时间相关函数的实现

1020阅读　0评论2018-01-08　Tay_linux
分类：LINUX


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  1. #include <stdio.h>

  3. /**/
  4. typedef struct _date_time_{
  5.     unsigned char year;/*[0,99]*/
  6.     unsigned char mon;/*[1,12]*/
  7.     unsigned char mday;/*[1,31]*/
  8.     unsigned char hour;/*[0,23]*/
  9.     unsigned char min;/*[0,59]*/
  10.     unsigned char sec;/*[0,59]*/
  11.     unsigned char week;/*[0,6]*/
  12. }date_time_t;


  15. /*
  16.     这个函数将时间转换成从1970年1月1日0小时0分0秒到你输入的时间所经过的秒数，溢出时间为2106-02-07 06:28:16。
  17.     若你要计算2000-02-07 06:28:16 到2000-02-09 06:28:16 所经过的秒数：
  18.     unsigned long time1 = Dt_mktime_abs(2000, 2, 7, 6, 28, 16) - Dt_mktime_abs(2000, 2, 9, 6, 28, 16)
  19. */
  20. unsigned long Dt_mktime_abs(date_time_t *pTm)
  21. {
  22.     int year/*[2000, 2099]*/ = pTm->tm_year/*[00, 99]*/ + 2000;
  23.     int mon/*[1, 12]*/ = pTm->tm_mon/*[1, 12]*/;
  24.     int day/*[1, 31]*/ = pTm->tm_mday/*[1, 31]*/;
  25.     int hour = pTm->tm_hour/*[0, 23]*/;
  26.     int min = pTm->tm_min/*[0, 59]*/;
  27.     int sec = pTm->tm_sec/*[0, 59]*/;

  29.     /**/
  30.     if(((int)(mon -= 2)) <= 0)
  31.     {
  32.         mon += 12;/* Puts Feb last since it has leap day */
  33.         year -= 1;
  34.     }

  36.     return ((((unsigned long)(year / 4 - year / 100 + year / 400 + 367 * mon / 12 + day) + year * 365 - 719499) * 24
  37.                     + hour /* now have hours */) * 60 + min /* now have minutes */)*60 + sec; /* finally seconds */
  38. }


  41. /**/
  42. int Calc_week_day( unsigned int year/*[2000, 2099]*/, unsigned int mon/*[1, 12]*/, unsigned int day/*[1, 31]*/)
  43. {
  44.     int week = 0;
  45.     unsigned int y = 0;
  46.     unsigned int c = 0;
  47.     unsigned int m = 0;
  48.     unsigned int d = 0;
  49.   
  50.     if((mon == 1) || (mon == 2))
  51.     {
  52.         c = ( year - 1 ) / 100;
  53.         y = ( year - 1 ) % 100;
  54.         m = mon + 12;
  55.         d = day;
  56.     }
  57.     else
  58.     {
  59.         c = year / 100;
  60.         y = year % 100;
  61.         m = mon;
  62.         d = day;
  63.     }
  64.       
  65.     week = y + y / 4 + c / 4 - 2 * c + 26 * ( m + 1 ) / 10 + d - 1; //蔡勒公式
  66.     week = week >= 0 ? ( week % 7 ) : ( week % 7 + 7 ); //week为负时取模
  67.   
  68.     return week;
  69. }


  72. /*
  73.     1:leap year
  74.     0:not leap year
  75. */
  76. int Is_leap_year(unsigned int year)
  77. {
  78.     if(((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0))
  79.     {
  80.         return 1;
  81.     }
  82.     else
  83.     {
  84.         return 0;
  85.     }
  86. }

  88. /*
  89.  The number of days in the month.
  90. */
  91. unsigned char Get_days_in_month(unsigned int year, unsigned int mon/*[1,12]*/)
  92. {
  93.     unsigned char day_table[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  94.     if((mon == 2) && (is_leap_year(year) == 1))
  95.     {
  96.         return 29;
  97.     }
  98.     else
  99.     {
  100.         return day_table[mon];
  101.     }
  102. }

