/* | |
* Copyright (c) 1986, 1993 | |
* The Regents of the University of California. All rights reserved. | |
* | |
* This code is derived from software contributed to Berkeley by | |
* J.Q. Johnson. | |
* | |
* Portions copyright (c) 1999, 2000 | |
* Intel Corporation. | |
* All rights reserved. | |
* | |
* Redistribution and use in source and binary forms, with or without | |
* modification, are permitted provided that the following conditions | |
* are met: | |
* | |
* 1. Redistributions of source code must retain the above copyright | |
* notice, this list of conditions and the following disclaimer. | |
* | |
* 2. Redistributions in binary form must reproduce the above copyright | |
* notice, this list of conditions and the following disclaimer in the | |
* documentation and/or other materials provided with the distribution. | |
* | |
* 3. All advertising materials mentioning features or use of this software | |
* must display the following acknowledgement: | |
* | |
* This product includes software developed by the University of | |
* California, Berkeley, Intel Corporation, and its contributors. | |
* | |
* 4. Neither the name of University, Intel Corporation, or their respective | |
* contributors may be used to endorse or promote products derived from | |
* this software without specific prior written permission. | |
* | |
* THIS SOFTWARE IS PROVIDED BY THE REGENTS, INTEL CORPORATION AND | |
* CONTRIBUTORS ``AS IS'' AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, | |
* BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS | |
* FOR A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS, | |
* INTEL CORPORATION OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, | |
* INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT | |
* NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, | |
* DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY | |
* THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT | |
* (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF | |
* THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE. | |
* | |
*/ | |
#if defined(LIBC_SCCS) && !defined(lint) | |
static char sccsid[] = "@(#)ns_addr.c 8.1 (Berkeley) 6/7/93"; | |
#endif /* LIBC_SCCS and not lint */ | |
#include <sys/param.h> | |
#include <netns/ns.h> | |
#include <stdio.h> | |
#include <string.h> | |
static struct ns_addr addr, zero_addr; | |
static void Field (char *buf, u_char *out, int len); | |
static void cvtbase (long oldbase, int newbase, int input[], int inlen, unsigned char result[], int reslen); | |
struct ns_addr | |
ns_addr( | |
const char *name | |
) | |
{ | |
char separator; | |
char *hostname, *socketname, *cp; | |
char buf[50]; | |
(void)strncpy(buf, name, sizeof(buf) - 1); | |
buf[sizeof(buf) - 1] = '\0'; | |
/* | |
* First, figure out what he intends as a field separtor. | |
* Despite the way this routine is written, the prefered | |
* form 2-272.AA001234H.01777, i.e. XDE standard. | |
* Great efforts are made to insure backward compatability. | |
*/ | |
if ((hostname = strchr(buf, '#')) != NULL) | |
separator = '#'; | |
else { | |
hostname = strchr(buf, '.'); | |
if ((cp = strchr(buf, ':')) && | |
((hostname && cp < hostname) || (hostname == 0))) { | |
hostname = cp; | |
separator = ':'; | |
} else | |
separator = '.'; | |
} | |
if (hostname) | |
*hostname++ = 0; | |
addr = zero_addr; | |
Field(buf, addr.x_net.c_net, 4); | |
if (hostname == 0) | |
return (addr); /* No separator means net only */ | |
socketname = strchr(hostname, separator); | |
if (socketname) { | |
*socketname++ = 0; | |
Field(socketname, (u_char *)&addr.x_port, 2); | |
} | |
Field(hostname, addr.x_host.c_host, 6); | |
return (addr); | |
} | |
static void | |
Field( | |
char *buf, | |
u_char *out, | |
int len | |
) | |
{ | |
register char *bp = buf; | |
int i, ibase, base16 = 0, base10 = 0, clen = 0; | |
int hb[6], *hp; | |
char *fmt; | |
/* | |
* first try 2-273#2-852-151-014#socket | |
*/ | |
if ((*buf != '-') && | |
(1 < (i = sscanf(buf, "%d-%d-%d-%d-%d", | |
&hb[0], &hb[1], &hb[2], &hb[3], &hb[4])))) { | |
cvtbase(1000L, 256, hb, i, out, len); | |
return; | |
} | |
/* | |
* try form 8E1#0.0.AA.0.5E.E6#socket | |
*/ | |
if (1 < (i = sscanf(buf,"%x.%x.%x.%x.%x.%x", | |
&hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { | |
cvtbase(256L, 256, hb, i, out, len); | |
return; | |
} | |
/* | |
* try form 8E1#0:0:AA:0:5E:E6#socket | |
*/ | |
if (1 < (i = sscanf(buf,"%x:%x:%x:%x:%x:%x", | |
&hb[0], &hb[1], &hb[2], &hb[3], &hb[4], &hb[5]))) { | |
cvtbase(256L, 256, hb, i, out, len); | |
return; | |
} | |
/* | |
* This is REALLY stretching it but there was a | |
* comma notation separting shorts -- definitely non standard | |
*/ | |
if (1 < (i = sscanf(buf,"%x,%x,%x", | |
&hb[0], &hb[1], &hb[2]))) { | |
hb[0] = htons(hb[0]); hb[1] = htons(hb[1]); | |
hb[2] = htons(hb[2]); | |
cvtbase(65536L, 256, hb, i, out, len); | |
return; | |
} | |
/* Need to decide if base 10, 16 or 8 */ | |
while (*bp) switch (*bp++) { | |
case '0': case '1': case '2': case '3': case '4': case '5': | |
case '6': case '7': case '-': | |
break; | |
case '8': case '9': | |
base10 = 1; | |
break; | |
case 'a': case 'b': case 'c': case 'd': case 'e': case 'f': | |
case 'A': case 'B': case 'C': case 'D': case 'E': case 'F': | |
base16 = 1; | |
break; | |
case 'x': case 'X': | |
*--bp = '0'; | |
base16 = 1; | |
break; | |
case 'h': case 'H': | |
base16 = 1; | |
/* fall into */ | |
default: | |
*--bp = 0; /* Ends Loop */ | |
} | |
if (base16) { | |
fmt = "%3x"; | |
ibase = 4096; | |
} else if (base10 == 0 && *buf == '0') { | |
fmt = "%3o"; | |
ibase = 512; | |
} else { | |
fmt = "%3d"; | |
ibase = 1000; | |
} | |
for (bp = buf; *bp++; ) clen++; | |
if (clen == 0) clen++; | |
if (clen > 18) clen = 18; | |
i = ((clen - 1) / 3) + 1; | |
bp = clen + buf - 3; | |
hp = hb + i - 1; | |
while (hp > hb) { | |
(void)sscanf(bp, fmt, hp); | |
bp[0] = 0; | |
hp--; | |
bp -= 3; | |
} | |
(void)sscanf(buf, fmt, hp); | |
cvtbase((long)ibase, 256, hb, i, out, len); | |
} | |
static void | |
cvtbase( | |
long oldbase, | |
int newbase, | |
int input[], | |
int inlen, | |
unsigned char result[], | |
int reslen | |
) | |
{ | |
int d, e; | |
long sum; | |
e = 1; | |
while (e > 0 && reslen > 0) { | |
d = 0; e = 0; sum = 0; | |
/* long division: input=input/newbase */ | |
while (d < inlen) { | |
sum = sum*oldbase + (long) input[d]; | |
e += (sum > 0); | |
input[d++] = sum / newbase; | |
sum %= newbase; | |
} | |
result[--reslen] = (u_char)sum; /* accumulate remainder */ | |
} | |
for (d=0; d < reslen; d++) | |
result[d] = 0; | |
} |