ipx_addr.c revision 17141
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 * Redistribution and use in source and binary forms, with or without 9 * modification, are permitted provided that the following conditions 10 * are met: 11 * 1. Redistributions of source code must retain the above copyright 12 * notice, this list of conditions and the following disclaimer. 13 * 2. Redistributions in binary form must reproduce the above copyright 14 * notice, this list of conditions and the following disclaimer in the 15 * documentation and/or other materials provided with the distribution. 16 * 3. All advertising materials mentioning features or use of this software 17 * must display the following acknowledgement: 18 * This product includes software developed by the University of 19 * California, Berkeley and its contributors. 20 * 4. Neither the name of the University nor the names of its contributors 21 * may be used to endorse or promote products derived from this software 22 * without specific prior written permission. 23 * 24 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 25 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 26 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 27 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 28 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 29 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 30 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 31 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 32 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 33 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 34 * SUCH DAMAGE. 35 */ 36 37#if defined(LIBC_SCCS) && !defined(lint) 38static char sccsid[] = "@(#)ipx_addr.c"; 39#endif /* LIBC_SCCS and not lint */ 40 41#include <sys/param.h> 42#include <netipx/ipx.h> 43#include <stdio.h> 44#include <string.h> 45 46static struct ipx_addr addr, zero_addr; 47 48static void Field(), cvtbase(); 49 50struct ipx_addr 51ipx_addr(name) 52 const char *name; 53{ 54 char separator; 55 char *hostname, *socketname, *cp; 56 char buf[50]; 57 58 (void)strncpy(buf, name, sizeof(buf) - 1); 59 buf[sizeof(buf) - 1] = '\0'; 60 61 /* 62 * First, figure out what he intends as a field separtor. 63 * Despite the way this routine is written, the prefered 64 * form 2-272.AA001234H.01777, i.e. XDE standard. 65 * Great efforts are made to insure backward compatability. 66 */ 67 if ( (hostname = strchr(buf, '#')) ) 68 separator = '#'; 69 else { 70 hostname = strchr(buf, '.'); 71 if ((cp = strchr(buf, ':')) && 72 ((hostname && cp < hostname) || (hostname == 0))) { 73 hostname = cp; 74 separator = ':'; 75 } else 76 separator = '.'; 77 } 78 if (hostname) 79 *hostname++ = 0; 80 81 addr = zero_addr; 82 Field(buf, addr.x_net.c_net, 4); 83 if (hostname == 0) 84 return (addr); /* No separator means net only */ 85 86 socketname = strchr(hostname, separator); 87 if (socketname) { 88 *socketname++ = 0; 89 Field(socketname, (u_char *)&addr.x_port, 2); 90 } 91 92 Field(hostname, addr.x_host.c_host, 6); 93 94 return (addr); 95} 96 97static void 98Field(buf, out, len) 99 char *buf; 100 u_char *out; 101 int len; 102{ 103 register char *bp = buf; 104 int i, ibase, base16 = 0, base10 = 0, clen = 0; 105 int hb[6], *hp; 106 char *fmt; 107 108 /* 109 * first try 2-273#2-852-151-014#socket 110 */ 111 if ((*buf != '-') && 112 (1 < (i = sscanf(buf, "%d-%d-%d-%d-%d", 113 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4])))) { 114 cvtbase(1000L, 256, hb, i, out, len); 115 return; 116 } 117 /* 118 * try form 8E1#0.0.AA.0.5E.E6#socket 119 */ 120 if (1 < (i = sscanf(buf,"%x.%x.%x.%x.%x.%x", 121 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { 122 cvtbase(256L, 256, hb, i, out, len); 123 return; 124 } 125 /* 126 * try form 8E1#0:0:AA:0:5E:E6#socket 127 */ 128 if (1 < (i = sscanf(buf,"%x:%x:%x:%x:%x:%x", 129 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { 130 cvtbase(256L, 256, hb, i, out, len); 131 return; 132 } 133 /* 134 * This is REALLY stretching it but there was a 135 * comma notation separting shorts -- definitely non standard 136 */ 137 if (1 < (i = sscanf(buf,"%x,%x,%x", 138 &hb[0], &hb[1], &hb[2]))) { 139 hb[0] = htons(hb[0]); hb[1] = htons(hb[1]); 140 hb[2] = htons(hb[2]); 141 cvtbase(65536L, 256, hb, i, out, len); 142 return; 143 } 144 145 /* Need to decide if base 10, 16 or 8 */ 146 while (*bp) switch (*bp++) { 147 148 case '0': case '1': case '2': case '3': case '4': case '5': 149 case '6': case '7': case '-': 150 break; 151 152 case '8': case '9': 153 base10 = 1; 154 break; 155 156 case 'a': case 'b': case 'c': case 'd': case 'e': case 'f': 157 case 'A': case 'B': case 'C': case 'D': case 'E': case 'F': 158 base16 = 1; 159 break; 160 161 case 'x': case 'X': 162 *--bp = '0'; 163 base16 = 1; 164 break; 165 166 case 'h': case 'H': 167 base16 = 1; 168 /* fall into */ 169 170 default: 171 *--bp = 0; /* Ends Loop */ 172 } 173 if (base16) { 174 fmt = "%3x"; 175 ibase = 4096; 176 } else if (base10 == 0 && *buf == '0') { 177 fmt = "%3o"; 178 ibase = 512; 179 } else { 180 fmt = "%3d"; 181 ibase = 1000; 182 } 183 184 for (bp = buf; *bp++; ) clen++; 185 if (clen == 0) clen++; 186 if (clen > 18) clen = 18; 187 i = ((clen - 1) / 3) + 1; 188 bp = clen + buf - 3; 189 hp = hb + i - 1; 190 191 while (hp > hb) { 192 (void)sscanf(bp, fmt, hp); 193 bp[0] = 0; 194 hp--; 195 bp -= 3; 196 } 197 (void)sscanf(buf, fmt, hp); 198 cvtbase((long)ibase, 256, hb, i, out, len); 199} 200 201static void 202cvtbase(oldbase,newbase,input,inlen,result,reslen) 203 long oldbase; 204 int newbase; 205 int input[]; 206 int inlen; 207 unsigned char result[]; 208 int reslen; 209{ 210 int d, e; 211 long sum; 212 213 e = 1; 214 while (e > 0 && reslen > 0) { 215 d = 0; e = 0; sum = 0; 216 /* long division: input=input/newbase */ 217 while (d < inlen) { 218 sum = sum*oldbase + (long) input[d]; 219 e += (sum > 0); 220 input[d++] = sum / newbase; 221 sum %= newbase; 222 } 223 result[--reslen] = sum; /* accumulate remainder */ 224 } 225 for (d=0; d < reslen; d++) 226 result[d] = 0; 227} 228