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#1 2020-06-27 22:52:30

elektronek
Member
Registered: 2020-06-27
Posts: 7

[Solved] Nested virtualization doesn't work (AMD)

I'm trying to run hyper-v machines in virtualized windows server 2012R2 but it keep throwing error that hypervizor is already running.
Nested parameter is enabled

systool -m kvm_amd -v | grep nested
    nested              = "1"

And I run qemu with -enable-kvm -cpu host

qemu-system-x86_64 -m 4096 -hda winsys2012r2.img -enable-kvm -cpu host 

Everything as arch wiki shows

Last edited by elektronek (2020-06-29 12:36:01)

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#2 2020-06-28 13:52:35

Lone_Wolf
Member
From: Netherlands, Europe
Registered: 2005-10-04
Posts: 11,911

Re: [Solved] Nested virtualization doesn't work (AMD)

https://docs.microsoft.com/en-us/virtua … ualization

Some more reading suggests WS 2016 is the first to support nested hyper-V on intel processors, while you need WS 2019 to use hyper-V nesting on amd procesoors.

Running Hyper-V on kvm appears to be completely unsupported.

If the host has multiple processors, you could maybe passthrough a full processor to the windows server VM .


Disliking systemd intensely, but not satisfied with alternatives so focusing on taming systemd.


(A works at time B)  && (time C > time B ) ≠  (A works at time C)

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#3 2020-06-29 00:37:20

elektronek
Member
Registered: 2020-06-27
Posts: 7

Re: [Solved] Nested virtualization doesn't work (AMD)

Lone_Wolf wrote:

https://docs.microsoft.com/en-us/virtua … ualization

Some more reading suggests WS 2016 is the first to support nested hyper-V on intel processors, while you need WS 2019 to use hyper-V nesting on amd procesoors.

Running Hyper-V on kvm appears to be completely unsupported.

If the host has multiple processors, you could maybe passthrough a full processor to the windows server VM .

I checked it out and on both server 2016 and 2019 it works. Thanks!

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#4 2020-06-29 10:45:00

Lone_Wolf
Member
From: Netherlands, Europe
Registered: 2005-10-04
Posts: 11,911

Re: [Solved] Nested virtualization doesn't work (AMD)

Great, please prepend [Solved] to the thread title (edit first post)


Disliking systemd intensely, but not satisfied with alternatives so focusing on taming systemd.


(A works at time B)  && (time C > time B ) ≠  (A works at time C)

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