blocks|key|742951|text|取决于您想要gmtime或localtime的时间，然后只需读取struct_tm|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|742952|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/clibrary/ctime/gmtime/|1|http://www.cplusplus.com/reference/clibrary/ctime/localtime/|2|http://www.cplusplus.com/reference/clibrary/ctime/tm/^0|6|6|0|D|9|1|W|9|2|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]|$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|-4|5|6|7|Z|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]|N|$5|G|H|I|C|$J|O]]]]

Depends on which time you want <a href="http://www.cplusplus.com/reference/clibrary/ctime/gmtime/" rel="nofollow">gmtime</a> or <a href="http://www.cplusplus.com/reference/clibrary/ctime/localtime/" rel="nofollow">localtime</a> then just read the <a href="http://www.cplusplus.com/reference/clibrary/ctime/tm/" rel="nofollow">struct_tm</a>

blocks|key|526563|text|首先，不要将秒存储为浮点数。如果您需要微纳秒，请将它们分开存储。你将需要整数来做这些计算。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|526564|这取决于你的时区(DST规则，闰年，闰秒)，但我想说的是，首先得到天数除以86400的整数。然后通过模除以86400找出剩余的部分。现在，您可以通过第一个整数除以365天数，然后从剩余天数中减去闰日数(通过将天数除以365计算得出)，计算出已经过去了多少年。您还需要从剩余秒数(已经计算)中减去闰秒的个数。如果减法使这些数字低于零，则从下一个最大面额中减去。然后，您可以使用日历的显式逻辑来计算月份的第几天。如果你在DST登陆，请确保添加一个小时(或DST偏移量)。|526565|就我个人而言，我只会使用Boost.Date_Time，因为它做了所有这些事情，而且做得更多(可能比你我在最初几次迭代中犯的错误更少)，但我想我应该尝试一下你的问题……|offset|length|526566|entityMap|0|LINK|mutability|MUTABLE|url|http://www.boost.org/doc/libs/1_47_0/doc/html/date_time.html^0|0|0|C|F|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@$F|S|G|T|1|U]]|A|$]]|$1|H|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|O]]]]

First of all, do not store your seconds as a float. If you need micro/nanoseconds, store them separately. You're going to need integers to do these calculations.

It depends on your time zone (DST rules, leap years, leap seconds), but I would say first get the number of days by integer dividing by 86400. Then find out what's left over, by modulo dividing by 86400. Now you can figure out how many years have passed by first integer dividing the number of days by 365, and then subtracting the number of leap days from the remaining days (calculated by modulo dividing the number of days by 365). You'll also want to subtract the number of leap seconds from the number of remaining seconds (already calculated). If that subtraction drives those numbers below zero, then subtract from the next biggest denomination. Then you can calculate the day of month using explicit logic for your calendar. Make sure to add an hour (or whatever the DST offset is) if you land in DST.

Personally, I would just use <a href="http://www.boost.org/doc/libs/1_47_0/doc/html/date_time.html" rel="nofollow">Boost.Date_Time</a>, since it does all this and more (probably with fewer mistakes than you or I would make in the first few iterations), but I figured I'd take a shot at your question...

blocks|key|526633|text|这段代码可以工作..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|526634|用法:+uint32_t+getSecsSinceEpoch(1970，月，日，years_since_epoch，时，分，秒)；|526635|示例:+timestamp+=+getSecsSinceEpoch(1970，6，12，(2014+-+1970)，15，29，0)|526636|回报率:+1402586940|526637|你可以在www.epochconverter.com上验证。|526638|写这篇文章花了大约20分钟，其中大部分时间都花在和一个朋友争论我是否应该包含leap-seconds，nano-seconds等等。|526639|玩得开心..。|526640|布莱恩·威尔卡特博士|526641|#define+DAYSPERWEEK+(7)
#define+DAYSPERNORMYEAR+(365U)
#define+DAYSPERLEAPYEAR+(366U)

