postgresql/contrib/pgcrypto/crypt-md5.c

170 lines
3.8 KiB
C

/*
* File imported from FreeBSD, original by Poul-Henning Kamp.
*
* $FreeBSD: src/lib/libcrypt/crypt-md5.c,v 1.5 1999/12/17 20:21:45 peter Exp $
*
* contrib/pgcrypto/crypt-md5.c
*/
#include "postgres.h"
#include "px-crypt.h"
#include "px.h"
#define MD5_SIZE 16
static const char _crypt_a64[] =
"./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
static void
_crypt_to64(char *s, unsigned long v, int n)
{
while (--n >= 0)
{
*s++ = _crypt_a64[v & 0x3f];
v >>= 6;
}
}
/*
* UNIX password
*/
char *
px_crypt_md5(const char *pw, const char *salt, char *passwd, unsigned dstlen)
{
static char *magic = "$1$"; /* This string is magic for this algorithm.
* Having it this way, we can get better later
* on */
static char *p;
static const char *sp,
*ep;
unsigned char final[MD5_SIZE];
int sl,
pl,
i;
PX_MD *ctx,
*ctx1;
int err;
unsigned long l;
if (!passwd || dstlen < 120)
return NULL;
/* Refine the Salt first */
sp = salt;
/* If it starts with the magic string, then skip that */
if (strncmp(sp, magic, strlen(magic)) == 0)
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;
/* we need two PX_MD objects */
err = px_find_digest("md5", &ctx);
if (err)
return NULL;
err = px_find_digest("md5", &ctx1);
if (err)
{
/* this path is possible under low-memory circumstances */
px_md_free(ctx);
return NULL;
}
/* The password first, since that is what is most unknown */
px_md_update(ctx, (const uint8 *) pw, strlen(pw));
/* Then our magic string */
px_md_update(ctx, (uint8 *) magic, strlen(magic));
/* Then the raw salt */
px_md_update(ctx, (const uint8 *) sp, sl);
/* Then just as many characters of the MD5(pw,salt,pw) */
px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
px_md_update(ctx1, (const uint8 *) sp, sl);
px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
px_md_finish(ctx1, final);
for (pl = strlen(pw); pl > 0; pl -= MD5_SIZE)
px_md_update(ctx, final, pl > MD5_SIZE ? MD5_SIZE : pl);
/* Don't leave anything around in vm they could use. */
px_memset(final, 0, sizeof final);
/* Then something really weird... */
for (i = strlen(pw); i; i >>= 1)
if (i & 1)
px_md_update(ctx, final, 1);
else
px_md_update(ctx, (const uint8 *) pw, 1);
/* Now make the output string */
strcpy(passwd, magic);
strncat(passwd, sp, sl);
strcat(passwd, "$");
px_md_finish(ctx, final);
/*
* 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++)
{
px_md_reset(ctx1);
if (i & 1)
px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
else
px_md_update(ctx1, final, MD5_SIZE);
if (i % 3)
px_md_update(ctx1, (const uint8 *) sp, sl);
if (i % 7)
px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
if (i & 1)
px_md_update(ctx1, final, MD5_SIZE);
else
px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
px_md_finish(ctx1, final);
}
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. */
px_memset(final, 0, sizeof final);
px_md_free(ctx1);
px_md_free(ctx);
return passwd;
}