8位单片机可用的 mktime localtime函数

8位单片机可用的 mktime  localtime函数及源码

最近在做一个8位单片机项目，其中用到了时间戳转换函数，这个在32位机上一个库函数就解决了问题，没想到在8位单片机中没有对应库（time.h），没有办法只有自己来写。

目标：1，满足和库函数mktime  localtime所计算出的数据一至；2，考虑8位单片机的处理能力慢软件效率问题。

分享给大家，方便有同样需求的朋友。

gcc 环境进行测试：

测试程序：

#include <stdio.h>
#include <stdint.h>
#include <time.h>
#include <string.h>

#if 0
struct tm {
    int tm_sec;   /* seconds after the minute, 0 to 60
                     (0 - 60 allows for the occasional leap second) */
    int tm_min;   /* minutes after the hour, 0 to 59 */
    int tm_hour;  /* hours since midnight, 0 to 23 */
    int tm_mday;  /* day of the month, 1 to 31 */
    int tm_mon;   /* months since January, 0 to 11 */
    int tm_year;  /* years since 1900 */
//    int tm_wday;  /* days since Sunday, 0 to 6 */
//    int tm_yday;  /* days since January 1, 0 to 365 */
//    int tm_isdst; /* Daylight Savings Time flag */
};
#endif
static const char mon_list[12]             = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
static const char leap_mon_list[12]        = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

/*******************************************************************************
* Function Name : fun_mktime
* Description   : 时间转为时间戳
* Input         : 
* Output        : 
* Other         : 
* Date          : 2016.11.14
*******************************************************************************/
int32_t fun_mktime(struct tm *pT)
{
    const char *pDays = NULL;
    int32_t tmp = 0;
    int16_t i = 0;
    
    //计算总共有多少个闰年
    tmp = (pT->tm_year  / 4 - pT->tm_year / 100 + pT->tm_year / 400) - (1970 / 4 - 1970 / 100 + 1970 / 400);

    //如果当年是闰年，需要减去当年的闰年
    if ((pT->tm_year % 4 == 0) && ((pT->tm_year  % 100 != 0) || (pT->tm_year  % 400 == 0)))
    {
        tmp = tmp - 1 + (pT->tm_year - 1970) * 365;
        pDays = leap_mon_list;
    }
    else
    {
        tmp = tmp + (pT->tm_year - 1970) * 365;
        pDays = mon_list;
    }

    for (i = 0; i < pT->tm_mon - 1; i++)
        tmp += pDays[i];

    tmp = tmp + pT->tm_mday - 1;

    tmp = tmp * 24 + pT->tm_hour;

    tmp = tmp * 60 + pT->tm_min;

    tmp = tmp * 60 + pT->tm_sec;
    
    return tmp;
}

/*******************************************************************************
* Function Name : fun_localtime
* Description   : 时间戳转为时间
* Input         :   struct tm *pT： 输出的时间缓冲区   uint32_t tim:当前时间戳
* Output        : 
* Other         : 
* Date          : 2016.11.14
*******************************************************************************/
void fun_localtime(struct tm *pT, int32_t tim)
{
  const char *pDays = NULL;

  uint16_t index = 0;

  memset(pT, 0, sizeof(*pT));

  //year initialization
  if (tim > 0x5685C180L)            // 2016-1-1 0:0:0
  {
    pT->tm_year = 2016;
    tim -= 0x5685C180L;
  } 
  else if (tim > 0x4B3D3B00L)       // 2010-1-1 0:0:0
  {
    pT->tm_year = 2010;  
    tim -= 0x4B3D3B00L;
  }
  else if (tim > 0x386D4380L)       // 2000-1-1 0:0:0
  {
    pT->tm_year = 2000;  
    tim -= 0x386D4380L;
  }
  else 
  {
    pT->tm_year = 1970;   
  }

  //now have year
  while (tim >= 366L * 24 * 60 * 60)
  {
    if ((pT->tm_year % 4 == 0) && ((pT->tm_year  % 100 != 0) || (pT->tm_year  % 400 == 0)))
      tim -= 366L * 24 * 60 * 60;
    else 
      tim -= 365L * 24 * 60 * 60;

    pT->tm_year++;
  }

  // then 365 * 24 * 60 * 60 < tim < 366 * 24 * 60 * 60
  if (!(((pT->tm_year % 4 == 0) && ((pT->tm_year  % 100 != 0) || (pT->tm_year  % 400 == 0)))) 
    && (tim > 365L * 24 * 60 * 60))
  {
    tim -= 365L * 24 * 60 * 60; 
    pT->tm_year++;
  }

  // this year is a leap year?
  if (((pT->tm_year % 4 == 0) && ((pT->tm_year  % 100 != 0) || (pT->tm_year  % 400 == 0))))
    pDays = leap_mon_list;
  else 
    pDays = mon_list;

  pT->tm_mon = 1;
  // now have mon
  while (tim > pDays[index] * 24L * 60 * 60)
  {
    tim -= pDays[index] * 24L * 60 * 60;
    index++;
    pT->tm_mon++;
  }
  
  // now have days
  pT->tm_mday = tim / (24L * 60 * 60) + 1;
  tim = tim % (24L * 60 * 60);

  // now have hour
  pT->tm_hour = tim / (60 * 60);
  tim = tim % (60 * 60);

  // now have min 
  pT->tm_min = tim / 60;
  tim = tim % 60;

  pT->tm_sec = tim;  
}


int main (void *parg)
{
  struct tm *pT = {0};
  time_t timep = 0;
  uint32_t cur_tim = 0;

  time(&timep);

  pT = localtime(&timep);

  printf("linux time 	= %d
", (int32_t)timep);
  pT->tm_year += 1900;
  pT->tm_mon += 1;
  printf("fun_mktime 	= %d
", cur_tim = (uint32_t)fun_mktime(pT));

  printf("localtime 	= %d-%d-%d %d:%d:%d
", pT->tm_year, pT->tm_mon, pT->tm_mday, pT->tm_hour, pT->tm_min, pT->tm_sec);
  memset(pT, 0, sizeof(*pT));
  fun_localtime(pT, cur_tim);
  printf("fun_localtime 	= %d-%d-%d %d:%d:%d
", pT->tm_year, pT->tm_mon, pT->tm_mday, pT->tm_hour, pT->tm_min, pT->tm_sec);
  return 0;
}

测试结果：

linux time      = 1480133002
fun_mktime      = 1480161802
localtime       = 2016-11-26 12:3:22
fun_localtime   = 2016-11-26 12:3:22

linux time 是库函数mktime计算结果，因为进行了时区处理，所以与fun_mktime计算出来刚好是8 * 3600 秒的差值

此函数在C51下进行过测试，符合要求。