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Re: [Qemu-devel] qemu-fast kernel patch question

From: Johannes Martin
Subject: Re: [Qemu-devel] qemu-fast kernel patch question
Date: Sun, 25 Jul 2004 12:24:16 +0200 (CEST)


On Sun, 25 Jul 2004, Mulyadi Santosa wrote:
> Do you mean:
> 1. applying qemu-fast patch to Win98 guest? --> Not a chance :-)
> 2. running qemu-fast on top of Win98 host? Well, I haven't tried it
> personally. But, if you can run qemu-fast on Win98 host, I doubt if qemu-fast
> can do direct MMU access. The reason is: to do direct MMU access, the
> developers need to know to request that into host kernel. On Linux or other
> open source kernel, we can find it out, but on closed kernel, this would be
> difficult.....
> The conclusion: just run the usual "qemu" on top of Win98 host

Here's a quote from qemu-tech.html:
  2.10 MMU emulation

  For system emulation, QEMU uses the mmap() system call to emulate the
  target CPU MMU. It works as long the emulated OS does not use an area
  reserved by the host OS (such as the area above 0xc0000000 on x86

As far as I understand this, the qemu kernel patch makes the guest kernel
use a different reserved area than the linux host. So it shouldn't matter
which kernel I apply the patch to, right? If I patch the (linux) host
kernel, I should be able to boot any unpatched (linux) guest kernel, if I
don't patch the (linux) host kernel, I have to patch my (linux) guest
kernels. Or do we need a patch-aware qemy-fast to run on a patched host?

Assuming that OS XYZ does not use the area around 0xc0000000, it should
work inside qemu-fast, right? And if it does use that area, it should work
inside qemu-fast on a patched host?

Maybe one of the wizards could clarify this...


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