1 /* 2 * Copyright (c) 1986 Regents of the University of California. 3 * All rights reserved. The Berkeley software License Agreement 4 * specifies the terms and conditions for redistribution. 5 * 6 * Includes material written at Cornell University, by J. Q. Johnson. 7 * Used by permission. 8 */ 9 10 #if defined(LIBC_SCCS) && !defined(lint) 11 static char sccsid[] = "@(#)ns_addr.c 6.3 (Berkeley) 09/30/87"; 12 #endif LIBC_SCCS and not lint 13 14 #include <sys/types.h> 15 #include <netns/ns.h> 16 17 static struct ns_addr addr, zero_addr; 18 19 struct ns_addr 20 ns_addr(name) 21 char *name; 22 { 23 char separator = '.'; 24 char *hostname, *socketname, *cp; 25 char buf[50]; 26 extern char *index(); 27 28 addr = zero_addr; 29 (void)strncpy(buf, name, 49); 30 31 /* 32 * First, figure out what he intends as a field separtor. 33 * Despite the way this routine is written, the prefered 34 * form 2-272.AA001234H.01777, i.e. XDE standard. 35 * Great efforts are made to insure backward compatability. 36 */ 37 if (hostname = index(buf, '#')) 38 separator = '#'; 39 else { 40 hostname = index(buf, '.'); 41 if ((cp = index(buf, ':')) && 42 ( (hostname && cp < hostname) || (hostname == 0))) { 43 hostname = cp; 44 separator = ':'; 45 } 46 } 47 if (hostname) 48 *hostname++ = 0; 49 Field(buf, addr.x_net.c_net, 4); 50 if (hostname == 0) 51 return (addr); /* No separator means net only */ 52 53 socketname = index(hostname, separator); 54 if (socketname) { 55 *socketname++ = 0; 56 Field(socketname, (u_char *)&addr.x_port, 2); 57 } 58 59 Field(hostname, addr.x_host.c_host, 6); 60 61 return (addr); 62 } 63 64 static 65 Field(buf, out, len) 66 char *buf; 67 u_char *out; 68 int len; 69 { 70 register char *bp = buf; 71 int i, ibase, base16 = 0, base10 = 0, clen = 0; 72 int hb[6], *hp; 73 char *fmt; 74 75 /* 76 * first try 2-273#2-852-151-014#socket 77 */ 78 if ((*buf != '-') && 79 (1 < (i = sscanf(buf, "%d-%d-%d-%d-%d", 80 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4])))) { 81 cvtbase(1000L, 256, hb, i, out, len); 82 return; 83 } 84 /* 85 * try form 8E1#0.0.AA.0.5E.E6#socket 86 */ 87 if (1 < (i = sscanf(buf,"%x.%x.%x.%x.%x.%x", 88 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { 89 cvtbase(256L, 256, hb, i, out, len); 90 return; 91 } 92 /* 93 * try form 8E1#0:0:AA:0:5E:E6#socket 94 */ 95 if (1 < (i = sscanf(buf,"%x:%x:%x:%x:%x:%x", 96 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { 97 cvtbase(256L, 256, hb, i, out, len); 98 return; 99 } 100 /* 101 * This is REALLY stretching it but there was a 102 * comma notation separting shorts -- definitely non standard 103 */ 104 if (1 < (i = sscanf(buf,"%x,%x,%x", 105 &hb[0], &hb[1], &hb[2]))) { 106 hb[0] = htons(hb[0]); hb[1] = htons(hb[1]); 107 hb[2] = htons(hb[2]); 108 cvtbase(65536L, 256, hb, i, out, len); 109 return; 110 } 111 112 /* Need to decide if base 10, 16 or 8 */ 113 while (*bp) switch (*bp++) { 114 115 case '0': case '1': case '2': case '3': case '4': case '5': 116 case '6': case '7': case '-': 117 break; 118 119 case '8': case '9': 120 base10 = 1; 121 break; 122 123 case 'a': case 'b': case 'c': case 'd': case 'e': case 'f': 124 case 'A': case 'B': case 'C': case 'D': case 'E': case 'F': 125 base16 = 1; 126 break; 127 128 case 'x': case 'X': 129 *--bp = '0'; 130 base16 = 1; 131 break; 132 133 case 'h': case 'H': 134 base16 = 1; 135 /* fall into */ 136 137 default: 138 *--bp = 0; /* Ends Loop */ 139 } 140 if (base16) { 141 fmt = "%3x"; 142 ibase = 4096; 143 } else if (base10 == 0 && *buf == '0') { 144 fmt = "%3o"; 145 ibase = 512; 146 } else { 147 fmt = "%3d"; 148 ibase = 1000; 149 } 150 151 for (bp = buf; *bp++; ) clen++; 152 if (clen == 0) clen++; 153 if (clen > 18) clen = 18; 154 i = ((clen - 1) / 3) + 1; 155 bp = clen + buf - 3; 156 hp = hb + i - 1; 157 158 while (hp > hb) { 159 (void)sscanf(bp, fmt, hp); 160 bp[0] = 0; 161 hp--; 162 bp -= 3; 163 } 164 (void)sscanf(buf, fmt, hp); 165 cvtbase((long)ibase, 256, hb, i, out, len); 166 } 167 168 static 169 cvtbase(oldbase,newbase,input,inlen,result,reslen) 170 long oldbase; 171 int newbase; 172 int input[]; 173 int inlen; 174 unsigned char result[]; 175 int reslen; 176 { 177 int d, e; 178 long sum; 179 180 e = 1; 181 while (e > 0 && reslen > 0) { 182 d = 0; e = 0; sum = 0; 183 /* long division: input=input/newbase */ 184 while (d < inlen) { 185 sum = sum*oldbase + (long) input[d]; 186 e += (sum > 0); 187 input[d++] = sum / newbase; 188 sum %= newbase; 189 } 190 result[--reslen] = sum; /* accumulate remainder */ 191 } 192 for (d=0; d < reslen; d++) 193 result[d] = 0; 194 } 195