xref: /freebsd/lib/libc/string/strverscmp.c (revision 61e21613)
1 /*-
2 * SPDX-License-Identifier: BSD-2-Clause
3 * Copyright (c) 2022 Aymeric Wibo <obiwac@gmail.com>
4 */
5 
6 #include <ctype.h>
7 #include <stddef.h>
8 
9 int
10 strverscmp(const char *s1, const char *s2)
11 {
12 	size_t digit_count_1, digit_count_2;
13 	size_t zeros_count_1, zeros_count_2;
14 	const unsigned char *num_1, *num_2;
15 	const unsigned char *u1 = __DECONST(const unsigned char *, s1);
16 	const unsigned char *u2 = __DECONST(const unsigned char *, s2);
17 
18 	/*
19 	 * If pointers are the same, no need to go through to process of
20 	 * comparing them.
21 	 */
22 	if (s1 == s2)
23 		return (0);
24 
25 	while (*u1 != '\0' && *u2 != '\0') {
26 		/* If either character is not a digit, act like strcmp(3). */
27 
28 		if (!isdigit(*u1) || !isdigit(*u2)) {
29 			if (*u1 != *u2)
30 				return (*u1 - *u2);
31 			u1++;
32 			u2++;
33 			continue;
34 		}
35 		if (*u1 == '0' || *u2 == '0') {
36 			/*
37 			 * Treat leading zeros as if they were the fractional
38 			 * part of a number, i.e. as if they had a decimal point
39 			 * in front. First, count the leading zeros (more zeros
40 			 * == smaller number).
41 			 */
42 			zeros_count_1 = 0;
43 			zeros_count_2 = 0;
44 			for (; *u1 == '0'; u1++)
45 				zeros_count_1++;
46 			for (; *u2 == '0'; u2++)
47 				zeros_count_2++;
48 			if (zeros_count_1 != zeros_count_2)
49 				return (zeros_count_2 - zeros_count_1);
50 
51 			/* Handle the case where 0 < 09. */
52 			if (!isdigit(*u1) && isdigit(*u2))
53 				return (1);
54 			if (!isdigit(*u2) && isdigit(*u1))
55 				return (-1);
56 		} else {
57 			/*
58 			 * No leading zeros; we're simply comparing two numbers.
59 			 * It is necessary to first count how many digits there
60 			 * are before going back to compare each digit, so that
61 			 * e.g. 7 is not considered larger than 60.
62 			 */
63 			num_1 = u1;
64 			num_2 = u2;
65 
66 			/* Count digits (more digits == larger number). */
67 			for (; isdigit(*u1); u1++)
68 				;
69 			for (; isdigit(*u2); u2++)
70 				;
71 			digit_count_1 = u1 - num_1;
72 			digit_count_2 = u2 - num_2;
73 			if (digit_count_1 != digit_count_2)
74 				return (digit_count_1 - digit_count_2);
75 
76 			/*
77 			 * If there are the same number of digits, go back to
78 			 * the start of the number.
79 			 */
80 			u1 = num_1;
81 			u2 = num_2;
82 		}
83 
84 		/* Compare each digit until there are none left. */
85 		for (; isdigit(*u1) && isdigit(*u2); u1++, u2++) {
86 			if (*u1 != *u2)
87 				return (*u1 - *u2);
88 		}
89 	}
90 	return (*u1 - *u2);
91 }
92