1、算法

時(shí)間是有周期規(guī)律的，4年一個(gè)周期(平年、平年、平年、閏年)共計(jì)1461天。Windows上C庫(kù)函數(shù)time(NULL)返回的是從1970年1月1日以來(lái)的毫秒數(shù)，我們最后算出來(lái)的年數(shù)一定要加上這個(gè)基數(shù)1970?？偟奶鞌?shù)除以1461就可以知道經(jīng)歷了多少個(gè)周期；總的天數(shù)對(duì)1461取余數(shù)就可以知道剩余的不足一個(gè)周期的天數(shù)，對(duì)這個(gè)余數(shù)進(jìn)行判斷也就可以得到月份和日了。

當(dāng)然了，C語(yǔ)言庫(kù)函數(shù)：localtime就可以獲得一個(gè)時(shí)間戳對(duì)應(yīng)的具體日期了，這里 主要說(shuō)的是實(shí)現(xiàn)的一種算法。

2、C語(yǔ)言代碼實(shí)現(xiàn)

int nTime = time(NULL);//得到當(dāng)前系統(tǒng)時(shí)間

int nDays = nTime/DAYMS + 1;//time函數(shù)獲取的是從1970年以來(lái)的毫秒數(shù)，因此需要先得到天數(shù)

int nYear4 = nDays/FOURYEARS;//得到從1970年以來(lái)的周期(4年)的次數(shù)

int nRemain = nDays%FOURYEARS;//得到不足一個(gè)周期的天數(shù)

int nDesYear = 1970 + nYear4*4;

int nDesMonth = 0,nDesDay = 0;

bool bLeapYear = false;

if ( nRemain<365 )//一個(gè)周期內(nèi)，第一年

{//平年

}

else if ( nRemain

{//平年

nDesYear += 1;

nRemain -= 365;

}

else if ( nRemain

{//平年

nDesYear += 2;

nRemain -= (365+365);

}

else//一個(gè)周期內(nèi)，第四年，這一年是閏年

{//潤(rùn)年

nDesYear += 3;

nRemain -= (365+365+365);

bLeapYear = true;

}

GetMonthAndDay(nRemain,nDesMonth,nDesDay,bLeapYear);

計(jì)算月份和日期的函數(shù)：

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

static const int MON2[12] = {31,29,31};//閏年

static const int FOURYEARS = (366 + 365 +365 +365);//每個(gè)四年的總天數(shù)

static const int DAYMS = 24*3600;//每天的毫秒數(shù)

void GetMonthAndDay(int nDays,int& nMonth,int& nDay,bool IsLeapYear)

{

int *pMonths = IsLeapYear?MON2:MON1;

//循環(huán)減去12個(gè)月中每個(gè)月的天數(shù)，直到剩余天數(shù)小于等于0，就找到了對(duì)應(yīng)的月份

for ( int i=0; i<12; ++i )

{

int nTemp = nDays - pMonths[i];

if ( nTemp<=0 )

{

nMonth = i+1;

if ( nTemp == 0 )//表示剛好是這個(gè)月的最后一天，那么天數(shù)就是這個(gè)月的總天數(shù)了

nDay = pMonths[i];

else

nDay = nDays;

break;

}

nDays = nTemp;

}

}

3、附上C語(yǔ)言庫(kù)函數(shù)的實(shí)現(xiàn)

/***

*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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