1 /* 2 * Copyright (c) 1986, 1993 3 * The Regents of the University of California. All rights reserved. 4 * 5 * This code is derived from software contributed to Berkeley by 6 * J.Q. Johnson. 7 * 8 * Portions copyright (c) 1999, 2000 9 * Intel Corporation. 10 * All rights reserved. 11 * 12 * Redistribution and use in source and binary forms, with or without 13 * modification, are permitted provided that the following conditions 14 * are met: 15 * 16 * 1. Redistributions of source code must retain the above copyright 17 * notice, this list of conditions and the following disclaimer. 18 * 19 * 2. Redistributions in binary form must reproduce the above copyright 20 * notice, this list of conditions and the following disclaimer in the 21 * documentation and/or other materials provided with the distribution. 22 * 23 * 3. All advertising materials mentioning features or use of this software 24 * must display the following acknowledgement: 25 * 26 * This product includes software developed by the University of 27 * California, Berkeley, Intel Corporation, and its contributors. 28 * 29 * 4. Neither the name of University, Intel Corporation, or their respective 30 * contributors may be used to endorse or promote products derived from 31 * this software without specific prior written permission. 32 * 33 * THIS SOFTWARE IS PROVIDED BY THE REGENTS, INTEL CORPORATION AND 34 * CONTRIBUTORS ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, 35 * BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS 36 * FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS, 37 * INTEL CORPORATION OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, 38 * INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT 39 * NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, 40 * DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY 41 * THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT 42 * (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF 43 * THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. 44 * 45 */ 46 47 #if defined(LIBC_SCCS) && !defined(lint) 48 static char sccsid[] = "@(#)ns_addr.c 8.1 (Berkeley) 6/7/93"; 49 #endif /* LIBC_SCCS and not lint */ 50 51 #include <sys/param.h> 52 #include <netns/ns.h> 53 #include <stdio.h> 54 #include <string.h> 55 56 static struct ns_addr addr, zero_addr; 57 58 static void Field (char *buf, u_char *out, int len); 59 static void cvtbase (long oldbase, int newbase, int input[], int inlen, unsigned char result[], int reslen); 60 61 struct ns_addr 62 ns_addr( 63 const char *name 64 ) 65 { 66 char separator; 67 char *hostname, *socketname, *cp; 68 char buf[50]; 69 70 (void)strncpy(buf, name, sizeof(buf) - 1); 71 buf[sizeof(buf) - 1] = '\0'; 72 73 /* 74 * First, figure out what he intends as a field separtor. 75 * Despite the way this routine is written, the prefered 76 * form 2-272.AA001234H.01777, i.e. XDE standard. 77 * Great efforts are made to insure backward compatability. 78 */ 79 if ((hostname = strchr(buf, '#')) != NULL) 80 separator = '#'; 81 else { 82 hostname = strchr(buf, '.'); 83 if ((cp = strchr(buf, ':')) && 84 ((hostname && cp < hostname) || (hostname == 0))) { 85 hostname = cp; 86 separator = ':'; 87 } else 88 separator = '.'; 89 } 90 if (hostname) 91 *hostname++ = 0; 92 93 addr = zero_addr; 94 Field(buf, addr.x_net.c_net, 4); 95 if (hostname == 0) 96 return (addr); /* No separator means net only */ 97 98 socketname = strchr(hostname, separator); 99 if (socketname) { 100 *socketname++ = 0; 101 Field(socketname, (u_char *)&addr.x_port, 2); 102 } 103 104 Field(hostname, addr.x_host.c_host, 6); 105 106 return (addr); 107 } 108 109 static void 110 Field( 111 char *buf, 112 u_char *out, 113 int len 114 ) 115 { 116 register char *bp = buf; 117 int i, ibase, base16 = 0, base10 = 0, clen = 0; 118 int hb[6], *hp; 119 char *fmt; 120 121 /* 122 * first try 2-273#2-852-151-014#socket 123 */ 124 if ((*buf != '-') && 125 (1 < (i = sscanf(buf, "%d-%d-%d-%d-%d", 126 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4])))) { 127 cvtbase(1000L, 256, hb, i, out, len); 128 return; 129 } 130 /* 131 * try form 8E1#0.0.AA.0.5E.E6#socket 132 */ 133 if (1 < (i = sscanf(buf,"%x.%x.%x.%x.%x.%x", 134 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { 135 cvtbase(256L, 256, hb, i, out, len); 136 return; 137 } 138 /* 139 * try form 8E1#0:0:AA:0:5E:E6#socket 140 */ 141 if (1 < (i = sscanf(buf,"%x:%x:%x:%x:%x:%x", 142 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { 143 cvtbase(256L, 256, hb, i, out, len); 144 return; 145 } 146 /* 147 * This is REALLY stretching it but there was a 148 * comma notation separting shorts -- definitely non standard 149 */ 150 if (1 < (i = sscanf(buf,"%x,%x,%x", 151 &hb[0], &hb[1], &hb[2]))) { 152 hb[0] = htons(hb[0]); hb[1] = htons(hb[1]); 153 hb[2] = htons(hb[2]); 154 cvtbase(65536L, 256, hb, i, out, len); 155 return; 156 } 157 158 /* Need to decide if base 10, 16 or 8 */ 159 while (*bp) switch (*bp++) { 160 161 case '0': case '1': case '2': case '3': case '4': case '5': 162 case '6': case '7': case '-': 163 break; 164 165 case '8': case '9': 166 base10 = 1; 167 break; 168 169 case 'a': case 'b': case 'c': case 'd': case 'e': case 'f': 170 case 'A': case 'B': case 'C': case 'D': case 'E': case 'F': 171 base16 = 1; 172 break; 173 174 case 'x': case 'X': 175 *--bp = '0'; 176 base16 = 1; 177 break; 178 179 case 'h': case 'H': 180 base16 = 1; 181 /* fall into */ 182 183 default: 184 *--bp = 0; /* Ends Loop */ 185 } 186 if (base16) { 187 fmt = "%3x"; 188 ibase = 4096; 189 } else if (base10 == 0 && *buf == '0') { 190 fmt = "%3o"; 191 ibase = 512; 192 } else { 193 fmt = "%3d"; 194 ibase = 1000; 195 } 196 197 for (bp = buf; *bp++; ) clen++; 198 if (clen == 0) clen++; 199 if (clen > 18) clen = 18; 200 i = ((clen - 1) / 3) + 1; 201 bp = clen + buf - 3; 202 hp = hb + i - 1; 203 204 while (hp > hb) { 205 (void)sscanf(bp, fmt, hp); 206 bp[0] = 0; 207 hp--; 208 bp -= 3; 209 } 210 (void)sscanf(buf, fmt, hp); 211 cvtbase((long)ibase, 256, hb, i, out, len); 212 } 213 214 static void 215 cvtbase( 216 long oldbase, 217 int newbase, 218 int input[], 219 int inlen, 220 unsigned char result[], 221 int reslen 222 ) 223 { 224 int d, e; 225 long sum; 226 227 e = 1; 228 while (e > 0 && reslen > 0) { 229 d = 0; e = 0; sum = 0; 230 /* long division: input=input/newbase */ 231 while (d < inlen) { 232 sum = sum*oldbase + (long) input[d]; 233 e += (sum > 0); 234 input[d++] = sum / newbase; 235 sum %= newbase; 236 } 237 result[--reslen] = (u_char)sum; /* accumulate remainder */ 238 } 239 for (d=0; d < reslen; d++) 240 result[d] = 0; 241 } 242
