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// SPDX-License-Identifier: GPL-2.0
#include "subsurface-time.h"
#include <string.h>
/*
* The date handling internally works in seconds since
* Jan 1, 1900. That avoids negative numbers which avoids
* some silly problems.
*
* But we then use the same base epoch base (Jan 1, 1970)
* that POSIX uses, so that we can use the normal date
* handling functions for getting current time etc.
*
* There's 25567 dats from Jan 1, 1900 to Jan 1, 1970.
*
* NOTE! The SEC_PER_DAY is not so much because the
* number is complicated, as to make sure we always
* expand the type to "timestamp_t" in the arithmetic.
*/
#define SEC_PER_DAY ((timestamp_t) 24*60*60)
#define EPOCH_OFFSET (25567 * SEC_PER_DAY)
/*
* Convert 64-bit timestamp to 'struct tm' in UTC.
*
* On 32-bit machines, only do 64-bit arithmetic for the seconds
* part, after that we do everything in 'long'. 64-bit divides
* are unnecessary once you're counting minutes (32-bit minutes:
* 8000+ years).
*/
void utc_mkdate(timestamp_t timestamp, struct tm *tm)
{
static const unsigned int mdays[] = {
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31,
};
static const unsigned int mdays_leap[] = {
31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31,
};
unsigned long val;
unsigned int leapyears;
int m;
const unsigned int *mp;
memset(tm, 0, sizeof(*tm));
// Midnight at Jan 1, 1970 means "no date"
if (!timestamp)
return;
/* Convert to seconds since 1900 */
timestamp += EPOCH_OFFSET;
/* minutes since 1900 */
tm->tm_sec = timestamp % 60;
val = timestamp /= 60;
/* Do the simple stuff */
tm->tm_min = val % 60;
val /= 60;
tm->tm_hour = val % 24;
val /= 24;
/* Jan 1, 1900 was a Monday (tm_wday=1) */
tm->tm_wday = (val + 1) % 7;
/*
* Now we're in "days since Jan 1, 1900". To make things easier,
* let's make it "days since Jan 1, 1904", since that's a leap-year.
* 1900 itself was not. The following logic will get 1900-1903
* wrong. If you were diving back then, you're kind of screwed.
*/
val -= 365*4;
/* This only works up until 2099 (2100 isn't a leap-year) */
leapyears = val / (365 * 4 + 1);
val %= (365 * 4 + 1);
tm->tm_year = 1904 + leapyears * 4;
/* Handle the leap-year itself */
mp = mdays_leap;
if (val > 365) {
tm->tm_year++;
val -= 366;
tm->tm_year += val / 365;
val %= 365;
mp = mdays;
}
for (m = 0; m < 12; m++) {
if (val < *mp)
break;
val -= *mp++;
}
tm->tm_mday = val + 1;
tm->tm_mon = m;
}
timestamp_t utc_mktime(struct tm *tm)
{
static const int mdays[] = {
0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334
};
int year = tm->tm_year;
int month = tm->tm_mon;
int day = tm->tm_mday;
int days_since_1900;
timestamp_t when;
/* First normalize relative to 1900 */
if (year < 50)
year += 100;
else if (year >= 1900)
year -= 1900;
if (year < 0 || year > 129) /* algo only works for 1900-2099 */
return 0;
if (month < 0 || month > 11) /* array bounds */
return 0;
if (month < 2 || (year && year % 4))
day--;
if (tm->tm_hour < 0 || tm->tm_min < 0 || tm->tm_sec < 0)
return 0;
/* This works until 2099 */
days_since_1900 = year * 365 + (year - 1) / 4;
/* Note the 'day' fixup for non-leapyears above */
days_since_1900 += mdays[month] + day;
/* Now add it all up, making sure to do this part in "timestamp_t" */
when = days_since_1900 * SEC_PER_DAY;
when += tm->tm_hour * 60 * 60 + tm->tm_min * 60 + tm->tm_sec;
return when - EPOCH_OFFSET;
}
/*
* Extract year from 64-bit timestamp.
*
* This looks inefficient, since it breaks down into a full
* struct tm. However, modern compilers are effective at throwing
* out unused calculations. If it turns out to be a bottle neck
* we will have to cache a struct tm per dive.
*/
int utc_year(timestamp_t timestamp)
{
struct tm tm;
utc_mkdate(timestamp, &tm);
return tm.tm_year;
}
/*
* Extract day of week from 64-bit timestamp.
* Returns 0-6, whereby 0 is Sunday and 6 is Saturday.
*
* Same comment as for utc_year(): Modern compilers are good
* at throwing out unused calculations, so this is more efficient
* than it looks.
*/
int utc_weekday(timestamp_t timestamp)
{
struct tm tm;
utc_mkdate(timestamp, &tm);
return tm.tm_wday;
}
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