SciFi (06.02.2013 21:56 - 21:58, просмотров: 77) ответил Гудвин на Никто не поделится функциями вычисления даты и времени из unix time (int32) для "Цэ"? Не монстрообразными, чтобы зарядить в мелкоконтроллер, с учетом високосных годов и пр...Я выпилил из newlib, ЕМНИП: #define SECSPERMIN 60L
#define MINSPERHOUR 60L
#define HOURSPERDAY 24L
#define SECSPERHOUR (SECSPERMIN * MINSPERHOUR)
#define SECSPERDAY (SECSPERHOUR * HOURSPERDAY)
#define DAYSPERWEEK 7
#define MONSPERYEAR 12
#define YEAR_BASE 1900
#define EPOCH_WDAY 1
#define EPOCH_YEARS_SINCE_LEAP 2
#define EPOCH_YEARS_SINCE_CENTURY 70
#define EPOCH_YEARS_SINCE_LEAP_CENTURY 370
#define isleap(y) ((((y) % 4) == 0 && ((y) % 100) != 0) || ((y) % 400) == 0)
static const uint8_t mon_lengths[2][MONSPERYEAR] =
{
{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
} ;
static const int year_lengths[2] = { 365, 366 } ;
void
seconds2tm(uint32_t sec, struct tm* res)
{
long days, rem;
int y;
int yleap;
const uint8_t *ip;
days = sec / SECSPERDAY;
rem = sec % SECSPERDAY;
/* compute hour, min, and sec */
res->tm_hour = (int) (rem / SECSPERHOUR);
rem %= SECSPERHOUR;
res->tm_min = (int) (rem / SECSPERMIN);
res->tm_sec = (int) (rem % SECSPERMIN);
/* compute day of week */
if ((res->tm_wday = ((EPOCH_WDAY + days) % DAYSPERWEEK)) < 0)
{
res->tm_wday += DAYSPERWEEK;
}
/* compute year & day of year */
y = YEAR_BASE;
if (days >= 40542) /* days in years 1900..2010 inclusive */
{
days -= 40542;
y += 111;
}
if (days >= 0)
{
for (;;)
{
yleap = isleap(y);
if (days < year_lengths[yleap])
{
break;
}
y++;
days -= year_lengths[yleap];
}
}
else
{
do
{
--y;
yleap = isleap(y);
days += year_lengths[yleap];
}
while (days < 0);
}
res->tm_year = y - YEAR_BASE;
res->tm_yday = days;
ip = mon_lengths[yleap];
for (res->tm_mon = 0; days >= ip[res->tm_mon]; ++res->tm_mon)
{
days -= ip[res->tm_mon];
}
res->tm_mday = days + 1;
}
ส็็็็็็็็็็็็็็็็็็็็็็็็็༼ ຈل͜ຈ༽ส้้้้้้้้้้้้้้้้้้้้้้้
-
- Спасибо. - Гудвин(07.02.2013 10:12)