Question

Is this legit? Im trying to get to a time_t as fast as possible given a string formatted like YYYYMMDDHHMMSS.

static time_t ConvertToSecSince1970(char *szYYYYMMDDHHMMSS)
{
struct tm    Tm;    

    memset(&Tm, 0, sizeof(Tm));
    Tm.tm_year = makeInt(szYYYYMMDDHHMMSS +  0, 4) - 1900;
    Tm.tm_mon  = makeInt(szYYYYMMDDHHMMSS +  4, 2) - 1;
    Tm.tm_mday = makeInt(szYYYYMMDDHHMMSS +  6, 2);
    Tm.tm_hour = makeInt(szYYYYMMDDHHMMSS +  8, 2);
    Tm.tm_min  = makeInt(szYYYYMMDDHHMMSS + 10, 2);
    Tm.tm_sec  = makeInt(szYYYYMMDDHHMMSS + 12, 2);
    return mktime(&Tm);
}

It seems to produce the same answer if I created TM using:

strptime(szYYYYMMDDHHMMSS, "%Y%m%d%H%M%S", &Tm);

I am worried that tm_yday, tm_wday, tm_isdst, tm_gmtoff, tm_zone are important. My dates are UTC so I figured gmtoff = 0 and tm_zone = 0 might work.

By the way, Here is makeInt:

inline int makeInt(const char *p, int size)
{
    const char *endp;
    int intval = 0;

    endp = p + size;
    while (p < endp)
    {
        intval = intval * 10 + *p - '0';
        p++;
    }
    return intval;
}

Answer 1

mktime() ignores the tm_wday and tm_yday fields, and calculates new values for them based on the other fields. The same applies to the BSD extensions tm_gmtoff and tm_zone, except that they are calculated from the local time zone.

Note however that mktime() uses local time, not UTC, so if your input dates are UTC then your timezone must be set to UTC.