#define+SECSPERDAY+(86400UL)+/*+==+(+24+*+60+*+60)+*/
#define+SECSPERHOUR+(3600UL)+/*+==+(+60+*+60)+*/
#define+SECSPERMIN+(60UL)+/*+==+(+60)+*/

#define+LEAPYEAR(year)++++++++++(!((year)+%25+4)+&&+(((year)+%25+100)+%7C%7C+!((year)+%25+400)))

const+int+_ytab[2][12]+=+{
{31,+28,+31,+30,+31,+30,+31,+31,+30,+31,+30,+31},
{31,+29,+31,+30,+31,+30,+31,+31,+30,+31,+30,+31}
};

/****************************************************
*+Class:Function++++:+getSecsSomceEpoch
*+Input+++++:+uint16_t+epoch+date+(ie,+1970)
*+Input+++++:+uint8+ptr+to+returned+month
*+Input+++++:+uint8+ptr+to+returned+day
*+Input+++++:+uint8+ptr+to+returned+years+since+Epoch
*+Input+++++:+uint8+ptr+to+returned+hour
*+Input+++++:+uint8+ptr+to+returned+minute
*+Input+++++:+uint8+ptr+to+returned+seconds
*+Output++++++++:+uint32_t+Seconds+between+Epoch+year+and+timestamp
*+Behavior++++++:
*
*+Converts+MM/DD/YY+HH:MM:SS+to+actual+seconds+since+epoch.
*+Epoch+year+is+assumed+at+Jan+1,+00:00:01am.
****************************************************/
uint32_t+getSecsSinceEpoch(uint16_t+epoch,+uint8_t+month,+uint8_t+day,+uint8_t+years,+uint8_t+hour,+uint8_t+minute,+uint8_t+second)
{
unsigned+long+secs+=+0;
int+countleap+=+0;
int+i;
int+dayspermonth;

secs+=+years+*+(SECSPERDAY+*+365);
for+(i+=+0;+i+<+(years+-+1);+i%2B%2B)
{+++
++++if+(LEAPYEAR((epoch+%2B+i)))
++++++countleap%2B%2B;
}
secs+%2B=+(countleap+*+SECSPERDAY);

secs+%2B=+second;
secs+%2B=+(hour+*+SECSPERHOUR);
secs+%2B=+(minute+*+SECSPERMIN);
secs+%2B=+((day+-+1)+*+SECSPERDAY);

if+(month+>+1)
{
++++dayspermonth+=+0;

++++if+(LEAPYEAR((epoch+%2B+years)))+//+Only+counts+when+we're+on+leap+day+or+past+it
++++{
++++++++if+(month+>+2)
++++++++{
++++++++++++dayspermonth+=+1;
++++++++}+else+if+(month+==+2+&&+day+>=+29)+{
++++++++++++dayspermonth+=+1;
++++++++}
++++}

++++for+(i+=+0;+i+<+month+-+1;+i%2B%2B)
++++{+++
++++++++secs+%2B=+(_ytab[dayspermonth][i]+*+SECSPERDAY);
++++}
}

return+secs;
}|code-block|syntax|javascript|526642|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|11|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|12|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|13|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|14|8|@]|9|@]|A|$S|T]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

This code works...

Usage:
uint32_t getSecsSinceEpoch(1970, month, day, years_since_epoch, hour, minute, second);

Example:
timestamp = getSecsSinceEpoch(1970, 6, 12, (2014 - 1970), 15, 29, 0)

Returns: 1402586940

You can verify at www.epochconverter.com.

Took about 20 mins to write it and most of that was spent arguing with a friend as to whether I should include leap-seconds, nano-seconds, etc. Blech.

Have fun...

