使用john破解ubuntu(linux)9.10密码

浏览数：14 / 时间：2015年06月20日

Title:使用john破解ubuntu(linux)9.10密码 --2011-11-23 15:00

ubuntu 9.10的账户密码加密方式改用sha512了，默认的john是破不了的，还好官方有补丁。

首先解压缩john1.75的源代码，vi编辑Makefile文件，添加我下面标注好的红色字体

LDFLAGS = -s -lcrypt
JOHN_OBJS_MINIMAL = \
DES_fmt.o DES_std.o DES_bs.o \
BSDI_fmt.o \
MD5_fmt.o MD5_std.o \
BF_fmt.o BF_std.o \
AFS_fmt.o \
LM_fmt.o \
batch.o bench.o charset.o common.o compiler.o config.o cracker.o \
crc32.o external.o formats.o getopt.o idle.o inc.o john.o list.o \
loader.o logger.o math.o memory.o misc.o options.o params.o path.o \
recovery.o rpp.o rules.o signals.o single.o status.o tty.o wordlist.o \
unshadow.o \
unafs.o \
unique.o \
crypt_fmt.o

然后新建一个crypt_fmt.c文件，代码如下/* public domain proof-of-concept code by Solar Designer */

#define _XOPEN_SOURCE /* for crypt(3) */
#include <string.h>
#include <unistd.h>

#include "arch.h"
#include "params.h"
#include "formats.h"

#define FORMAT_LABEL "crypt"
#define FORMAT_NAME "generic crypt(3)"
#define ALGORITHM_NAME "?/" ARCH_BITS_STR

#define BENCHMARK_COMMENT ""
#define BENCHMARK_LENGTH 0

#define PLAINTEXT_LENGTH 72

#define BINARY_SIZE 128
#define SALT_SIZE BINARY_SIZE

#define MIN_KEYS_PER_CRYPT 1
#define MAX_KEYS_PER_CRYPT 1

static struct fmt_tests tests[] = {
{"CCNf8Sbh3HDfQ", "U*U*U*U*"},
{"CCX.K.MFy4Ois", "U*U***U"},
{"CC4rMpbg9AMZ.", "U*U***U*"},
{"XXxzOu6maQKqQ", "*U*U*U*U"},
{"SDbsugeBiC58A", ""},
{NULL}
};

static char saved_key[PLAINTEXT_LENGTH + 1];
static char saved_salt[SALT_SIZE];
static char *crypt_out;

static int valid(char *ciphertext)
{
#if 1
int l = strlen(ciphertext);
return l >= 13 && l < BINARY_SIZE;
#else
/* Poor load time, but more effective at rejecting bad/unsupported hashes */
char *r = crypt("", ciphertext);
int l = strlen(r);
return
!strncmp(r, ciphertext, 2) &&
l == strlen(ciphertext) &&
l >= 13 && l < BINARY_SIZE;
#endif
}

static void *binary(char *ciphertext)
{
static char out[BINARY_SIZE];
strncpy(out, ciphertext, sizeof(out)); /* NUL padding is required */
return out;
}

static void *salt(char *ciphertext)
{
static char out[SALT_SIZE];
int cut = sizeof(out);

#if 1
/* This piece is optional, but matching salts are not detected without it */
switch (strlen(ciphertext)) {
case 13:
case 24:
cut = 2;
break;

case 20:
if (ciphertext[0] == ‘_‘) cut = 9;
break;

case 34:
if (!strncmp(ciphertext, "$1$", 3)) {
char *p = strchr(ciphertext + 3, ‘$‘);
if (p) cut = p - ciphertext;
}
break;

case 59:
if (!strncmp(ciphertext, "$2$", 3)) cut = 28;
break;

case 60:
if (!strncmp(ciphertext, "$2a$", 4)) cut = 29;
break;
}
#endif

/* NUL padding is required */
memset(out, 0, sizeof(out));
memcpy(out, ciphertext, cut);

return out;
}

static int binary_hash_0(void *binary)
{
return ((unsigned char *)binary)[12] & 0xF;
}

static int binary_hash_1(void *binary)
{
return ((unsigned char *)binary)[12] & 0xFF;
}

static int binary_hash_2(void *binary)
{
return
(((unsigned char *)binary)[12] & 0xFF) |
((int)(((unsigned char *)binary)[11] & 0xF) << 8);
}

static int get_hash_0(int index)
{
return (unsigned char)crypt_out[12] & 0xF;
}

static int get_hash_1(int index)
{
return (unsigned char)crypt_out[12] & 0xFF;
}

static int get_hash_2(int index)
{
return
((unsigned char)crypt_out[12] & 0xFF) |
((int)((unsigned char)crypt_out[11] & 0xF) << 8);
}

static int salt_hash(void *salt)
{
int pos = strlen((char *)salt) - 2;

return
(((unsigned char *)salt)[pos] & 0xFF) |
((int)(((unsigned char *)salt)[pos + 1] & 3) << 8);
}

static void set_salt(void *salt)
{
strcpy(saved_salt, salt);
}

static void set_key(char *key, int index)
{
strcpy(saved_key, key);
}

static char *get_key(int index)
{
return saved_key;
}

static void crypt_all(int count)
{
crypt_out = crypt(saved_key, saved_salt);
}

static int cmp_all(void *binary, int count)
{
return !strcmp((char *)binary, crypt_out);
}

static int cmp_exact(char *source, int index)
{
return 1;
}

struct fmt_main fmt_crypt = {
{
FORMAT_LABEL,
FORMAT_NAME,
ALGORITHM_NAME,
BENCHMARK_COMMENT,
BENCHMARK_LENGTH,
PLAINTEXT_LENGTH,
BINARY_SIZE,
SALT_SIZE,
MIN_KEYS_PER_CRYPT,
MAX_KEYS_PER_CRYPT,
FMT_CASE | FMT_8_BIT,
tests
}, {
fmt_default_init,
valid,
fmt_default_split,
binary,
salt,
{
binary_hash_0,
binary_hash_1,
binary_hash_2
},
salt_hash,
set_salt,
set_key,
get_key,
fmt_default_clear_keys,
crypt_all,
{
get_hash_0,
get_hash_1,
get_hash_2
},
cmp_all,
cmp_all,
cmp_exact
}
};
最后修改john.c文件，添加我下面标注的红色字体
extern struct fmt_main fmt_DES, fmt_BSDI, fmt_MD5, fmt_BF;
extern struct fmt_main fmt_AFS, fmt_LM;
extern struct fmt_main fmt_crypt;

john_register_one(&fmt_DES);
john_register_one(&fmt_BSDI);
john_register_one(&fmt_MD5);
john_register_one(&fmt_BF);
john_register_one(&fmt_AFS);
john_register_one(&fmt_LM);
john_register_one(&fmt_crypt);

现在可以编译了，选择好你的平台和CPU类型，能够提高破解速度，我这里用的是linux,X86架构，所以选择的是
linux-x86-sse2 Linux, x86 with SSE2 (best if 32-bit)
如果你和我一样，输入下面的红色字体
make linux-x86-sse2

现在实践下，可以发现能够破解了

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