// SPDX-License-Identifier: GPL-2.0 #include "subsurface-time.h" #include /* * 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; }