C语言实现时间戳转日期的算法(推荐)

int nTime = time(NULL);//得到当前系统时间
int nDays = nTime/DAYMS + 1;//time函数获取的是从1970年以来的毫秒数，因此需要先得到天数
int nYear4 = nDays/FOURYEARS;//得到从1970年以来的周期（4年）的次数
int nRemain = nDays%FOURYEARS;//得到不足一个周期的天数
int nDesYear = 1970 + nYear4*4;
int nDesMonth = 0, nDesDay = 0;
bool bLeapYear = false;
if ( nRemain<365 )//一个周期内，第一年
{//平年

}
else if ( nRemain<(365+365) )//一个周期内，第二年
{//平年
nDesYear += 1;
nRemain -= 365;
}
else if ( nRemain<(365+365+365) )//一个周期内，第三年
{//平年
nDesYear += 2;
nRemain -= (365+365);
}
else//一个周期内，第四年，这一年是闰年
{//润年
nDesYear += 3;
nRemain -= (365+365+365);
bLeapYear = true;
}
GetMonthAndDay(nRemain, nDesMonth, nDesDay, bLeapYear);

static const int MON1[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};	//平年
static const int MON2[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};	//闰年
static const int FOURYEARS = (366 + 365 +365 +365);	//每个四年的总天数
static const int DAYMS = 24*3600;	//每天的毫秒数

void GetMonthAndDay(int nDays, int& nMonth, int& nDay, bool IsLeapYear)
{
	int *pMonths = IsLeapYear?MON2:MON1;
	//循环减去12个月中每个月的天数，直到剩余天数小于等于0，就找到了对应的月份
	for ( int i=0; i<12; ++i )
	{
		int nTemp = nDays - pMonths[i];
		if ( nTemp<=0 )
		{
			nMonth = i+1;
			if ( nTemp == 0 )//表示刚好是这个月的最后一天，那么天数就是这个月的总天数了
				nDay = pMonths[i];
			else
				nDay = nDays;
			break;
		}
		nDays = nTemp;
	}
}

<pre name="code" class="cpp">/***
*errno_t _gmtime32_s(ptm, timp) - convert *timp to a structure (UTC)
*
*Purpose:
*    Converts the calendar time value, in 32 bit internal format, to
*    broken-down time (tm structure) with the corresponding UTC time.
*
*Entry:
*    const time_t *timp - pointer to time_t value to convert
*
*Exit:
*    errno_t = 0 success
* tm members filled-in
*    errno_t = non zero
* tm members initialized to -1 if ptm != NULL
*
*Exceptions:
*
*******************************************************************************/

errno_t __cdecl _gmtime32_s (
struct tm *ptm,
const __time32_t *timp
)
{
__time32_t caltim;/* = *timp; *//* calendar time to convert */
int islpyr = 0; /* is-current-year-a-leap-year flag */
REG1 int tmptim;
REG3 int *mdays;/* pointer to days or lpdays */
struct tm *ptb = ptm;

_VALIDATE_RETURN_ERRCODE( ( ptm != NULL ), EINVAL )
memset( ptm, 0xff, sizeof( struct tm ) );

_VALIDATE_RETURN_ERRCODE( ( timp != NULL ), EINVAL )

caltim = *timp;
_VALIDATE_RETURN_ERRCODE_NOEXC( ( caltim >= _MIN_LOCAL_TIME ), EINVAL )

/*
 * Determine years since 1970. First, identify the four-year interval
 * since this makes handling leap-years easy (note that 2000 IS a
 * leap year and 2100 is out-of-range).
 */
tmptim = (int)(caltim / _FOUR_YEAR_SEC);
caltim -= ((__time32_t)tmptim * _FOUR_YEAR_SEC);

/*
 * Determine which year of the interval
 */
tmptim = (tmptim * 4) + 70; /* 1970, 1974, 1978,...,etc. */

if ( caltim >= _YEAR_SEC ) {

  tmptim++;    /* 1971, 1975, 1979,...,etc. */
  caltim -= _YEAR_SEC;

  if ( caltim >= _YEAR_SEC ) {

tmptim++;  /* 1972, 1976, 1980,...,etc. */
caltim -= _YEAR_SEC;

/*
 * Note, it takes 366 days-worth of seconds to get past a leap
 * year.
 */
if ( caltim >= (_YEAR_SEC + _DAY_SEC) ) {

tmptim++;  /* 1973, 1977, 1981,...,etc. */
caltim -= (_YEAR_SEC + _DAY_SEC);
}
else {
/*
 * In a leap year after all, set the flag.
 */
islpyr++;
}
  }
}

/*
 * tmptim now holds the value for tm_year. caltim now holds the
 * number of elapsed seconds since the beginning of that year.
 */
ptb->tm_year = tmptim;

/*
 * Determine days since January 1 (0 - 365). This is the tm_yday value.
 * Leave caltim with number of elapsed seconds in that day.
 */
ptb->tm_yday = (int)(caltim / _DAY_SEC);
caltim -= (__time32_t)(ptb->tm_yday) * _DAY_SEC;

/*
 * Determine months since January (0 - 11) and day of month (1 - 31)
 */
if ( islpyr )
  mdays = _lpdays;
else
  mdays = _days;


for ( tmptim = 1 ; mdays[tmptim] < ptb->tm_yday ; tmptim++ ) ;

ptb->tm_mon = --tmptim;

ptb->tm_mday = ptb->tm_yday - mdays[tmptim];

/*
 * Determine days since Sunday (0 - 6)
 */
ptb->tm_wday = ((int)(*timp / _DAY_SEC) + _BASE_DOW) % 7;

/*
 * Determine hours since midnight (0 - 23), minutes after the hour
 * (0 - 59), and seconds after the minute (0 - 59).
 */
ptb->tm_hour = (int)(caltim / 3600);
caltim -= (__time32_t)ptb->tm_hour * 3600L;

ptb->tm_min = (int)(caltim / 60);
ptb->tm_sec = (int)(caltim - (ptb->tm_min) * 60);

ptb->tm_isdst = 0;
return 0;

}