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Re: VMWare, 98 and Linux



Thus spake Kent West (westk@acu.edu):

> "Robert L. Harris" wrote:
> 
> > Ok,
> >   Here's the scenario, I have a 6Gig windows drive and 4 Gig Linux.  The
> > 6 is divided into 1 and 6 (/dev/hda1 and /dev/hda2 respectivly).  I have
> > 98 on /dev/hda1.  I run this one nativly(sp?) and don't want to run it
> > under vmware as it will trash my drivers, etc.  I can't blow it away as
> > Rogue Spear won't run under it vmware yet.  I'd like to put 98 on /dev/hda2
> > and hide /dev/hda1 from windows so it won't even know it exists (causes
> > problems otherwise).
> >
> >   Has anyone found a way to "hide" a partition or another workaround
> > like this yet?
> >
> > Robert
> 
> So you want to put Win98 on both /dev/hda1 (already there) and /dev/hda2, and not let
> the Win98 on /dev/hda2 see the /dev/hda1 partition (and it's corresponding Win98
> installation)?
> 
> And presumably you meant that the 6GB drive is partitioned into 1GB and 5GB (which
> equals 6), rather than 1GB and 6GB (which equals 7).
> 
> What I'd do is convert the /dev/hda2 partition to ext2 and mount it under /win, then
> use VMWARE to create a new virtual machine in /win, and install Win98 to it. This
> Win98 won't see the Win98 on /dev/hda1 since it's in a totally different (virtual)
> computer, which is "powered off" when Linux/VMWARE is up and running. This assumes
> that Rogue Spear is run on the /dev/hda1 copy of Win98, and you don't have any such
> apps on the /dev/hda2 copy of Win98 that won't run under VMWARE.
> 
> 

Yes I meant 1 and 5.

I can't convert /dev/hda2 to ext2 or all the software, apps, etc on
/dev/hda2 won't be visible to the windows partition on /dev/hda1 and
I won't be able to "share" a single install instead of multiple installs
of the same softwre.

Robert


:wq!
---------------------------------------------------------------------------
Robert L. Harris                |  Low quality in a product happens.
Senior System Engineer          |    That doesn't mean it's right and
  at RnD Consulting.		|      definitely doesn't mean it should
                                \_       be accepted.  Require quality.

http://www.rnd-consulting.com/~nomad

DISCLAIMER:
      These are MY OPINIONS ALONE.  I speak for no-one else.

FYI:
 perl -e 'print $i=pack(c5,(41*2),sqrt(7056),(unpack(c,H)-2),oct(115),10);'


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