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9fe55259fd
Specifically, on-stack memset() might be removed, so: * Replace memset() with px_memset() * Add px_memset to copy_crlf() * Add px_memset to pgp-s2k.c Patch by Marko Kreen Report by PVS-Studio Backpatch through 8.4.
164 lines
3.6 KiB
C
164 lines
3.6 KiB
C
/*
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* File imported from FreeBSD, original by Poul-Henning Kamp.
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*
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* $FreeBSD: src/lib/libcrypt/crypt-md5.c,v 1.5 1999/12/17 20:21:45 peter Exp $
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*
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* contrib/pgcrypto/crypt-md5.c
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*/
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#include "postgres.h"
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#include "px.h"
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#include "px-crypt.h"
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#define MD5_SIZE 16
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static const char _crypt_a64[] =
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"./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
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static void
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_crypt_to64(char *s, unsigned long v, int n)
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{
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while (--n >= 0)
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{
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*s++ = _crypt_a64[v & 0x3f];
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v >>= 6;
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}
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}
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/*
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* UNIX password
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*/
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char *
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px_crypt_md5(const char *pw, const char *salt, char *passwd, unsigned dstlen)
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{
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static char *magic = "$1$"; /* This string is magic for this algorithm.
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* Having it this way, we can get better later
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* on */
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static char *p;
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static const char *sp,
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*ep;
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unsigned char final[MD5_SIZE];
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int sl,
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pl,
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i;
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PX_MD *ctx,
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*ctx1;
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int err;
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unsigned long l;
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if (!passwd || dstlen < 120)
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return NULL;
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/* Refine the Salt first */
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sp = salt;
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/* If it starts with the magic string, then skip that */
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if (strncmp(sp, magic, strlen(magic)) == 0)
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sp += strlen(magic);
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/* It stops at the first '$', max 8 chars */
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for (ep = sp; *ep && *ep != '$' && ep < (sp + 8); ep++)
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continue;
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/* get the length of the true salt */
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sl = ep - sp;
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/* */
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err = px_find_digest("md5", &ctx);
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if (err)
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return NULL;
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err = px_find_digest("md5", &ctx1);
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/* The password first, since that is what is most unknown */
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px_md_update(ctx, (const uint8 *) pw, strlen(pw));
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/* Then our magic string */
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px_md_update(ctx, (uint8 *) magic, strlen(magic));
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/* Then the raw salt */
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px_md_update(ctx, (const uint8 *) sp, sl);
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/* Then just as many characters of the MD5(pw,salt,pw) */
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px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
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px_md_update(ctx1, (const uint8 *) sp, sl);
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px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
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px_md_finish(ctx1, final);
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for (pl = strlen(pw); pl > 0; pl -= MD5_SIZE)
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px_md_update(ctx, final, pl > MD5_SIZE ? MD5_SIZE : pl);
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/* Don't leave anything around in vm they could use. */
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px_memset(final, 0, sizeof final);
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/* Then something really weird... */
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for (i = strlen(pw); i; i >>= 1)
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if (i & 1)
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px_md_update(ctx, final, 1);
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else
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px_md_update(ctx, (const uint8 *) pw, 1);
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/* Now make the output string */
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strcpy(passwd, magic);
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strncat(passwd, sp, sl);
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strcat(passwd, "$");
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px_md_finish(ctx, final);
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/*
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* and now, just to make sure things don't run too fast On a 60 Mhz
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* Pentium this takes 34 msec, so you would need 30 seconds to build a
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* 1000 entry dictionary...
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*/
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for (i = 0; i < 1000; i++)
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{
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px_md_reset(ctx1);
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if (i & 1)
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px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
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else
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px_md_update(ctx1, final, MD5_SIZE);
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if (i % 3)
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px_md_update(ctx1, (const uint8 *) sp, sl);
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if (i % 7)
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px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
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if (i & 1)
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px_md_update(ctx1, final, MD5_SIZE);
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else
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px_md_update(ctx1, (const uint8 *) pw, strlen(pw));
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px_md_finish(ctx1, final);
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}
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p = passwd + strlen(passwd);
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l = (final[0] << 16) | (final[6] << 8) | final[12];
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_crypt_to64(p, l, 4);
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p += 4;
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l = (final[1] << 16) | (final[7] << 8) | final[13];
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_crypt_to64(p, l, 4);
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p += 4;
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l = (final[2] << 16) | (final[8] << 8) | final[14];
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_crypt_to64(p, l, 4);
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p += 4;
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l = (final[3] << 16) | (final[9] << 8) | final[15];
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_crypt_to64(p, l, 4);
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p += 4;
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l = (final[4] << 16) | (final[10] << 8) | final[5];
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_crypt_to64(p, l, 4);
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p += 4;
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l = final[11];
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_crypt_to64(p, l, 2);
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p += 2;
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*p = '\0';
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/* Don't leave anything around in vm they could use. */
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px_memset(final, 0, sizeof final);
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px_md_free(ctx1);
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px_md_free(ctx);
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return passwd;
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}
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