/* * SPDX-FileCopyrightText: 1991 - 1994, Julianne Frances Haugh * SPDX-FileCopyrightText: 1996 - 1999, Marek Michałkiewicz * SPDX-FileCopyrightText: 2003 - 2005, Tomasz Kłoczko * SPDX-FileCopyrightText: 2008 , Nicolas François * * SPDX-License-Identifier: BSD-3-Clause */ #if !defined(__GLIBC__) #define _XOPEN_SOURCE 500 #endif #include #include #ident "$Id$" #include "defines.h" #include "prototypes.h" #ifndef USE_GETDATE #define USE_GETDATE 1 #endif #if USE_GETDATE #include "getdate.h" /* * strtoday() now uses get_date() (borrowed from GNU shellutils) * which can handle many date formats, for example: * 1970-09-17 # ISO 8601. * 70-9-17 # This century assumed by default. * 70-09-17 # Leading zeros are ignored. * 9/17/72 # Common U.S. writing. * 24 September 1972 * 24 Sept 72 # September has a special abbreviation. * 24 Sep 72 # Three-letter abbreviations always allowed. * Sep 24, 1972 * 24-sep-72 * 24sep72 */ long strtoday (const char *str) { time_t t; bool isnum = true; const char *s = str; /* * get_date() interprets an empty string as the current date, * which is not what we expect, unless you're a BOFH :-). * (useradd sets sp_expire = current date for new lusers) */ if ((NULL == str) || ('\0' == *str)) { return -1; } /* If a numerical value is provided, this is already a number of * days since EPOCH. */ if ('-' == *s) { s++; } while (' ' == *s) { s++; } while (isnum && ('\0' != *s)) { if (!isdigit (*s)) { isnum = false; } s++; } if (isnum) { long retdate; if (getlong (str, &retdate) == 0) { return -2; } return retdate; } t = get_date (str, NULL); if ((time_t) - 1 == t) { return -2; } /* convert seconds to days since 1970-01-01 */ return (long) (t + DAY / 2) / DAY; } #else /* !USE_GETDATE */ /* * Old code, just in case get_date() doesn't work as expected... */ #include #ifdef HAVE_STRPTIME /* * for now we allow just one format, but we can define more later * (we try them all until one succeeds). --marekm */ static const char *const date_formats[] = { "%Y-%m-%d", (char *) 0 }; #else /* * days and juldays are used to compute the number of days in the * current month, and the cumulative number of days in the preceding * months. they are declared so that january is 1, not 0. */ static const short days[13] = { 0, 31, 28, 31, 30, 31, 30, /* JAN - JUN */ 31, 31, 30, 31, 30, 31 }; /* JUL - DEC */ static const short juldays[13] = { 0, 0, 31, 59, 90, 120, 151, /* JAN - JUN */ 181, 212, 243, 273, 304, 334 }; /* JUL - DEC */ #endif /* * strtoday - compute the number of days since 1970. * * the total number of days prior to the current date is * computed. january 1, 1970 is used as the origin with * it having a day number of 0. */ long strtoday (const char *str) { #ifdef HAVE_STRPTIME struct tm tp; const char *const *fmt; char *cp; time_t result; memzero (&tp, sizeof tp); for (fmt = date_formats; *fmt; fmt++) { cp = strptime ((char *) str, *fmt, &tp); if ((NULL == cp) || ('\0' != *cp)) { continue; } result = mktime (&tp); if ((time_t) - 1 == result) { continue; } return (long) (result / DAY); /* success */ } return -1; #else char slop[2]; int month; int day; int year; long total; /* * start by separating the month, day and year. the order * is compiled in ... */ if (sscanf (str, "%d/%d/%d%c", &year, &month, &day, slop) != 3) { return -1; } /* * the month, day of the month, and year are checked for * correctness and the year adjusted so it falls between * 1970 and 2069. */ if ((month < 1) || (month > 12)) { return -1; } if (day < 1) { return -1; } if ( ((2 != month) || ((year % 4) != 0)) && (day > days[month])) { return -1; } else if ((month == 2) && ((year % 4) == 0) && (day > 29)) { return -1; } if (year < 0) { return -1; } else if (year <= 69) { year += 2000; } else if (year <= 99) { year += 1900; } /* * On systems with 32-bit signed time_t, time wraps around in 2038 * - for now we just limit the year to 2037 (instead of 2069). * This limit can be removed once no one is using 32-bit systems * anymore :-). --marekm */ if ((year < 1970) || (year > 2037)) { return -1; } /* * the total number of days is the total number of days in all * the whole years, plus the number of leap days, plus the * number of days in the whole months preceding, plus the number * of days so far in the month. */ total = (long) ((year - 1970) * 365L) + (((year + 1) - 1970) / 4); total += (long) juldays[month] + (month > 2 && (year % 4) == 0 ? 1 : 0); total += (long) day - 1; return total; #endif /* HAVE_STRPTIME */ } #endif /* !USE_GETDATE */