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9f9276c34c
Rename this function so that it is clear that it is provided by the RTC. Also return an error when it cannot function as expected. This is unlikely to occur since it works for dates since 1752 and many RTCs do not support such old dates. Still it is better to be accurate. Signed-off-by: Simon Glass <sjg@chromium.org> Acked-by: Heiko Schocher <hs@denx.de>
150 lines
3.5 KiB
C
150 lines
3.5 KiB
C
/*
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* (C) Copyright 2001
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* Wolfgang Denk, DENX Software Engineering, wd@denx.de.
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*
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* SPDX-License-Identifier: GPL-2.0+
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*/
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/*
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* Date & Time support for Philips PCF8563 RTC
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*/
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#include <common.h>
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#include <command.h>
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#include <errno.h>
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#include <rtc.h>
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#if defined(CONFIG_CMD_DATE) || defined(CONFIG_TIMESTAMP)
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#define FEBRUARY 2
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#define STARTOFTIME 1970
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#define SECDAY 86400L
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#define SECYR (SECDAY * 365)
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#define leapyear(year) ((year) % 4 == 0)
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#define days_in_year(a) (leapyear(a) ? 366 : 365)
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#define days_in_month(a) (month_days[(a) - 1])
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static int month_days[12] = {
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31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
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};
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/*
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* This only works for the Gregorian calendar - i.e. after 1752 (in the UK)
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*/
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int rtc_calc_weekday(struct rtc_time *tm)
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{
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int leapsToDate;
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int lastYear;
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int day;
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int MonthOffset[] = { 0,31,59,90,120,151,181,212,243,273,304,334 };
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if (tm->tm_year < 1753)
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return -EINVAL;
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lastYear=tm->tm_year-1;
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/*
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* Number of leap corrections to apply up to end of last year
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*/
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leapsToDate = lastYear/4 - lastYear/100 + lastYear/400;
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/*
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* This year is a leap year if it is divisible by 4 except when it is
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* divisible by 100 unless it is divisible by 400
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*
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* e.g. 1904 was a leap year, 1900 was not, 1996 is, and 2000 will be
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*/
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if((tm->tm_year%4==0) &&
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((tm->tm_year%100!=0) || (tm->tm_year%400==0)) &&
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(tm->tm_mon>2)) {
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/*
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* We are past Feb. 29 in a leap year
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*/
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day=1;
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} else {
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day=0;
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}
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day += lastYear*365 + leapsToDate + MonthOffset[tm->tm_mon-1] + tm->tm_mday;
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tm->tm_wday=day%7;
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return 0;
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}
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int rtc_to_tm(int tim, struct rtc_time *tm)
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{
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register int i;
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register long hms, day;
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day = tim / SECDAY;
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hms = tim % SECDAY;
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/* Hours, minutes, seconds are easy */
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tm->tm_hour = hms / 3600;
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tm->tm_min = (hms % 3600) / 60;
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tm->tm_sec = (hms % 3600) % 60;
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/* Number of years in days */
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for (i = STARTOFTIME; day >= days_in_year(i); i++) {
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day -= days_in_year(i);
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}
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tm->tm_year = i;
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/* Number of months in days left */
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if (leapyear(tm->tm_year)) {
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days_in_month(FEBRUARY) = 29;
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}
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for (i = 1; day >= days_in_month(i); i++) {
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day -= days_in_month(i);
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}
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days_in_month(FEBRUARY) = 28;
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tm->tm_mon = i;
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/* Days are what is left over (+1) from all that. */
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tm->tm_mday = day + 1;
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/* Zero unused fields */
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tm->tm_yday = 0;
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tm->tm_isdst = 0;
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/*
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* Determine the day of week
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*/
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return rtc_calc_weekday(tm);
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}
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/* Converts Gregorian date to seconds since 1970-01-01 00:00:00.
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* Assumes input in normal date format, i.e. 1980-12-31 23:59:59
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* => year=1980, mon=12, day=31, hour=23, min=59, sec=59.
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*
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* [For the Julian calendar (which was used in Russia before 1917,
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* Britain & colonies before 1752, anywhere else before 1582,
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* and is still in use by some communities) leave out the
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* -year/100+year/400 terms, and add 10.]
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*
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* This algorithm was first published by Gauss (I think).
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*
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* WARNING: this function will overflow on 2106-02-07 06:28:16 on
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* machines were long is 32-bit! (However, as time_t is signed, we
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* will already get problems at other places on 2038-01-19 03:14:08)
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*/
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unsigned long
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mktime (unsigned int year, unsigned int mon,
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unsigned int day, unsigned int hour,
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unsigned int min, unsigned int sec)
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{
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if (0 >= (int) (mon -= 2)) { /* 1..12 -> 11,12,1..10 */
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mon += 12; /* Puts Feb last since it has leap day */
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year -= 1;
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}
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return (((
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(unsigned long) (year/4 - year/100 + year/400 + 367*mon/12 + day) +
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year*365 - 719499
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)*24 + hour /* now have hours */
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)*60 + min /* now have minutes */
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)*60 + sec; /* finally seconds */
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}
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#endif
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