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Message-ID: <CACZ9PQUO4cBsTdO37n4UWeHk=26g_WqWo-cVsDCf8E1gkq2Zkg@mail.gmail.com>
Date: Tue, 17 Mar 2015 17:22:46 +0900
From: Roman Peniaev <r.peniaev@...il.com>
To: Joonsoo Kim <iamjoonsoo.kim@....com>
Cc: Andrew Morton <akpm@...ux-foundation.org>,
Eric Dumazet <edumazet@...gle.com>,
David Rientjes <rientjes@...gle.com>,
WANG Chao <chaowang@...hat.com>,
Fabian Frederick <fabf@...net.be>,
Christoph Lameter <cl@...ux.com>, Gioh Kim <gioh.kim@....com>,
Rob Jones <rob.jones@...ethink.co.uk>, linux-mm@...ck.org,
"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>,
"stable@...r.kernel.org" <stable@...r.kernel.org>
Subject: Re: [PATCH 1/3] mm/vmalloc: fix possible exhaustion of vmalloc space
caused by vm_map_ram allocator
On Tue, Mar 17, 2015 at 4:29 PM, Joonsoo Kim <iamjoonsoo.kim@....com> wrote:
> On Tue, Mar 17, 2015 at 02:12:14PM +0900, Roman Peniaev wrote:
>> On Tue, Mar 17, 2015 at 1:56 PM, Joonsoo Kim <iamjoonsoo.kim@....com> wrote:
>> > On Fri, Mar 13, 2015 at 09:12:55PM +0900, Roman Pen wrote:
>> >> If suitable block can't be found, new block is allocated and put into a head
>> >> of a free list, so on next iteration this new block will be found first.
>> >>
>> >> That's bad, because old blocks in a free list will not get a chance to be fully
>> >> used, thus fragmentation will grow.
>> >>
>> >> Let's consider this simple example:
>> >>
>> >> #1 We have one block in a free list which is partially used, and where only
>> >> one page is free:
>> >>
>> >> HEAD |xxxxxxxxx-| TAIL
>> >> ^
>> >> free space for 1 page, order 0
>> >>
>> >> #2 New allocation request of order 1 (2 pages) comes, new block is allocated
>> >> since we do not have free space to complete this request. New block is put
>> >> into a head of a free list:
>> >>
>> >> HEAD |----------|xxxxxxxxx-| TAIL
>> >>
>> >> #3 Two pages were occupied in a new found block:
>> >>
>> >> HEAD |xx--------|xxxxxxxxx-| TAIL
>> >> ^
>> >> two pages mapped here
>> >>
>> >> #4 New allocation request of order 0 (1 page) comes. Block, which was created
>> >> on #2 step, is located at the beginning of a free list, so it will be found
>> >> first:
>> >>
>> >> HEAD |xxX-------|xxxxxxxxx-| TAIL
>> >> ^ ^
>> >> page mapped here, but better to use this hole
>> >>
>> >> It is obvious, that it is better to complete request of #4 step using the old
>> >> block, where free space is left, because in other case fragmentation will be
>> >> highly increased.
>> >>
>> >> But fragmentation is not only the case. The most worst thing is that I can
>> >> easily create scenario, when the whole vmalloc space is exhausted by blocks,
>> >> which are not used, but already dirty and have several free pages.
>> >>
>> >> Let's consider this function which execution should be pinned to one CPU:
>> >>
>> >> ------------------------------------------------------------------------------
>> >> /* Here we consider that our block is equal to 1MB, thus 256 pages */
>> >> static void exhaust_virtual_space(struct page *pages[256], int iters)
>> >> {
>> >> /* Firstly we have to map a big chunk, e.g. 16 pages.
>> >> * Then we have to occupy the remaining space with smaller
>> >> * chunks, i.e. 8 pages. At the end small hole should remain.
>> >> * So at the end of our allocation sequence block looks like
>> >> * this:
>> >> * XX big chunk
>> >> * |XXxxxxxxx-| x small chunk
>> >> * - hole, which is enough for a small chunk,
>> >> * but not for a big chunk
>> >> */
>> >> unsigned big_allocs = 1;
>> >> /* -1 for hole, which should be left at the end of each block
>> >> * to keep it partially used, with some free space available */
>> >> unsigned small_allocs = (256 - 16) / 8 - 1;
>> >> void *vaddrs[big_allocs + small_allocs];
>> >>
>> >> while (iters--) {
>> >> int i = 0, j;
>> >>
>> >> /* Map big chunk */
>> >> vaddrs[i++] = vm_map_ram(pages, 16, -1, PAGE_KERNEL);
>> >>
>> >> /* Map small chunks */
>> >> for (j = 0; j < small_allocs; j++)
>> >> vaddrs[i++] = vm_map_ram(pages + 16 + j * 8, 8, -1,
>> >> PAGE_KERNEL);
>> >>
>> >> /* Unmap everything */
>> >> while (i--)
>> >> vm_unmap_ram(vaddrs[i], (i ? 8 : 16));
>> >> }
>> >> }
>> >> ------------------------------------------------------------------------------
>> >>
>> >> On every iteration new block (1MB of vm area in my case) will be allocated and
>> >> then will be occupied, without attempt to resolve small allocation request
>> >> using previously allocated blocks in a free list.
>> >>
>> >> In current patch I simply put newly allocated block to the tail of a free list,
>> >> thus reduce fragmentation, giving a chance to resolve allocation request using
>> >> older blocks with possible holes left.
>> >
>> > Hello,
>> >
>> > I think that if you put newly allocated block to the tail of a free
>> > list, below example would results in enormous performance degradation.
>> >
>> > new block: 1MB (256 pages)
>> >
>> > while (iters--) {
>> > vm_map_ram(3 or something else not dividable for 256) * 85
>> > vm_unmap_ram(3) * 85
>> > }
>> >
>> > On every iteration, it needs newly allocated block and it is put to the
>> > tail of a free list so finding it consumes large amount of time.
>> >
>> > Is there any other solution to prevent your problem?
>>
>> Hello.
>>
>> My second patch fixes this problem.
>> I occupy the block on allocation and avoid jumping to the search loop.
>
> I'm not sure that this fixes above case.
> 'vm_map_ram (3) * 85' means 85 times vm_map_ram() calls.
>
> First vm_map_ram(3) caller could get benefit from your second patch.
> But, second caller and the other callers in each iteration could not
> get benefit and should iterate whole list to find suitable free block,
> because this free block is put to the tail of the list. Am I missing
> something?
You are missing the fact that we occupy blocks in 2^n.
So in your example 4 page slots will be occupied (order is 2), not 3.
The maximum size of allocation is 32 pages for 32-bit system
(if you try to map more, original alloc_vmap_area will be called).
So the maximum order is 5. That means that worst case, before we make
the decision
to allocate new block, is to iterate 6 blocks:
HEAD
1st block - has 1 page slot free (order 0)
2nd block - has 2 page slots free (order 1)
3rd block - has 4 page slots free (order 2)
4th block - has 8 page slots free (order 3)
5th block - has 16 page slots free (order 4)
6th block - has 32 page slots free (order 5)
TAIL
So the worst scenario is that each CPU queue can have 6 blocks in a free list.
This can happen only and only if you allocate blocks increasing the order.
(as I did in the function written in the comment of the first patch)
This is weird and rare case, but still it is possible.
Afterwards you will get 6 blocks in a list.
All further requests should be placed in a newly allocated block or
some free slots
should be found in a free list. Seems it does not look dramatically awful.
--
Roman
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