  104. /*
  105.  * Convert seconds since 01-01-1970 00:00:00 to date.
  106.  */
  107. void Dt_time_to_tm(unsigned long time, date_time_t *pTm)
  108. {
  109.     unsigned int month = 0;
  110.     unsigned int year = 0;
  111.     int days = 0;
  112.     int tm_year = 0;
  113.     int tm_mon = 0;
  114.     int tm_mday = 0;
  115.     int tm_hour = 0;
  116.     int tm_min = 0;
  117.     int tm_sec = 0;
  118.     int newdays = 0;

  120.     /**/
  121. #define LEAPS_THRU_END_OF(y) ((y)/4 - (y)/100 + (y)/400)

  123.     /**/
  124.     days = time / 86400;
  125.     time -= (unsigned int)days * 86400;
  126.     year = 1970 + days / 365;
  127.     days -= (year - 1970) * 365 + LEAPS_THRU_END_OF(year - 1) - LEAPS_THRU_END_OF(1970 - 1);
  128.     if (days < 0)
  129.     {
  130.         year -= 1;
  131.         days += 365 + is_leap_year(year);
  132.     }
  133.     tm_year = year - 1900;
  134.     for (month = 0; month < 11; month++)
  135.     {
  136.         newdays = days - rtc_month_days(month, year);
  137.         if (newdays < 0)
  138.         {
  139.             break;
  140.         }
  141.         days = newdays;
  142.     }
  143.     tm_mon = month;
  144.     tm_mday = days + 1;
  145.     tm_hour = time / 3600;
  146.     time -= tm_hour * 3600;
  147.     tm_min = time / 60;
  148.     tm_sec = time - tm_min * 60;

  150.     /**/
  151.     pTm->year = (unsigned char)(tm_year - 100);
  152.     pTm->mon = (unsigned char)(tm_mon + 1);
  153.     pTm->mday = (unsigned char)(tm_mday);
  154.     pTm->hour = (unsigned char)(tm_hour);
  155.     pTm->min = (unsigned char)(tm_min);
  156.     pTm->sec = (unsigned char)(tm_sec);

  158.     /**/
  159.     pTm->week = (unsigned char)Calc_week_day(pTm->year + 2000, pTm->mon, pTm->mday);    
  160. }


  163. /**/
  164. int Dt_get_day_by_week(unsigned char year/*[0, 99]*/, unsigned char month/*[1, 12]*/,
  165.                                 unsigned char week_index/*>= 1*/, unsigned char week_day/*[0, 6]*/)
  166. {
  167.     /**/
  168.     unsigned char week_day_ = 0;
  169.     unsigned char day = 0;

  171.     /**/
  172.     if(!((year <= 99) && ((month >= 1) && (month <= 12)) && (week_index >= 1) &&
  173.                                 (week_day <= 6)))
  174.     {
  175.         printf("Invalid parameter(%d, %d, %d, %d)\r\n", year, month, week_index, week_day);
  176.         return -1;
  177.     }
  178.     /*计算这个月的1号是星期几*/
  179.     week_day_ = Calc_week_day(year + 2000, month, 1);
  180.     /**/
  181.     if(week_day >= week_day_)
  182.     {
  183.         day = week_day - week_day_ + 1;
  184.     }
  185.     else
  186.     {
  187.         day = 8 - week_day_ + week_day;
  188.     }
  189.     /**/
  190.     day += (week_index - 1) * 7;

  192.     /**/
  193.     if(day > Get_days_in_month(year, month))
  194.     {
  195.         printf("Invalid result(%d, %d, %d, %d, %d)\r\n", year, month, week_index, week_day, day);
  196.         return -1;
  197.     }

  199.     /**/
  200.     return day;
  201. }

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