C语言实现时间戳转日期的算法(推荐)

概述1、算法时间是有周期规律的，4年一个周期（平年、平年、平年、闰年）共计1461天。Windows上C库函数time(NULL)返回的是从1970年1月1日以来的毫秒数，我们最后算出来的年数一定要加上这个基数1970。总的天数除以1461就

1、算法

时间是有周期规律的，4年一个周期（平年、平年、平年、闰年）共计1461天。windows上C库函数time(NulL)返回的是从1970年1月1日以来的毫秒数，我们最后算出来的年数一定要加上这个基数1970。总的天数除以1461就可以知道经历了多少个周期；总的天数对1461取余数就可以知道剩余的不足一个周期的天数，对这个余数进行判断也就可以得到月份和日了。

当然了，C语言库函数：localtime就可以获得一个时间戳对应的具体日期了，这里 主要说的是实现的一种算法。

2、C语言代码实现

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);

计算月份和日期的函数：

static const int MON1[12] = {31,28,31,30,31};	//平年static const int MON2[12] = {31,29,31};	//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;	}}

3、附上C语言库函数的实现

<pre name="code" >/****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;}

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