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 * %sccs.include.redist.c%
9 */
10
11 #if defined(LIBC_SCCS) && !defined(lint)
12 static char sccsid[] = "@(#)ns_addr.c 8.1 (Berkeley) 06/07/93";
13 #endif /* LIBC_SCCS and not lint */
14
15 #include <sys/param.h>
16 #include <netns/ns.h>
17 #include <stdio.h>
18 #include <string.h>
19
20 static struct ns_addr addr, zero_addr;
21
22 static void Field(), cvtbase();
23
24 struct ns_addr
ns_addr(name)25 ns_addr(name)
26 const char *name;
27 {
28 char separator;
29 char *hostname, *socketname, *cp;
30 char buf[50];
31
32 (void)strncpy(buf, name, sizeof(buf) - 1);
33 buf[sizeof(buf) - 1] = '\0';
34
35 /*
36 * First, figure out what he intends as a field separtor.
37 * Despite the way this routine is written, the prefered
38 * form 2-272.AA001234H.01777, i.e. XDE standard.
39 * Great efforts are made to insure backward compatability.
40 */
41 if (hostname = strchr(buf, '#'))
42 separator = '#';
43 else {
44 hostname = strchr(buf, '.');
45 if ((cp = strchr(buf, ':')) &&
46 ((hostname && cp < hostname) || (hostname == 0))) {
47 hostname = cp;
48 separator = ':';
49 } else
50 separator = '.';
51 }
52 if (hostname)
53 *hostname++ = 0;
54
55 addr = zero_addr;
56 Field(buf, addr.x_net.c_net, 4);
57 if (hostname == 0)
58 return (addr); /* No separator means net only */
59
60 socketname = strchr(hostname, separator);
61 if (socketname) {
62 *socketname++ = 0;
63 Field(socketname, (u_char *)&addr.x_port, 2);
64 }
65
66 Field(hostname, addr.x_host.c_host, 6);
67
68 return (addr);
69 }
70
71 static void
Field(buf,out,len)72 Field(buf, out, len)
73 char *buf;
74 u_char *out;
75 int len;
76 {
77 register char *bp = buf;
78 int i, ibase, base16 = 0, base10 = 0, clen = 0;
79 int hb[6], *hp;
80 char *fmt;
81
82 /*
83 * first try 2-273#2-852-151-014#socket
84 */
85 if ((*buf != '-') &&
86 (1 < (i = sscanf(buf, "%d-%d-%d-%d-%d",
87 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4])))) {
88 cvtbase(1000L, 256, hb, i, out, len);
89 return;
90 }
91 /*
92 * try form 8E1#0.0.AA.0.5E.E6#socket
93 */
94 if (1 < (i = sscanf(buf,"%x.%x.%x.%x.%x.%x",
95 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) {
96 cvtbase(256L, 256, hb, i, out, len);
97 return;
98 }
99 /*
100 * try form 8E1#0:0:AA:0:5E:E6#socket
101 */
102 if (1 < (i = sscanf(buf,"%x:%x:%x:%x:%x:%x",
103 &hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) {
104 cvtbase(256L, 256, hb, i, out, len);
105 return;
106 }
107 /*
108 * This is REALLY stretching it but there was a
109 * comma notation separting shorts -- definitely non standard
110 */
111 if (1 < (i = sscanf(buf,"%x,%x,%x",
112 &hb[0], &hb[1], &hb[2]))) {
113 hb[0] = htons(hb[0]); hb[1] = htons(hb[1]);
114 hb[2] = htons(hb[2]);
115 cvtbase(65536L, 256, hb, i, out, len);
116 return;
117 }
118
119 /* Need to decide if base 10, 16 or 8 */
120 while (*bp) switch (*bp++) {
121
122 case '0': case '1': case '2': case '3': case '4': case '5':
123 case '6': case '7': case '-':
124 break;
125
126 case '8': case '9':
127 base10 = 1;
128 break;
129
130 case 'a': case 'b': case 'c': case 'd': case 'e': case 'f':
131 case 'A': case 'B': case 'C': case 'D': case 'E': case 'F':
132 base16 = 1;
133 break;
134
135 case 'x': case 'X':
136 *--bp = '0';
137 base16 = 1;
138 break;
139
140 case 'h': case 'H':
141 base16 = 1;
142 /* fall into */
143
144 default:
145 *--bp = 0; /* Ends Loop */
146 }
147 if (base16) {
148 fmt = "%3x";
149 ibase = 4096;
150 } else if (base10 == 0 && *buf == '0') {
151 fmt = "%3o";
152 ibase = 512;
153 } else {
154 fmt = "%3d";
155 ibase = 1000;
156 }
157
158 for (bp = buf; *bp++; ) clen++;
159 if (clen == 0) clen++;
160 if (clen > 18) clen = 18;
161 i = ((clen - 1) / 3) + 1;
162 bp = clen + buf - 3;
163 hp = hb + i - 1;
164
165 while (hp > hb) {
166 (void)sscanf(bp, fmt, hp);
167 bp[0] = 0;
168 hp--;
169 bp -= 3;
170 }
171 (void)sscanf(buf, fmt, hp);
172 cvtbase((long)ibase, 256, hb, i, out, len);
173 }
174
175 static void
cvtbase(oldbase,newbase,input,inlen,result,reslen)176 cvtbase(oldbase,newbase,input,inlen,result,reslen)
177 long oldbase;
178 int newbase;
179 int input[];
180 int inlen;
181 unsigned char result[];
182 int reslen;
183 {
184 int d, e;
185 long sum;
186
187 e = 1;
188 while (e > 0 && reslen > 0) {
189 d = 0; e = 0; sum = 0;
190 /* long division: input=input/newbase */
191 while (d < inlen) {
192 sum = sum*oldbase + (long) input[d];
193 e += (sum > 0);
194 input[d++] = sum / newbase;
195 sum %= newbase;
196 }
197 result[--reslen] = sum; /* accumulate remainder */
198 }
199 for (d=0; d < reslen; d++)
200 result[d] = 0;
201 }
202