I have VirtualBox, and it requires to disable kvm to run properly. Removing the module works fine. But the question is why it’s loaded by default if it prevents VirtualBox from running.
What kind of software depends on kvm? I want to know when I should load it again.
The only software that use I know that may require something like this is Docker, but it works fine without kvm module.
Ditto. I’ve ran VirtualBox alongside kvm on the same system many times (although not very recently), as well as VMWare, too. I’ve never had to disable one hypervisor’s kernel module to use the other. Why would VirtualBox have anything to do with the kvm module? That doesn’t seem right.
I just tested the problem reported here.
I did not have VB installed so I installed it on my host.
I then rebooted so the kernel modules were properly loaded.
I opened both VMM and VB Manager.
I created a new VM (f41) on VB and completed the config and setup of the new VM.
Now the actual test.
first with a VB VM running,
I attempted to launch a VM with VMM and I get the VM window opened, but when I attempt to power it on I get the below message
These messages are clearly different than I have ever seen previously and definitely prevent having a VM running in one of the hypervisors and then starting another VM in the other hypervisor.
Removing or disabling the modules is not necessary, but it certainly means that a user cannot run VMs in both hypervisors at the same time.
@jcubic
I suggest you file a bug on this at bugzilla.redhat.com with the component being KVM & with Oracle with the component being VirtualBox
You can link or quote the test I described here.
The expected results would be that both VMs should be able to run simultaneously but with this kernel it does not happen.
The above test was done with the 6.13.5 kernel.
I then rebooted with the 6.12.15 kernel and repeated the same tests. I received identical results.
Conslusion:
At present the user may have run VMs using either VMM or VB, but cannot use both hypervisors simultaneously. The same limitation applies regardless of the kernel used.