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.
Not sure whether I get it clarified.
>> >>
>> >> if (nid != NUMA_NO_NODE) {
>> >> alloc = memblock_find_in_range_node(size, align, min_addr,
>> >> max_addr, NUMA_NO_NODE,
>> >> flags);
>> >> - if (alloc)
>> >> + if (alloc && !memblock_reserve(alloc, size))
>> >> goto done;
>> >> }
>> >
>> >This doesn't look right. You can end up leaking the first allocated
>> >range.
>> >
>>
>> Hmm... why?
>>
>> If first memblock_reserve() succeed, it will jump to done, so that no 2nd
>> allocation.
>> If the second executes, it means the first allocation failed.
>> memblock_find_in_range_node() doesn't modify the memblock, it just tell you
>> there is a proper memory region available.
>
>yes, my bad. I have missed this. Sorry about the confusion
So do you agree with my patch now?
>--
>Michal Hocko
>SUSE Labs
--
Wei Yang
Help you, Help me
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