On Fri, Jun 13, 2014 at 04:30:41PM +0200, Michal Privoznik wrote:
On 13.06.2014 10:28, Daniel P. Berrange wrote:
> On Thu, Jun 12, 2014 at 07:21:47PM +0200, Martin Kletzander wrote:
>> On Thu, Jun 12, 2014 at 02:30:50PM +0100, Daniel P. Berrange wrote:
>>> On Tue, Jun 10, 2014 at 07:21:12PM +0200, Michal Privoznik wrote:
>>>> There are two places where you'll find info on huge pages. The first
>>>> one is under <cpu/> element, where all supported huge page sizes
are
>>>> listed. Then the second one is under each <cell/> element which
refers
>>>> to concrete NUMA node. At this place, the size of huge page's pool
is
>>>> reported. So the capabilities XML looks something like this:
>>>>
>>>> <capabilities>
>>>>
>>>> <host>
>>>> <uuid>01281cda-f352-cb11-a9db-e905fe22010c</uuid>
>>>> <cpu>
>>>> <arch>x86_64</arch>
>>>> <model>Westmere</model>
>>>> <vendor>Intel</vendor>
>>>> <topology sockets='1' cores='1'
threads='1'/>
>>>> ...
>>>> <pages unit='KiB' size='1048576'/>
>>>> <pages unit='KiB' size='2048'/>
>>>
>>> Should have normal sized pages (ie 4k on x86) too, to avoid
>>> apps having to special case small pages.
>>>
>>
>> Since we have to special-case small pages and kernel (at least to my
>> knowledge) doesn't expose that information by classic means, I think
>> reporting only hugepages is actually what we want here. For normal
>> memory there are existing APIs already.
>>
>> Hugepages are different mainly because of one thing. The fact that
>> there are some hugepages allocated is known by the user of the machine
>> (be it mgmt app or an admin) and these hugepages were allocated for
>> some purpose. It is fairly OK to presume that the number of hugepages
>> (free or total) will change only when and if the user wants to
>> (e.g. running a machine with specified size and hugepages). That
>> cannot be said about small pages, though, and I think it is fair
>> reason to special-case normal pages like this.
>
> That difference is something that's only relevant to the person who
> is provisioning the machine though. For applications consuming the
> libvirt APIs it is not relevant. For OpenStack we really want to have
> normal size pages dealt with the in the same way as huge pages since
> it will simplify our schedular/placement logic. So I really want these
> APIs to do this in libvirt so that OpenStack doesn't have to reverse
> engineer this itself.
But if we go this way, there are black holes hidden. For instance, the
sizeof(ordinary pages pool). This is not accessible anywhere and the
only algorithm I can think of is to take [(MemTotal on NODE #i) -
sum(mem taken by all huge pages)] / PAGE_SIZE. So for instance on my
machine where I have 1GB huge page per NUMA node, and 3 2MB per NUMA node:
# grep MemTotal /sys/devices/system/node/node0/meminfo
Node 0 MemTotal: 4054408 kB
# cat
/sys/devices/system/node/node0/hugepages/hugepages-1048576kB/nr_hugepages
1
# cat
/sys/devices/system/node/node0/hugepages/hugepages-2048kB/nr_hugepages
3
# getconf PAGESIZE
4096
(4054408 - (1*1048576 + 3*2048)) / 4 = 2999688 / 4 = 749922 ordinary
pages. But it's not that simple as not all pages are available. Some are
reserved for DMA transfers, some for kernel itself, etc. Without
overcommit it's impossible to allocate that nearly 3GB. Is this
something we really want to do?
I've found one other way to get the number of free normal pages. It
looks like nr_free_pages in /proc/zoneinfo is what Daniel wants to
report probably. But given the fact that this is something that might
not be true even in the time of parsing the file, I'm still not
convinced it's something you want to report. Bit more accurate would
be having the amount of memory that might be available to the
machine. Although with overcommit settings and file caches this might
not be feasible.
Martin