up vote 2 down vote favorite
share [g+] share [fb]

Possible Duplicate:
Virtualization: Guest in guest?

Is it possible to run a virtual machine under a virtual machine ?

link|improve this question
I thought of this the other day. I wonder if you can measure the performance loss the deeper the VM's go. A VMware inside a Hyper-V inside a VMWare? Hmmm... inside virtualbox? – Jason Taylor Sep 29 '11 at 17:31
3  
2 words: VM Inception! – TylerShads Sep 29 '11 at 17:40
feedback

closed as exact duplicate by Shane Madden, Chris S♦, Iain♦, sysadmin1138♦ Sep 29 '11 at 17:56

This question covers exactly the same ground as earlier questions on this topic; its answers may be merged with another identical question. See the FAQ for guidance on how to improve it.

5 Answers

up vote 2 down vote

Possibly, but not with ahrdware acceleration - the processor only has one ring 0. For the hyper-visor.

The sense of it is academic, though - it makes NO sense from an admin perspective.

link|improve this answer
...but Linux kernel 3.1 is going to add this functionality. With all the problems. – mailq Sep 29 '11 at 17:02
feedback
up vote 2 down vote

esxi 4.1 can virtualize itself.

link|improve this answer
As can Linux's KVM. avikivity.blogspot.com/2008/09/… – sciurus Sep 29 '11 at 18:01
feedback
up vote 2 down vote

Yes it is known a nested virtualization. Here is a paper on same http://www.xen.org/files/xensummit_intel09/xensummit-nested-virt.pdf relative to Xen from Intel.

link|improve this answer
feedback
up vote 1 down vote

Yes, it is possible with bare-metal hypervizors, and is very common for virtualization lab environments. Not generally recommended for production though.

http://www.vcritical.com/2009/05/vmware-esx-4-can-even-virtualize-itself/

link|improve this answer
feedback
up vote 0 down vote

Sure. For some reason not long ago I was running QEMU as an ARM emulator inside Fedora inside VMWare Server on a Windows XP desktop.

It worked well enough.

link|improve this answer
feedback

Not the answer you're looking for? Browse other questions tagged or ask your own question.