> > Details: I have glibc-2.1.1, and linux-pam-0.66 (that was what was
> > supplied with Redhat-6.0). Can anyone confirm this incompatibility?
> > One of its consequences is that if you have a system with these
> > versions of glibc and linux-pam, and a lot of users with
> > md5-passwords, and you decide to uninstall PAM, or replace the
> > pam_pwdb module with a module that uses the crypt()-function from
> > glibc, those users will no longer be able to log in. If the root
> > password was encrypted with md5, you may have to dig out your boot
> > floppies (which would be a little difficult for me; I have no floppy
> > and no CD in the machine).
>
> I was bitten by this, on a remote machine. Using Debian.
> It seems the md5s really are incompatible. It may be
> worthwhile to look at the corresponding crypting method
> in both sources, and compare.
You should also check out the FreeBSD md5-based crypt - I believe this
is where it showed up first. See src/lib/libcrypt/crypt.c, available
from lots of FreeBSD mirrors around the world. It's short enough (and
readable enough) that I'm posting it in its entirety below.
Steinar Haug, Nethelp consulting, [EMAIL PROTECTED]
----------------------------------------------------------------------
/*
* ----------------------------------------------------------------------------
* "THE BEER-WARE LICENSE" (Revision 42):
* <[EMAIL PROTECTED]> 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
* ----------------------------------------------------------------------------
*
* $Id: crypt.c,v 1.9 1999/01/23 08:27:36 markm Exp $
*
*/
#if defined(LIBC_SCCS) && !defined(lint)
static char rcsid[] = "$Header: /home/ncvs/src/lib/libcrypt/crypt.c,v 1.9 1999/01/23
08:27:36 markm Exp $";
#endif /* LIBC_SCCS and not lint */
#include <unistd.h>
#include <stdio.h>
#include <string.h>
#include <md5.h>
#include <string.h>
static unsigned char itoa64[] = /* 0 ... 63 => ascii - 64 */
"./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
static void to64 __P((char *, unsigned long, int));
static void
to64(s, v, n)
char *s;
unsigned long v;
int n;
{
static void to64 __P((char *, unsigned long, int));
while (--n >= 0) {
*s++ = itoa64[v&0x3f];
v >>= 6;
}
}
/*
* UNIX password
*
* Use MD5 for what it is best at...
*/
char *
crypt(pw, salt)
register const char *pw;
register const char *salt;
{
static char *magic = "$1$"; /*
* This string is magic for
* this algorithm. Having
* it this way, we can get
* get better later on
*/
static char passwd[120], *p;
static const char *sp,*ep;
unsigned char final[16];
int sl,pl,i,j;
MD5_CTX ctx,ctx1;
unsigned long l;
/* 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,pw,strlen(pw));
/* Then our magic string */
MD5Update(&ctx,magic,strlen(magic));
/* Then the raw salt */
MD5Update(&ctx,sp,sl);
/* Then just as many characters of the MD5(pw,salt,pw) */
MD5Init(&ctx1);
MD5Update(&ctx1,pw,strlen(pw));
MD5Update(&ctx1,sp,sl);
MD5Update(&ctx1,pw,strlen(pw));
MD5Final(final,&ctx1);
for(pl = strlen(pw); pl > 0; pl -= 16)
MD5Update(&ctx,final,pl>16 ? 16 : 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, final, 1);
else
MD5Update(&ctx, pw, 1);
/* Now make the output string */
strcpy(passwd,magic);
strncat(passwd,sp,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,pw,strlen(pw));
else
MD5Update(&ctx1,final,16);
if(i % 3)
MD5Update(&ctx1,sp,sl);
if(i % 7)
MD5Update(&ctx1,pw,strlen(pw));
if(i & 1)
MD5Update(&ctx1,final,16);
else
MD5Update(&ctx1,pw,strlen(pw));
MD5Final(final,&ctx1);
}
p = passwd + strlen(passwd);
l = (final[ 0]<<16) | (final[ 6]<<8) | final[12]; to64(p,l,4); p += 4;
l = (final[ 1]<<16) | (final[ 7]<<8) | final[13]; to64(p,l,4); p += 4;
l = (final[ 2]<<16) | (final[ 8]<<8) | final[14]; to64(p,l,4); p += 4;
l = (final[ 3]<<16) | (final[ 9]<<8) | final[15]; to64(p,l,4); p += 4;
l = (final[ 4]<<16) | (final[10]<<8) | final[ 5]; to64(p,l,4); p += 4;
l = final[11] ; 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;
}
