将日期/时间转换为时间戳，反之亦然 [英] Convert date/time to time-stamp and vice versa

问题描述

我试图在C两个简单的转换器中实现日期/时间到时间戳，反之亦然，而不依赖于时间库例程，如mktime等。

时间戳以秒为单位，日期/时间结构采用以下格式：

• unsigned char年份：0到99（代表范围2000到2099）




• unsigned char month：1到12




• unsigned char day：1到31




• unsigned char hour：0到23




• unsigned char minute：0到59




• unsigned char second：0到59

我想对dt2ts转换器发表第二个意见（假设输入是合法的）：

  unsigned int dt2ts（const dt_t * dt）
 {
 static unsigned short days [] = {0,31,59,90,120,151,181,212,243,273,304,334}; 
 return（（（dt-> year * 365 + dt-> year / 4）+ days [dt-> month-1] + dt-> day）* 24 + dt->小时）* 60 + dt->分钟）* 60 + dt->秒; 
} 
  

除此之外，我希望有一些帮助完成ts2dt转换器： / p>

  void ts2dt（unsigned int ts，dt_t * dt）
 {
 dt-> second = ts ％60; ts / = 60; 
 dt-> minute = ts％60; ts / = 60; 
 dt-> hour = ts％24; ts / = 24; 
 dt-> day = 
 dt-> month = 
 dt-> year = 
} 
  

感谢

解决方案

OP全部准备好处理小时，分，秒。注意：从2000年1月1日到2099年12月31日的天数至少需要一个16位整数。即使 unsigned 是2个字节，以下内容应该可以工作。

  unsigned DivRem无符号股息，无符号除数，无符号*余数）{
 unsigned Quotient =股息/除数; 
 *剩余=股息 - 商数*除数; 
 return Quotient; 
} 
 
 void Day2000ToYMD（unsigned DaySince2000Jan1，unsigned * Y，unsigned * M，unsigned * D）{
 unsigned OlympiadDay; //每4年是一个奥运会
 * Y = 4 * DivRem（DaySince2000Jan1,365 * 4 + 1，& OlympiadDay）; 
 * D = 1; 
 if（OlympiadDay> =（31 + 29-1））{//处理2月29日和之后
 OlympiadDay--; 
 if（OlympiadDay ==（31 + 29-1））{
（* D）++; 
} 
} 
 unsigned YearDay; //年的日子0到364 
 * Y + = DivRem（OlympiadDay，365和& YearDay）; 
 static const unsigned short days [] = {0,31,59,90,120,151,181,212,243,273,304,334,365}; 
 * M = 1; 
 while（days [* M] <= YearDay）（* M）++; 
 * D + = YearDay  -  days [* M  -  1]; 
} 
  

---

答案提供的尝试将年度的概念保持在1月1日至12月31日。由于这个答案不是需要处理大约100和400年的闰年，所以我一直保持着这种风格。一般来说，一旦添加了这2条规则，如果将年初的时间从3月1日开始转移到28/29，则数学变得更加容易。 FWIW，这个是对Julian / Gregorian日历古代发展的更一致的观点。因此* 10月10日 * ober是第8个月，而* 12月 * ember是第10个月。

I am trying to implement in C two simple convertors, date/time to time-stamp and vice versa, without any dependencies on time library routines, such as mktime, etc.

The time-stamp is in seconds, and the date/time structure is in the following format:

• unsigned char year: 0 to 99 (representing the range 2000 to 2099)

• unsigned char month: 1 to 12

• unsigned char day: 1 to 31

• unsigned char hour: 0 to 23

• unsigned char minute: 0 to 59

• unsigned char second: 0 to 59

I would like to have a second opinion on the dt2ts convertor (assuming that the input is legal):

unsigned int dt2ts(const dt_t* dt)
{
    static unsigned short days[] = {0,31,59,90,120,151,181,212,243,273,304,334};
    return ((((dt->year*365+dt->year/4)+days[dt->month-1]+dt->day)*24+dt->hour)*60+dt->minute)*60+dt->second;
}

In addition to that, I would appreciate some help completing the ts2dt convertor:

void ts2dt(unsigned int ts,dt_t* dt)
{
    dt->second = ts%60; ts /= 60;
    dt->minute = ts%60; ts /= 60;
    dt->hour   = ts%24; ts /= 24;
    dt->day    = ?????;
    dt->month  = ?????;
    dt->year   = ?????;
}

Thanks

解决方案

OP is all ready well handling the hours, minutes, seconds. Just a bit of assist on Y,M,D.

Note: The number of days from Jan 1, 2000 to Dec 31, 2099 needs at least a 16 bit integer. Following should work even if unsigned is 2 bytes.

unsigned DivRem(unsigned Dividend, unsigned Divisor, unsigned *Remainder) {
  unsigned Quotient = Dividend/Divisor;
  *Remainder = Dividend - Quotient*Divisor;
  return Quotient;
}

void Day2000ToYMD(unsigned DaySince2000Jan1, unsigned *Y, unsigned *M, unsigned *D) {
  unsigned OlympiadDay;  // Every 4 years is an Olympiad
  *Y = 4*DivRem(DaySince2000Jan1, 365*4+1, &OlympiadDay);
  *D = 1;
  if (OlympiadDay >= (31+29-1)) {  // deal with Feb 29th and after
    OlympiadDay--;
    if (OlympiadDay == (31+29-1)) {
      (*D)++;
    }
  }
  unsigned YearDay;      // Day of the year 0 to 364
  *Y += DivRem(OlympiadDay, 365, &YearDay);
  static const unsigned short days[] = {0,31,59,90,120,151,181,212,243,273,304,334,365};
  *M = 1;
  while (days[*M] <= YearDay) (*M)++;
  *D += YearDay - days[*M - 1];
}

---

[Edit] The answer provided tries to keep the concept of the year as Jan 1 to Dec 31. As this answer does not need to handle the leap year years about 100 and 400 years, I've kept with this style.

In general, once those 2 rules are added, the math becomes easier to if one shifts the beginning of the year to March 1 and ending on Feb 28/29. FWIW, this is a more consistent view of the ancient development of the Julian/Gregorian calendar. Thus *Oct*ober is then the 8th month and *Dec*ember is the 10th month.

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