[SOLVED] Function for conversion of GMT to Unix epoch

NevemTeve · 10-29-2020, 02:41 PM

I've just found an older code of mine that contains this bit:

Code:

        const ASN1_TIME *at;

        at= X509_get0_notBefore (cert);
        if (at) {
            struct tm tm;

            memset (&tm, 0, sizeof *&tm);
            ASN1_TIME_to_tm (at, &tm);
            into->tNotBefore= timegm (&tm);
        }

Also I created a `timegm` implementation for AIX (which doesn't have it):

Code:

/* aix-timegm.c */

#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>

time_t timegm (struct tm *tm);
time_t timegm (struct tm *tm)
{
    int year, ytmp, dtmp, ytmpe, dtmpe;
    int isleapyear;
    static int msum [2][12] = {
        { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334},       /* normal years */
        { 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335}        /* leap years */
    };
    static int mlen [2][12] = {
        { 31, 28, 31, 30,  31,  30,  31,  31,  30,  31,  30, 31},
        { 31, 29, 31, 30,  31,  30,  31,  31,  30,  31,  30, 31}
    };
    long long t;

    if (!tm) return -1;

    year = tm->tm_year+1900;
    isleapyear= (year%4==0) - (year%100==0) + (year%400==0);

    if (year<1600  || year>2400  ||
        tm->tm_mon<0  || tm->tm_mon>11 ||
        tm->tm_mday<1 || tm->tm_mday>mlen[isleapyear][tm->tm_mon] ||
        tm->tm_hour<0 || tm->tm_hour>23 ||
        tm->tm_min<0  || tm->tm_min>59 ||
        tm->tm_min<0  || tm->tm_sec>59) return -1;

    ytmp = year - 1601;
    dtmp = ytmp*365 + ytmp/4 - ytmp/100 + ytmp/400;

    ytmpe = 1970 - 1601;
    dtmpe = ytmpe*365 + ytmpe/4 - ytmpe/100 + ytmpe/400; /* = 134685 + 92 - 3 + 0 = 134774 */

    t  = dtmp - dtmpe;
    t += msum[isleapyear][tm->tm_mon];
    t += tm->tm_mday-1;

    tm->tm_wday = (t+4)%7;/* 0..6=Sun..Sat; adding 4 to adjust 1970.01.01.=Thursday; */
    if (tm->tm_wday<0) tm->tm_wday += 7;
    tm->tm_yday = msum[isleapyear][tm->tm_mon] + tm->tm_mday-1;
    tm->tm_isdst= 0;

    t  = t*24 + tm->tm_hour;
    t  = t*60 + tm->tm_min;
    t  = t*60 + tm->tm_sec;

#if LONG_MAX == 2147483647L
    if (t<LONG_MIN || t>LONG_MAX) {
        t= -1;
    }
#endif

    return t;
}

GazL · 10-29-2020, 03:12 PM

How about declaring extern long timezone and then adjusting t_of_day by subtracting adding the value in timezone?

edit: in fact, it looks like time.h defines it already so you don't even need to declare.

edot2: actually, mktime() will return a t_of_day that will have the assumed timezone subtracted, so you need to put it back on if the original values were already specified in GMT0, so adding, not subtracting when you do it this way around.

Something like this maybe?

Code:

#include <time.h>
// extern long timezone;


time_t mktimegm(struct tm *tm_gmt)
{
    time_t t;
    t = mktime(tm_gmt);
    
    return ( t == -1 ) ? t : t + timezone ;
}

edit3: Ah nutz! Apparently mktime() will still adjust for DST, even if you set isdst to 0 before calling, so looks like this is too simplistic and one also needs to adjust for dst. Maybe temporarily setting the timezone to GMT was the better idea after all. Or, just write your own routines like Neve' did.

linxbee · 10-29-2020, 03:25 PM

cannot thank enough @NevemTeve , looks like this is what i need , have to do some tests with different dates.

looks like you are calculating exact number days that are elapsed from 1900 and then multiplying with 24*60*60.

Well i dont get some part of the code, what is the purpose of this part?

Code:

// is it number of days in nonleap years ??
ytmp = year - 1601;
dtmp = ytmp*365 + ytmp/4 - ytmp/100 + ytmp/400;

// is it number of days in leap years ??
ytmpe = 1970 - 1601;
dtmpe = ytmpe*365 + ytmpe/4 - ytmpe/100 + ytmpe/400; /* = 134685 + 92 - 3 + 0 = 134774 */

https://www.linuxquestions.org/quest...s_lq/icon7.gif

NevemTeve · 10-29-2020, 03:41 PM

If I remember correctly (file dated 2018-07-24), this program uses 1601-01-01 as base, as it is the start of an 400-year period of leap-years, so the 'days passed before the actual year (since the base)' can be calculated as (year*365)+(year/4)-(year/100)+(year/400).

Then this has to be converted to the unix-epoch, that's where the number of days between 1601-01-01 and 1970-01-01 matters (dtmpe=134774)

NevemTeve · 10-29-2020, 03:42 PM

If I remember correctly (file dated 2018-07-24), this program uses 1601-01-01 as base, as it is the start of an 400-year period of leap-years, so the 'days passed before the actual year (since the base)' can be calculated as (year*365)+(year/4)-(year/100)+(year/400).

Then this has to be converted to the unix-epoch, that's where the number of days between 1601-01-01 and 1970-01-01 matters (dtmpe=134774)

NevemTeve · 01-02-2021, 01:07 AM

uploaded here:
https://github.com/lzsiga/timegm