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add timestampts to logger #8
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Seems like the linux kernel only offers syscalls to get the timezone or epoch. So we would have to convert that to year/month/day which is not that straightforward. I yoinked the musl libc code without safety checks and timezone into a sample program that works. #include <stdio.h>
#include <time.h>
#define LEAPOCH (946684800LL + 86400*(31+29))
#define DAYS_PER_400Y (365*400 + 97)
#define DAYS_PER_100Y (365*100 + 24)
#define DAYS_PER_4Y (365*4 + 1)
int __secs_to_tm(long long t, struct tm *tm)
{
long long days, secs, years;
int remdays, remsecs, remyears;
int qc_cycles, c_cycles, q_cycles;
int months;
int wday, yday, leap;
static const char days_in_month[] = {31,30,31,30,31,31,30,31,30,31,31,29};
secs = t - LEAPOCH;
days = secs / 86400;
remsecs = secs % 86400;
if (remsecs < 0) {
remsecs += 86400;
days--;
}
wday = (3+days)%7;
if (wday < 0) wday += 7;
qc_cycles = days / DAYS_PER_400Y;
remdays = days % DAYS_PER_400Y;
if (remdays < 0) {
remdays += DAYS_PER_400Y;
qc_cycles--;
}
c_cycles = remdays / DAYS_PER_100Y;
if (c_cycles == 4) c_cycles--;
remdays -= c_cycles * DAYS_PER_100Y;
q_cycles = remdays / DAYS_PER_4Y;
if (q_cycles == 25) q_cycles--;
remdays -= q_cycles * DAYS_PER_4Y;
remyears = remdays / 365;
if (remyears == 4) remyears--;
remdays -= remyears * 365;
leap = !remyears && (q_cycles || !c_cycles);
yday = remdays + 31 + 28 + leap;
if (yday >= 365+leap) yday -= 365+leap;
years = remyears + 4*q_cycles + 100*c_cycles + 400LL*qc_cycles;
for (months=0; days_in_month[months] <= remdays; months++)
remdays -= days_in_month[months];
if (months >= 10) {
months -= 12;
years++;
}
tm->tm_year = years + 100;
tm->tm_mon = months + 2;
tm->tm_mday = remdays + 1;
tm->tm_wday = wday;
tm->tm_yday = yday;
tm->tm_hour = remsecs / 3600;
tm->tm_min = remsecs / 60 % 60;
tm->tm_sec = remsecs % 60;
return 0;
}
int main() {
struct tm dt = {};
__secs_to_tm(1734062161, &dt);
printf("year=%d\n", dt.tm_year + 1900);
} So porting that to assembly should do. But that would be a lot of effort and would need lots of unit tests. |
Needed for #8 in a few years (2038 problem)
No description provided.
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