postgresql/contrib/pgcrypto/crypt-md5.c

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/*
* ----------------------------------------------------------------------------
* "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
* ----------------------------------------------------------------------------
*
* $FreeBSD: src/lib/libcrypt/crypt-md5.c,v 1.5 1999/12/17 20:21:45 peter Exp $
*
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* $PostgreSQL: pgsql/contrib/pgcrypto/crypt-md5.c,v 1.6 2005/10/15 02:49:06 momjian Exp $
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*/
#include "postgres.h"
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#include "px.h"
#include "px-crypt.h"
#define MD5_SIZE 16
/*
* UNIX password
*/
char *
px_crypt_md5(const char *pw, const char *salt, char *passwd, unsigned dstlen)
{
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static char *magic = "$1$"; /* This string is magic for this algorithm.
* Having it this way, we can get get better
* later on */
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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;
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/* 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;
/* */
err = px_find_digest("md5", &ctx);
if (err)
return NULL;
err = px_find_digest("md5", &ctx1);
/* The password first, since that is what is most unknown */
px_md_update(ctx, (uint8 *) pw, strlen(pw));
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/* Then our magic string */
px_md_update(ctx, (uint8 *) magic, strlen(magic));
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/* Then the raw salt */
px_md_update(ctx, (uint8 *) sp, sl);
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/* Then just as many characters of the MD5(pw,salt,pw) */
px_md_update(ctx1, (uint8 *) pw, strlen(pw));
px_md_update(ctx1, (uint8 *) sp, sl);
px_md_update(ctx1, (uint8 *) pw, strlen(pw));
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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. */
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, (uint8 *) pw, 1);
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/* 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...
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*/
for (i = 0; i < 1000; i++)
{
px_md_reset(ctx1);
if (i & 1)
px_md_update(ctx1, (uint8 *) pw, strlen(pw));
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else
px_md_update(ctx1, final, MD5_SIZE);
if (i % 3)
px_md_update(ctx1, (uint8 *) sp, sl);
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if (i % 7)
px_md_update(ctx1, (uint8 *) pw, strlen(pw));
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if (i & 1)
px_md_update(ctx1, final, MD5_SIZE);
else
px_md_update(ctx1, (uint8 *) pw, strlen(pw));
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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. */
memset(final, 0, sizeof final);
px_md_free(ctx1);
px_md_free(ctx);
return passwd;
}