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[Qemu-devel] Re: Best way to get 6GB = 3 x 2GB + soft raid?

From: Garth Dahlstrom
Subject: [Qemu-devel] Re: Best way to get 6GB = 3 x 2GB + soft raid?
Date: Mon, 2 Aug 2004 11:52:11 -0400

Let me explain how I became confused here... 

I read Readme-EN.txt bundled with QEMU-Windows-0.6 which says:

"6. Notes

    A hard disk image file is up to 2GB."

I took that to mean that QEMU was only capable of open HDD images of
up to 2GB in size, so I was thinking that I could do 3x2GB disks and
software JBOD/Raid0 them together under a linux guest.

However it turns out, as Ronald pointed out, QEMU on Windows can read
whatever size image the host OS can muster, meaning in my case (using
NTFS) I can create a 6GB disk image file straight up using Cygwin and
'dd of=oe-hda.img bs=1M count=6000 seek=6000' (followed quickly by

That limitation would appear to apply to FAT based host OS's such as Win9x.

QEMU has no trouble with that size guest image file at all and it's
easier then trying to build a softRAID on the guest.

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