Dr. Bryan Wilcutt

<pre><code>#define DAYSPERWEEK (7)
#define DAYSPERNORMYEAR (365U)
#define DAYSPERLEAPYEAR (366U)

#define SECSPERDAY (86400UL) /* == ( 24 * 60 * 60) */
#define SECSPERHOUR (3600UL) /* == ( 60 * 60) */
#define SECSPERMIN (60UL) /* == ( 60) */

#define LEAPYEAR(year) (!((year) % 4) &amp;&amp; (((year) % 100) || !((year) % 400)))

const int _ytab[2][12] = {
{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31},
{31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
};

/****************************************************
* Class:Function : getSecsSomceEpoch
* Input : uint16_t epoch date (ie, 1970)
* Input : uint8 ptr to returned month
* Input : uint8 ptr to returned day
* Input : uint8 ptr to returned years since Epoch
* Input : uint8 ptr to returned hour
* Input : uint8 ptr to returned minute
* Input : uint8 ptr to returned seconds
* Output : uint32_t Seconds between Epoch year and timestamp
* Behavior :
*
* Converts MM/DD/YY HH:MM:SS to actual seconds since epoch.
* Epoch year is assumed at Jan 1, 00:00:01am.
****************************************************/
uint32_t getSecsSinceEpoch(uint16_t epoch, uint8_t month, uint8_t day, uint8_t years, uint8_t hour, uint8_t minute, uint8_t second)
{
unsigned long secs = 0;
int countleap = 0;
int i;
int dayspermonth;

secs = years * (SECSPERDAY * 365);
for (i = 0; i &lt; (years - 1); i++)
{ 
 if (LEAPYEAR((epoch + i)))
 countleap++;
}
secs += (countleap * SECSPERDAY);

secs += second;
secs += (hour * SECSPERHOUR);
secs += (minute * SECSPERMIN);
secs += ((day - 1) * SECSPERDAY);

if (month &gt; 1)
{
 dayspermonth = 0;

 if (LEAPYEAR((epoch + years))) // Only counts when we're on leap day or past it
 {
 if (month &gt; 2)
 {
 dayspermonth = 1;
 } else if (month == 2 &amp;&amp; day &gt;= 29) {
 dayspermonth = 1;
 }
 }

 for (i = 0; i &lt; month - 1; i++)
 { 
 secs += (_ytab[dayspermonth][i] * SECSPERDAY);
 }
}

return secs;
}
</code></pre>

blocks|key|743165|text|在此之前|type|unstyled|depth|inlineStyleRanges|entityRanges|data|743166|++++for+(i+=+0;+i+<+(years+-+1);+i%2B%2B)
++++{+++
++++++++if+(LEAPYEAR((epoch+%2B+i)))
++++++++countleap%2B%2B;
++++}|code-block|syntax|javascript|743167|稍后：|743168|++++for+(i+=+0;+i+<+years;+i%2B%2B)
+{+++
+++if+(LEAPYEAR((epoch+%2B+i)))
++++countleap%2B%2B;
+}|743169|经过修正后，代码对我来说是有效的。|743170|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

BEFORE

<pre><code> for (i = 0; i &lt; (years - 1); i++)
 { 
 if (LEAPYEAR((epoch + i)))
 countleap++;
 }
</code></pre>

LATER:

<pre><code> for (i = 0; i &lt; years; i++)
 { 
 if (LEAPYEAR((epoch + i)))
 countleap++;
 }
</code></pre>

After the correction the code worked for me.

I have seconds since Jan 1 1970 00:00 as an int64 in nanoseconds and I'm trying to convert it into month/day/year/day of week.

It's easy to do this iteratively, I have that working but I want to do it formulaically. I'm looking for the actual math.

Math to convert seconds since 1970 into date and vice versa

从1970年1月1日00:00开始，我有以纳秒为单位的int64秒数，我正在尝试将其转换为月/日/年/星期几。迭代地做这件事很容易，我已经做到了，但我想用公式化的方式来做。我要找的是真正的数学。

问将1970年以来的秒转换为日期的数学方法
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