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#include <string.h>
#include "dive.h"
/*
* 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));
/* seconds since 1970 -> minutes since 1970 */
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, 1970 was a Thursday (tm_wday=4) */
tm->tm_wday = (val + 4) % 7;
/*
* Now we're in "days since Jan 1, 1970". To make things easier,
* let's make it "days since Jan 1, 1968", since that's a leap-year
*/
val += 365 + 366;
/* This only works up until 2099 (2100 isn't a leap-year) */
leapyears = val / (365 * 4 + 1);
val %= (365 * 4 + 1);
tm->tm_year = 68 + 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;
/* First normalize relative to 1900 */
if (year < 70)
year += 100;
else if (year > 1900)
year -= 1900;
/* Normalized to Jan 1, 1970: unix time */
year -= 70;
if (year < 0 || year > 129) /* algo only works for 1970-2099 */
return -1;
if (month < 0 || month > 11) /* array bounds */
return -1;
if (month < 2 || (year + 2) % 4)
day--;
if (tm->tm_hour < 0 || tm->tm_min < 0 || tm->tm_sec < 0)
return -1;
return (year * 365 + (year + 1) / 4 + mdays[month] + day) * 24 * 60 * 60UL +
tm->tm_hour * 60 * 60 + tm->tm_min * 60 + tm->tm_sec;
}
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