u can use diskpart utility.
diskpart command.. if running win xp...
or you can boot into dos mode and with disk manager u can partition it...
2007-04-24 00:29:56
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answer #1
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answered by amaR . 2
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Bear in mind that if your current hard drive is a single partition and you want to create another partition/logical drive, that you will lose everything on you hard drive.
If your intention is to create a new partition and thereby resize your current partion and create a new partition then you will need to get hold of a program like Partition Magic which does not destroy data on your current partiton.
2007-04-24 00:44:51
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answer #2
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answered by Anonymous
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U can use a free DOS utility called "FDisk". Just run this from a Windows 98 boot disk. BE CAREFUL though, this will destroy any data on the disk. So backup first if you want any existing data
2007-04-24 00:29:41
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answer #3
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answered by Cocodmonkey 1
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You can use fdisk... part of most versions of DOS, certainly there in MS-DOS up to version 7 (the one under win98).
BUT
BUT
BUT
the MS-DOS fdisk program is pretty limited because it's old. It has problems with big disks, the older the version the more limited it is.. The MS-DOS fdisk program also has problems with partitions that are not recognised (because they belong to foreign OS versions).
2007-04-24 00:34:44
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answer #4
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answered by bambamitsdead 6
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DO NOT USE FDISK!!!
Today's hard drives can be severely damaged by this dinosaur utility.
Go to the hard drive manufacturer's web site and get their free Disk Management software.
2007-04-24 00:54:53
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answer #5
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answered by ELfaGeek 7
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If you are using pure DOS, i.e. something like DOS 5 or DOS 6.2, the command you use is
fdisk
However, do not try this while you are in Windows, spawning DOS through 'cmd'
2007-04-24 00:36:32
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answer #6
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answered by StarChaser 5
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You can use FDISK command to partition your harddisk in DOS.
2007-04-24 00:31:57
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answer #7
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answered by nick 2
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f disk, just need to make a boot floppy disk. or you could go to the manufacture too ether is good.
2007-04-24 01:05:31
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answer #8
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answered by Anonymous
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