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发表于 2008-12-18 15:50:34
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This program runs as a command under the system’s DOS operating system. It permanently erases all data on one or more hard disk drives that are installed on the system computer.
Command line:
SCRUB3 [/?] | [[/Q=NO] [/S=NO] { /D=<drive> | /D=ALL } { /L=<level> | /W=<writes> [/P=<string>] } ]
/Q=NO
This parameter causes the program to display a maximal number of messages on STDOUT. It is intended to be used for in-the-field debugging only, and a customer should normally not use this parameter. If present, this parameter should be the first (i.e., leftmost) parameter.
/S=NO
This parameter causes the program not to write the scrub signature to the disk(s) that will be erased.
/D=<drive>
Use this form of the /D parameter if you want to erase only one hard drive that is installed on the system computer. The value <drive> is the hard disk drive number of the drive that you want to erase. 1 is the first hard disk drive, 2 is the second hard disk drive, etc. There is no default value for this parameter. The /D parameter is required.
/D=ALL
Use this form of the /D parameter if you want to erase all hard disk drives that are installed on the system computer. There is no default value for this parameter. The /D parameter is required.
/L=<level>
The value <level> is the security level of the disposal operation. There is no default value for this parameter. Exactly one of the /L parameter or the /W parameter is required. It must be one of the following values:
/L=1
Limited security. The first 63 sectors on the drive (includes Master Boot Record), the last 2 sectors on the drive, and the first 100 sectors on each partition are overwritten with a 0x0000 pattern (i.e., each pair of bytes on the sector is overwritten with this pattern). This operation is very fast. The hard disk drive will not be usable via standard I/O methods. However, this is not a secure operation in an absolute sense, since it leaves most of the partitions on the hard drive unchanged.
/L=2
Medium security. All sectors on the drive are overwritten 1 time with a 0x0000 pattern (i.e., each pair of bytes on the sector is overwritten with this pattern). This operation is relatively slow, since it involves many write operations. Actual speed depends on the size and speed of the target hard disk drives.
/L=3
High security. All sectors on the drive are overwritten 4 times with the following patterns (in this order): a random pattern, the bit-wise complement of that random pattern, a different random pattern, and a 0x0000 pattern (i.e., each pair of bytes on the sector is overwritten with these patterns). This operation is quite slow, and it takes 4 times as long as a /L=2 operation.
/L=4
DOD-compliant security. All sectors on the drive are overwritten 7 times with the following patterns (in this order): a random pattern, the bit-wise complement of that random pattern (3 times, each with a different random pattern), and a 0x0000 pattern (i.e., each pair of bytes on the sector is overwritten with these patterns). This operation is quite slow, and it takes 7 times as long as a /L=2 operation.
/W=<writes>
The value <writes> is the number of times each sector is overwritten (done <writes> - 1 times with a random pattern, before a final write with a 0x0000 pattern). <writes> is an integer from 1 to 9999. There is no default value for this parameter. Exactly one of the /L parameter or the /W parameter is required.
/P=<string>
The <string> value can contain any character except for quotes due to command line restrictions.
The length of the string can be 1 to 64 characters in length.
This parameter must be used in conjunction with the /W parameter.
When used, it will replace the final overwrite pattern of 0x0000 with the specified string.
/?
This parameter causes the program to display a concise description of its execution syntax on STDOUT and then to terminate execution. If you run SCRUB.EXE with no parameters, it will display this same output. /? causes all other parameters to be ignored.
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发表于 2008-12-18 15:17:03
