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1、算法
時光是有周期紀律的,4年一個周期(閏年、閏年、閏年、閏年)合計1461天。Windows上C庫函數time(NULL)前往的是從1970年1月1日以來的毫秒數,我們最初算出來的年數必定要加上這個基數1970。總的天數除以1461便可以曉得閱歷了若干個周期;總的天數對1461取余數便可以曉得殘剩的缺乏一個周期的天數,對這個余數停止斷定也便可以獲得月份和日了。
固然了,C說話庫函數:localtime便可以取得一個時光戳對應的詳細日期了,這裡 重要說的是完成的一種算法。
2、C說話代碼完成
int nTime = time(NULL);//獲得以後體系時光 int nDays = nTime/DAYMS + 1;//time函數獲得的是從1970年以來的毫秒數,是以須要先獲得天數 int nYear4 = nDays/FOURYEARS;//獲得從1970年以來的周期(4年)的次數 int nRemain = nDays%FOURYEARS;//獲得缺乏一個周期的天數 int nDesYear = 1970 + nYear4*4; int nDesMonth = 0, nDesDay = 0; bool bLeapYear = false; if ( nRemain<365 )//一個周期內,第一年 {//閏年 } else if ( nRemain<(365+365) )//一個周期內,第二年 {//閏年 nDesYear += 1; nRemain -= 365; } else if ( nRemain<(365+365+365) )//一個周期內,第三年 {//閏年 nDesYear += 2; nRemain -= (365+365); } else//一個周期內,第四年,這一年是閏年 {//潤年 nDesYear += 3; nRemain -= (365+365+365); bLeapYear = true; } GetMonthAndDay(nRemain, nDesMonth, nDesDay, bLeapYear);
盤算月份和日期的函數:
static const int MON1[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; //閏年 static const int MON2[12] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; //閏年 static const int FOURYEARS = (366 + 365 +365 +365); //每一個四年的總天數 static const int DAYMS = 24*3600; //天天的毫秒數 void GetMonthAndDay(int nDays, int& nMonth, int& nDay, bool IsLeapYear) { int *pMonths = IsLeapYear?MON2:MON1; //輪回減去12個月中每一個月的天數,直到殘剩天數小於等於0,就找到了對應的月份 for ( int i=0; i<12; ++i ) { int nTemp = nDays - pMonths[i]; if ( nTemp<=0 ) { nMonth = i+1; if ( nTemp == 0 )//表現恰好是這個月的最初一天,那末天數就是這個月的總天數了 nDay = pMonths[i]; else nDay = nDays; break; } nDays = nTemp; } }
3、附上C說話庫函數的完成
<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; }
以上這篇C說話完成時光戳轉日期的算法(推舉)就是小編分享給年夜家的全體內容了,願望能給年夜家一個參考,也願望年夜家多多支撐。