77 lines
3.1 KiB
C
77 lines
3.1 KiB
C
/*-*- mode:c;indent-tabs-mode:nil;c-basic-offset:2;tab-width:8;coding:utf-8 -*-│
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│vi: set net ft=c ts=2 sts=2 sw=2 fenc=utf-8 :vi│
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╞══════════════════════════════════════════════════════════════════════════════╡
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│ Copyright 2020 Justine Alexandra Roberts Tunney │
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│ │
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│ This program is free software; you can redistribute it and/or modify │
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│ it under the terms of the GNU General Public License as published by │
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│ the Free Software Foundation; version 2 of the License. │
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│ │
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│ This program is distributed in the hope that it will be useful, but │
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│ WITHOUT ANY WARRANTY; without even the implied warranty of │
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│ MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU │
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│ General Public License for more details. │
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│ │
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│ You should have received a copy of the GNU General Public License │
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│ along with this program; if not, write to the Free Software │
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│ Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA │
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│ 02110-1301 USA │
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╚─────────────────────────────────────────────────────────────────────────────*/
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#include "libc/bits/safemacros.h"
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#include "libc/dns/dns.h"
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#include "libc/str/str.h"
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forceinline void FindDnsLabel(const char *A, size_t *i, size_t *n) {
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while (*i) {
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if (A[*i - 1] == '.') {
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if (*i == *n) {
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--(*n);
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} else {
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break;
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}
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}
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--(*i);
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}
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}
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/**
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* Compares DNS hostnames in reverse lexicographical asciibetical order.
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* @return <0, 0, or >0
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* @see test/libc/dns/dnsnamecmp_test.c (the code that matters)
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*/
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int dnsnamecmp(const char *A, const char *B) {
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if (A == B) return 0;
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size_t n = strlen(A);
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size_t m = strlen(B);
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if (!n || !m || ((A[n - 1] == '.') ^ (B[m - 1] == '.'))) {
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if (n && m && A[n - 1] == '.' && strchr(B, '.')) {
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--m;
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} else if (n && m && B[m - 1] == '.' && strchr(A, '.')) {
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--n;
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} else {
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return A[n ? n - 1 : 0] - B[m ? m - 1 : 0];
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}
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}
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size_t i = n;
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size_t j = m;
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bool first = true;
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for (;;) {
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FindDnsLabel(A, &i, &n);
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FindDnsLabel(B, &j, &m);
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if (first) {
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first = false;
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if (!i && j) return 1;
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if (!j && i) return -1;
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}
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int res;
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if ((res = strncasecmp(&A[i], &B[j], min(n - i + 1, m - j + 1)))) {
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return res;
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}
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if (!i || !j) {
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return i - j;
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}
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n = i;
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m = j;
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}
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}
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