计算当前时间距离1970年的秒数

#define is_leap_year(y) (((y) % 4  == 0 && (y) % 100 != 0) || (y) % 400 == 0)
/* 计算某个日期距1970年1月1日0时0分0秒的秒数 */
time_t calSecSince1970(int Y, int M, int D, int h, int m, int s)
{
    int i = 0;
    int sec = 0;
    int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    /* 年计算 */
    for(i = 1970; i < Y; i++)
    {
        if(is_leap_year(i))
            sec += 366 * 24 * 60 * 60;
        else
            sec += 365 * 24 * 60 * 60;
    }
    /* 月计算 */
    for(i = 1; i < M; i++)
    {
        sec += days[i] * 24 * 60 * 60;
        if(i == 2 && is_leap_year(Y))
        {
            sec += 24 * 60 * 60;
        }
    }
    /* 天计算 */
    sec += (D - 1) * 24 * 60 * 60;
    /* 时分秒计算 */
    sec += h * 60 * 60 + m * 60 + s;
    return sec;
}

unsigned long mktime (unsigned int year, unsigned int mon,
    unsigned int day, unsigned int hour,
    unsigned int min, unsigned int sec)
{
    if (0 >= (int) (mon -= 2)){    /**//* 1..12 -> 11,12,1..10 */
         mon += 12;      /**//* Puts Feb last since it has leap day */
         year -= 1;
    }
    return (((
             (unsigned long) (year/4 - year/100 + year/400 + 367*mon/12 + day) +
             year*365 - 719499
          )*24 + hour /**//* now have hours */
       )*60 + min /**//* now have minutes */
    )*60 + sec; /**//* finally seconds */
}