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/*
* This file is part of the coreboot project.
*
* (C) Copyright 2001 Wolfgang Denk, DENX Software Engineering, wd@denx.de.
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; version 2 of the License or (at your
* option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* From U-Boot 2016.05
*/
#include <console/console.h>
#include <rtc.h>
#define FEBRUARY 2
#define STARTOFTIME 1970
#define SECDAY 86400L
#define SECYR (SECDAY * 365)
#define LEAP_YEAR(year) ((year) % 4 == 0)
#define DAYS_IN_YEAR(a) (LEAP_YEAR(a) ? 366 : 365)
#define DAYS_IN_MONTH(a) (month_days[(a) - 1])
static const char *const weekdays[] = {
"Sun", "Mon", "Tues", "Wednes", "Thurs", "Fri", "Satur"
};
/* Zeller's rule */
static int rtc_calc_weekday(struct rtc_time *tm)
{
if (tm->year < 1971)
return -1;
/* In Zeller's rule, January and February are treated as if they
are months 13 and 14 of the previous year (March is still month 3) */
const int zyear = ((tm->mon < 3) ? tm->year - 1 : tm->year);
const int q = tm->mday;
const int m = (tm->mon < 3) ? tm->mon + 12 : tm->mon;
const int K = zyear % 100;
const int J = zyear / 100;
/*
* Because of the way the modulo operator works with negative numbers,
* the traditional formulation of Zeller's rule must be modified
* slightly to make the numerator positive (i.e., add 5J instead of
* subtracting 2J). Also subtract 1 so that Sunday is day 0.
*/
const int h = (q + (13 * (m + 1)) / 5
+ K + (K / 4) + (J / 4) + (5 * J) - 1) % 7;
tm->wday = h;
return 0;
}
int rtc_to_tm(int tim, struct rtc_time *tm)
{
int month_days[12] = {
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};
register int i;
register long hms, day;
day = tim / SECDAY;
hms = tim % SECDAY;
/* Hours, minutes, seconds are easy */
tm->hour = hms / 3600;
tm->min = (hms % 3600) / 60;
tm->sec = (hms % 3600) % 60;
/* Number of years in days */
for (i = STARTOFTIME; day >= DAYS_IN_YEAR(i); i++)
day -= DAYS_IN_YEAR(i);
tm->year = i;
/* Number of months in days left */
if (LEAP_YEAR(tm->year))
DAYS_IN_MONTH(FEBRUARY) = 29;
for (i = 1; day >= DAYS_IN_MONTH(i); i++)
day -= DAYS_IN_MONTH(i);
DAYS_IN_MONTH(FEBRUARY) = 28;
tm->mon = i;
/* Days are what is left over (+1) from all that */
tm->mday = day + 1;
/* Determine the day of week */
return rtc_calc_weekday(tm);
}
/*
* Converts Gregorian date to seconds since 1970-01-01 00:00:00.
* Assumes input in normal date format, i.e. 1980-12-31 23:59:59
* => year=1980, mon=12, day=31, hour=23, min=59, sec=59.
*
* [For the Julian calendar (which was used in Russia before 1917,
* Britain & colonies before 1752, anywhere else before 1582,
* and is still in use by some communities) leave out the
* -year / 100 + year / 400 terms, and add 10.]
*
* This algorithm was first published by Gauss (I think).
*
* WARNING: this function will overflow on 2106-02-07 06:28:16 on
* machines where long is 32-bit! (However, as time_t is signed, we
* will already get problems at other places on 2038-01-19 03:14:08)
*/
unsigned long rtc_mktime(const struct rtc_time *tm)
{
int mon = tm->mon;
int year = tm->year;
int days, hours;
mon -= 2;
if (0 >= (int)mon) { /* 1..12 -> 11, 12, 1..10 */
mon += 12; /* Puts Feb last since it has leap day */
year -= 1;
}
days = (unsigned long)(year / 4 - year / 100 + year / 400 +
367 * mon / 12 + tm->mday) +
year * 365 - 719499;
hours = days * 24 + tm->hour;
return (hours * 60 + tm->min) * 60 + tm->sec;
}
void rtc_display(const struct rtc_time *tm)
{
printk(BIOS_INFO, "Date: %5d-%02d-%02d (%sday) Time: %2d:%02d:%02d\n",
tm->year, tm->mon, tm->mday,
(tm->wday < 0 || tm->wday > 6) ? "unknown " : weekdays[tm->wday],
tm->hour, tm->min, tm->sec);
}
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