05c9a0158f
Add a strverscmp(3) function to libc, a GNU extension I implemented by reading its glibc manual page. It orders strings following a much more natural ordering (e.g. "ent1 < ent2 < ent10" as opposed to "ent1 < ent10 < ent2" with strcmp(3)'s lexicographic ordering). Also add versionsort(3) for use as scandir(3)'s compar argument. Update manual page for scandir(3) and add one for strverscmp(3). Reviewed by: pstef, gbe, kib MFC after: 1 week Differential Revision: https://reviews.freebsd.org/D35807
92 lines
2.2 KiB
C
92 lines
2.2 KiB
C
/*-
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* SPDX-License-Identifier: BSD-2-Clause
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* Copyright (c) 2022 Aymeric Wibo <obiwac@gmail.com>
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*/
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#include <ctype.h>
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#include <stddef.h>
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int
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strverscmp(const char *s1, const char *s2)
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{
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size_t digit_count_1, digit_count_2;
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size_t zeros_count_1, zeros_count_2;
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const unsigned char *num_1, *num_2;
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const unsigned char *u1 = __DECONST(const unsigned char *, s1);
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const unsigned char *u2 = __DECONST(const unsigned char *, s2);
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/*
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* If pointers are the same, no need to go through to process of
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* comparing them.
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*/
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if (s1 == s2)
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return (0);
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while (*u1 != '\0' && *u2 != '\0') {
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/* If either character is not a digit, act like strcmp(3). */
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if (!isdigit(*u1) || !isdigit(*u2)) {
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if (*u1 != *u2)
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return (*u1 - *u2);
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u1++;
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u2++;
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continue;
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}
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if (*u1 == '0' || *u2 == '0') {
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/*
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* Treat leading zeros as if they were the fractional
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* part of a number, i.e. as if they had a decimal point
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* in front. First, count the leading zeros (more zeros
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* == smaller number).
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*/
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zeros_count_1 = 0;
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zeros_count_2 = 0;
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for (; *u1 == '0'; u1++)
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zeros_count_1++;
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for (; *u2 == '0'; u2++)
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zeros_count_2++;
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if (zeros_count_1 != zeros_count_2)
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return (zeros_count_2 - zeros_count_1);
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/* Handle the case where 0 < 09. */
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if (!isdigit(*u1) && isdigit(*u2))
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return (1);
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if (!isdigit(*u2) && isdigit(*u1))
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return (-1);
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} else {
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/*
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* No leading zeros; we're simply comparing two numbers.
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* It is necessary to first count how many digits there
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* are before going back to compare each digit, so that
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* e.g. 7 is not considered larger than 60.
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*/
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num_1 = u1;
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num_2 = u2;
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/* Count digits (more digits == larger number). */
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for (; isdigit(*u1); u1++)
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;
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for (; isdigit(*u2); u2++)
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;
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digit_count_1 = u1 - num_1;
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digit_count_2 = u2 - num_2;
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if (digit_count_1 != digit_count_2)
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return (digit_count_1 - digit_count_2);
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/*
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* If there are the same number of digits, go back to
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* the start of the number.
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*/
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u1 = num_1;
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u2 = num_2;
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}
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/* Compare each digit until there are none left. */
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for (; isdigit(*u1) && isdigit(*u2); u1++, u2++) {
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if (*u1 != *u2)
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return (*u1 - *u2);
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}
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}
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return (*u1 - *u2);
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}
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