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.
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.
--
Cheers,
David / dhildenb
