On 2/4/25 11:04, David Hildenbrand wrote:
> On 04.02.25 10:57, Boris Fiuczynski wrote:
>> On 2/4/25 10:06, David Hildenbrand wrote:
>>> On 03.02.25 10:55, Michal Privoznik wrote:
>>>> v2 of:
>>>>
>>>> https://lists.libvirt.org/archives/list/devel@lists.libvirt.org/
>>>> thread/KA2DGRIY7DAMNMYM4MBKLOJCB7YYEUKU/
>>>>
>>>
>>> I should probably play with this myself, but due to lack of a test
>>> system where I can mess with systemd:
>>>
>>> Assuming we're on a s390x host (using KVM) and we specify:
>>>
>>> + <memory model='virtio-mem'>
>>> + <target dynamicMemslots='yes'>
>>> + <size unit='KiB'>2097152</size>
>>> + <node>0</node>
>>> + <requested unit='KiB'>1048576</requested>
>>> + </target>
>>> + </memory>
>>>
>>> In particular, without
>>>
>>> + <block unit='KiB'>2048</block>
>>>
>>> What would be the default value on a s390x host?
>>>
>>> In QEMU, we'd be using the default of 1 MiB on a s390x host, which
>>> corresponds to the THP size. Note that on an x86_64 host, the default
>>> will be 2 MiB.
>>>
>>> Ideally, we'd also be using the default of 1 MiB on an s390x host.
>>>
>>> How is the default value determined here?
>>>
>>> Thanks!
>>>
>>
>> Current libvirt implementation does not generate a default.
>> When not providing the element block an error occurs:
>> error: block size must be a power of two
>> When providing the element block the required minimum value is 1024KiB.
>
> Right, the 1 MiB minimum value independent of the memory backed is
> currently enforced by QEMU.
The validation is done by libvirt in the domain validation in which QEMU
is not involved.
>
> Interesting that no defaults are provided. I assume it might be an
> interesting idea to query the default from QEMU, instead of re-
> implementing that logic in libvirt.
>
Usually the later is done when a default is set by libvirt and is not
required to be provided by the user.
--
Mit freundlichen Grüßen/Kind regards
Boris Fiuczynski
IBM Deutschland Research & Development GmbH
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