>On Wed 21-12-16 13:13:32, Wei Yang wrote:
>> On Wed, Dec 21, 2016 at 08:51:16AM +0100, Michal Hocko wrote:
>> >On Tue 20-12-16 16:48:23, Wei Yang wrote:
>> >> On Mon, Dec 19, 2016 at 04:21:57PM +0100, Michal Hocko wrote:
>> >> >On Sun 18-12-16 14:47:50, Wei Yang wrote:
>> >> >> memblock_reserve() may fail in case there is not enough regions.
>> >> >
>> >> >Have you seen this happenning in the real setups or this is a by-review
>> >> >driven change?
>> >>
>> >> This is a by-review driven change.
>> >>
>> >> >[...]
>> >> >> again:
>> >> >> alloc = memblock_find_in_range_node(size, align, min_addr, max_addr,
>> >> >> nid, flags);
>> >> >> - if (alloc)
>> >> >> + if (alloc && !memblock_reserve(alloc, size))
>> >> >> goto done;
>> >
>> >So how exactly does the reserve fail when memblock_find_in_range_node
>> >found a suitable range for the given size?
>> >
>>
>> Even memblock_find_in_range_node() gets a suitable range, memblock_reserve()
>> still could fail. And the case just happens when memblock can't resize.
>> memblock_reserve() reserve a range by adding a range to memblock.reserved. In
>> case the memblock.reserved is full and can't resize, this fails.
>
>Sorry for being dense but what does it mean that the reserved will get
>full? Also how probable is such a situation? Is it even real? In other
>words does this fix a real or only a theoretical problem?
>

This is a theoretical problem. While if happens, it is hard to detect. Future
allocator will think this range is still available.

>Anyway this all should be part of the changelog.

Ok, let me add this in changelog in next version.

>--
>Michal Hocko
>SUSE Labs

--
Wei Yang
Help you, Help me

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