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Re: Can a VM machine ...

 

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Tony,

Sorry for the top post. I don¡¯t believe there are many people that understand I64/AMD64. No one codes in assembler for Windows. I don¡¯t believe the architecture is virtualizable in the same why 360/370 architecture is.

I say this because most modern hypervisors won¡¯t run without the virtualization extensions enabled. Reading Wikipedia it appears there is some debate on this topic¡­

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I do know that with the 68010 Motorola modified the instruction set from the 68000 to make it virtualizable, but I don¡¯t know if any OS uses this¡­

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Dave

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From: [email protected] <[email protected]> On Behalf Of Tony Harminc
Sent: 20 November 2022 04:22
To: [email protected]
Subject: Re: [h390-vm] Can a VM machine ...

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On Sat, 19 Nov 2022 at 22:44, Fish Fish <david.b.trout@...> wrote:

Tony Harminc wrote:

> What it does do is make the architecture no longer virtualizable
> (without some feature like SIE).

Eh? How so?


> This 100% virtualizable feature (even for privileged states,
> and for that matter I/O) was to my knowledge unique among the
> architectures of the day, and may be still true.

I'm not following you.


> Look at the early kluges with i86 and VMWare and such, that
> had to resort to paravirtualization, where the guest has to
> cooperate (or the host zaps the guest code on the fly so that
> it does).

What does that have to do with your claim that the introduction of new instructions that weren't defined in the original published version of the architecture's manuals somehow makes the architecture "longer virtualizable without some feature like SIE"? I'm not understanding your argument/reasoning.

The architecture should still be virtualizable (even *without* SIE) with or without the introduction of the new instructions. And if none of the new instructions are privileged instructions, the architecture should still be virtualizable (without SIE) without any modification at all to the hypervisor. Only if support for a new privileged instruction was needed would the hypervisor then need to be changed (to add support to simulate the privileged instruction).

Is that maybe what you're referring to?

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No - not at all.

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I simply refer to the fact that part of ("full") virtualization is that the guest cannot discover using architected features whether or not it is running in a virtual environment. Note that the DIAG interface to CP (or Hercules, or an LPAR, for that matter) is not part of the architecture. Yes, of course the guest can probably, but not deterministically, discover that is running virtualized - most likely using timing tests. But not by finding a different result from an architected instruction. This is discussed in the VM books from the earliest days.

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> I'm not sure that even today i64 is 100% virtualizable, is it?

I have no idea. I don't know i64.

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Oh, sorry - I thought all Windows people did.

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Tony H.

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