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