1 /*- 2 * SPDX-License-Identifier: BSD-2-Clause 3 * 4 * Copyright (c) 2003 Poul-Henning Kamp 5 * All rights reserved. 6 * 7 * Redistribution and use in source and binary forms, with or without 8 * modification, are permitted provided that the following conditions 9 * are met: 10 * 1. Redistributions of source code must retain the above copyright 11 * notice, this list of conditions and the following disclaimer. 12 * 2. Redistributions in binary form must reproduce the above copyright 13 * notice, this list of conditions and the following disclaimer in the 14 * documentation and/or other materials provided with the distribution. 15 * 16 * THIS SOFTWARE IS PROVIDED BY THE AUTHOR AND CONTRIBUTORS ``AS IS'' AND 17 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 18 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 19 * ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHOR OR CONTRIBUTORS BE LIABLE 20 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 21 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 22 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 23 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 24 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 25 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 26 * SUCH DAMAGE. 27 */ 28 29 #include <sys/types.h> 30 31 #include <err.h> 32 #include <md5.h> 33 #include <stdio.h> 34 #include <string.h> 35 #include <unistd.h> 36 37 #include "crypt.h" 38 39 /* 40 * UNIX password 41 */ 42 43 int 44 crypt_md5(const char *pw, const char *salt, char *buffer) 45 { 46 MD5_CTX ctx,ctx1; 47 unsigned long l; 48 int sl, pl; 49 u_int i; 50 u_char final[MD5_SIZE]; 51 const char *ep; 52 static const char *magic = "$1$"; 53 54 /* If the salt starts with the magic string, skip that. */ 55 if (!strncmp(salt, magic, strlen(magic))) 56 salt += strlen(magic); 57 58 /* It stops at the first '$', max 8 chars */ 59 for (ep = salt; *ep && *ep != '$' && ep < salt + 8; ep++) 60 continue; 61 62 /* get the length of the true salt */ 63 sl = ep - salt; 64 65 MD5Init(&ctx); 66 67 /* The password first, since that is what is most unknown */ 68 MD5Update(&ctx, (const u_char *)pw, strlen(pw)); 69 70 /* Then our magic string */ 71 MD5Update(&ctx, (const u_char *)magic, strlen(magic)); 72 73 /* Then the raw salt */ 74 MD5Update(&ctx, (const u_char *)salt, (u_int)sl); 75 76 /* Then just as many characters of the MD5(pw,salt,pw) */ 77 MD5Init(&ctx1); 78 MD5Update(&ctx1, (const u_char *)pw, strlen(pw)); 79 MD5Update(&ctx1, (const u_char *)salt, (u_int)sl); 80 MD5Update(&ctx1, (const u_char *)pw, strlen(pw)); 81 MD5Final(final, &ctx1); 82 for(pl = (int)strlen(pw); pl > 0; pl -= MD5_SIZE) 83 MD5Update(&ctx, (const u_char *)final, 84 (u_int)(pl > MD5_SIZE ? MD5_SIZE : pl)); 85 86 /* Don't leave anything around in vm they could use. */ 87 memset(final, 0, sizeof(final)); 88 89 /* Then something really weird... */ 90 for (i = strlen(pw); i; i >>= 1) 91 if(i & 1) 92 MD5Update(&ctx, (const u_char *)final, 1); 93 else 94 MD5Update(&ctx, (const u_char *)pw, 1); 95 96 /* Now make the output string */ 97 buffer = stpcpy(buffer, magic); 98 buffer = stpncpy(buffer, salt, (u_int)sl); 99 *buffer++ = '$'; 100 101 MD5Final(final, &ctx); 102 103 /* 104 * and now, just to make sure things don't run too fast 105 * On a 60 Mhz Pentium this takes 34 msec, so you would 106 * need 30 seconds to build a 1000 entry dictionary... 107 */ 108 for(i = 0; i < 1000; i++) { 109 MD5Init(&ctx1); 110 if(i & 1) 111 MD5Update(&ctx1, (const u_char *)pw, strlen(pw)); 112 else 113 MD5Update(&ctx1, (const u_char *)final, MD5_SIZE); 114 115 if(i % 3) 116 MD5Update(&ctx1, (const u_char *)salt, (u_int)sl); 117 118 if(i % 7) 119 MD5Update(&ctx1, (const u_char *)pw, strlen(pw)); 120 121 if(i & 1) 122 MD5Update(&ctx1, (const u_char *)final, MD5_SIZE); 123 else 124 MD5Update(&ctx1, (const u_char *)pw, strlen(pw)); 125 MD5Final(final, &ctx1); 126 } 127 128 l = (final[ 0]<<16) | (final[ 6]<<8) | final[12]; 129 _crypt_to64(buffer, l, 4); buffer += 4; 130 l = (final[ 1]<<16) | (final[ 7]<<8) | final[13]; 131 _crypt_to64(buffer, l, 4); buffer += 4; 132 l = (final[ 2]<<16) | (final[ 8]<<8) | final[14]; 133 _crypt_to64(buffer, l, 4); buffer += 4; 134 l = (final[ 3]<<16) | (final[ 9]<<8) | final[15]; 135 _crypt_to64(buffer, l, 4); buffer += 4; 136 l = (final[ 4]<<16) | (final[10]<<8) | final[ 5]; 137 _crypt_to64(buffer, l, 4); buffer += 4; 138 l = final[11]; 139 _crypt_to64(buffer, l, 2); buffer += 2; 140 *buffer = '\0'; 141 142 /* Don't leave anything around in vm they could use. */ 143 memset(final, 0, sizeof(final)); 144 145 return (0); 146 } 147