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dd33e89f | 1 | /* Return basename of given pathname according to the weird XPG specification. |
581c785b | 2 | Copyright (C) 1997-2022 Free Software Foundation, Inc. |
dd33e89f | 3 | This file is part of the GNU C Library. |
dd33e89f UD |
4 | |
5 | The GNU C Library is free software; you can redistribute it and/or | |
41bdb6e2 AJ |
6 | modify it under the terms of the GNU Lesser General Public |
7 | License as published by the Free Software Foundation; either | |
8 | version 2.1 of the License, or (at your option) any later version. | |
dd33e89f UD |
9 | |
10 | The GNU C Library is distributed in the hope that it will be useful, | |
11 | but WITHOUT ANY WARRANTY; without even the implied warranty of | |
12 | MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU | |
41bdb6e2 | 13 | Lesser General Public License for more details. |
dd33e89f | 14 | |
41bdb6e2 | 15 | You should have received a copy of the GNU Lesser General Public |
59ba27a6 | 16 | License along with the GNU C Library; if not, see |
5a82c748 | 17 | <https://www.gnu.org/licenses/>. */ |
dd33e89f UD |
18 | |
19 | #include <string.h> | |
20 | #include <libgen.h> | |
21 | ||
22 | ||
23 | char * | |
24 | __xpg_basename (char *filename) | |
25 | { | |
26 | char *p; | |
27 | ||
28 | if (filename == NULL || filename[0] == '\0') | |
29 | /* We return a pointer to a static string containing ".". */ | |
30 | p = (char *) "."; | |
31 | else | |
32 | { | |
33 | p = strrchr (filename, '/'); | |
34 | ||
35 | if (p == NULL) | |
36 | /* There is no slash in the filename. Return the whole string. */ | |
37 | p = filename; | |
38 | else | |
39 | { | |
40 | if (p[1] == '\0') | |
41 | { | |
42 | /* We must remove trailing '/'. */ | |
43 | while (p > filename && p[-1] == '/') | |
44 | --p; | |
45 | ||
46 | /* Now we can be in two situations: | |
47 | a) the string only contains '/' characters, so we return | |
48 | '/' | |
49 | b) p points past the last component, but we have to remove | |
50 | the trailing slash. */ | |
51 | if (p > filename) | |
52 | { | |
53 | *p-- = '\0'; | |
54 | while (p > filename && p[-1] != '/') | |
55 | --p; | |
56 | } | |
57 | else | |
58 | /* The last slash we already found is the right position | |
59 | to return. */ | |
60 | while (p[1] != '\0') | |
61 | ++p; | |
62 | } | |
64ad0de2 UD |
63 | else |
64 | /* Go to the first character of the name. */ | |
65 | ++p; | |
dd33e89f UD |
66 | } |
67 | } | |
68 | ||
69 | return p; | |
70 | } |