140 lines
3.6 KiB
C
140 lines
3.6 KiB
C
/*
|
|
* ----------------------------------------------------------------------------
|
|
* "THE BEER-WARE LICENSE" (Revision 42):
|
|
* <phk@login.dknet.dk> wrote this file. As long as you retain this notice you
|
|
* can do whatever you want with this stuff. If we meet some day, and you think
|
|
* this stuff is worth it, you can buy me a beer in return. Poul-Henning Kamp
|
|
* ----------------------------------------------------------------------------
|
|
*/
|
|
|
|
#include <sys/cdefs.h>
|
|
__FBSDID("$FreeBSD$");
|
|
|
|
#include <unistd.h>
|
|
#include <stdio.h>
|
|
#include <string.h>
|
|
#include <md5.h>
|
|
#include <err.h>
|
|
#include "crypt.h"
|
|
|
|
/*
|
|
* UNIX password
|
|
*/
|
|
|
|
char *
|
|
crypt_md5(const char *pw, const char *salt)
|
|
{
|
|
MD5_CTX ctx,ctx1;
|
|
unsigned long l;
|
|
int sl, pl;
|
|
u_int i;
|
|
u_char final[MD5_SIZE];
|
|
static const char *sp, *ep;
|
|
static char passwd[120], *p;
|
|
static const char *magic = "$1$"; /*
|
|
* This string is magic for
|
|
* this algorithm. Having
|
|
* it this way, we can get
|
|
* get better later on
|
|
*/
|
|
|
|
/* Refine the Salt first */
|
|
sp = salt;
|
|
|
|
/* If it starts with the magic string, then skip that */
|
|
if(!strncmp(sp, magic, strlen(magic)))
|
|
sp += strlen(magic);
|
|
|
|
/* It stops at the first '$', max 8 chars */
|
|
for(ep = sp; *ep && *ep != '$' && ep < (sp + 8); ep++)
|
|
continue;
|
|
|
|
/* get the length of the true salt */
|
|
sl = ep - sp;
|
|
|
|
MD5Init(&ctx);
|
|
|
|
/* The password first, since that is what is most unknown */
|
|
MD5Update(&ctx, (const u_char *)pw, strlen(pw));
|
|
|
|
/* Then our magic string */
|
|
MD5Update(&ctx, (const u_char *)magic, strlen(magic));
|
|
|
|
/* Then the raw salt */
|
|
MD5Update(&ctx, (const u_char *)sp, (u_int)sl);
|
|
|
|
/* Then just as many characters of the MD5(pw,salt,pw) */
|
|
MD5Init(&ctx1);
|
|
MD5Update(&ctx1, (const u_char *)pw, strlen(pw));
|
|
MD5Update(&ctx1, (const u_char *)sp, (u_int)sl);
|
|
MD5Update(&ctx1, (const u_char *)pw, strlen(pw));
|
|
MD5Final(final, &ctx1);
|
|
for(pl = (int)strlen(pw); pl > 0; pl -= MD5_SIZE)
|
|
MD5Update(&ctx, (const u_char *)final,
|
|
(u_int)(pl > MD5_SIZE ? MD5_SIZE : pl));
|
|
|
|
/* Don't leave anything around in vm they could use. */
|
|
memset(final, 0, sizeof(final));
|
|
|
|
/* Then something really weird... */
|
|
for (i = strlen(pw); i; i >>= 1)
|
|
if(i & 1)
|
|
MD5Update(&ctx, (const u_char *)final, 1);
|
|
else
|
|
MD5Update(&ctx, (const u_char *)pw, 1);
|
|
|
|
/* Now make the output string */
|
|
strcpy(passwd, magic);
|
|
strncat(passwd, sp, (u_int)sl);
|
|
strcat(passwd, "$");
|
|
|
|
MD5Final(final, &ctx);
|
|
|
|
/*
|
|
* and now, just to make sure things don't run too fast
|
|
* On a 60 Mhz Pentium this takes 34 msec, so you would
|
|
* need 30 seconds to build a 1000 entry dictionary...
|
|
*/
|
|
for(i = 0; i < 1000; i++) {
|
|
MD5Init(&ctx1);
|
|
if(i & 1)
|
|
MD5Update(&ctx1, (const u_char *)pw, strlen(pw));
|
|
else
|
|
MD5Update(&ctx1, (const u_char *)final, MD5_SIZE);
|
|
|
|
if(i % 3)
|
|
MD5Update(&ctx1, (const u_char *)sp, (u_int)sl);
|
|
|
|
if(i % 7)
|
|
MD5Update(&ctx1, (const u_char *)pw, strlen(pw));
|
|
|
|
if(i & 1)
|
|
MD5Update(&ctx1, (const u_char *)final, MD5_SIZE);
|
|
else
|
|
MD5Update(&ctx1, (const u_char *)pw, strlen(pw));
|
|
MD5Final(final, &ctx1);
|
|
}
|
|
|
|
p = passwd + strlen(passwd);
|
|
|
|
l = (final[ 0]<<16) | (final[ 6]<<8) | final[12];
|
|
_crypt_to64(p, l, 4); p += 4;
|
|
l = (final[ 1]<<16) | (final[ 7]<<8) | final[13];
|
|
_crypt_to64(p, l, 4); p += 4;
|
|
l = (final[ 2]<<16) | (final[ 8]<<8) | final[14];
|
|
_crypt_to64(p, l, 4); p += 4;
|
|
l = (final[ 3]<<16) | (final[ 9]<<8) | final[15];
|
|
_crypt_to64(p, l, 4); p += 4;
|
|
l = (final[ 4]<<16) | (final[10]<<8) | final[ 5];
|
|
_crypt_to64(p, l, 4); p += 4;
|
|
l = final[11] ;
|
|
_crypt_to64(p, l, 2); p += 2;
|
|
*p = '\0';
|
|
|
|
/* Don't leave anything around in vm they could use. */
|
|
memset(final, 0, sizeof(final));
|
|
|
|
return passwd;
|
|
}
|
|
|