Question

How can I convert ASN1_TIME to time_t format? I wanted to convert the return value of X509_get_notAfter() to seconds.

Answer 1

Times are stored as a string internally, on the format YYmmddHHMMSS or YYYYmmddHHMMSS.

At the end of the string there is room for fractions of seconds and timezone, but let’s ignore that for now, and have some (untested) code.

Note: also see Bryan Olson’s answer below, which discusses the undefined behavior due to the i++‘s. Also see Seak’s answer which removes the undefined behavior.

static time_t ASN1_GetTimeT(ASN1_TIME* time)
{
    struct tm t;
    const char* str = (const char*) time->data;
    size_t i = 0;

    memset(&t, 0, sizeof(t));

    if (time->type == V_ASN1_UTCTIME) /* two digit year */
    {
        t.tm_year = (str[i++] - '0') * 10 + (str[++i] - '0');
        if (t.tm_year < 70)
        t.tm_year += 100;
    }
    else if (time->type == V_ASN1_GENERALIZEDTIME) /* four digit year */
    {
        t.tm_year = (str[i++] - '0') * 1000 + (str[++i] - '0') * 100 + (str[++i] - '0') * 10 + (str[++i] - '0');
        t.tm_year -= 1900;
    }
    t.tm_mon = ((str[i++] - '0') * 10 + (str[++i] - '0')) - 1; // -1 since January is 0 not 1.
    t.tm_mday = (str[i++] - '0') * 10 + (str[++i] - '0');
    t.tm_hour = (str[i++] - '0') * 10 + (str[++i] - '0');
    t.tm_min  = (str[i++] - '0') * 10 + (str[++i] - '0');
    t.tm_sec  = (str[i++] - '0') * 10 + (str[++i] - '0');

    /* Note: we did not adjust the time based on time zone information */
    return mktime(&t);
}