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, 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; } }
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; }
以上这篇C语言实现时间戳转日期的算法(推荐)就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持查字典教程网。
【C语言实现时间戳转日期的算法(推荐)】相关文章：
★ 用C实现PHP扩展 Fetch_Url 类数据抓取的方法