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@xxxxxxxxxxxxxxxxx/
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
Vorsitzender des Aufsichtsrats: Wolfgang Wendt
Geschäftsführung: David Faller
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