blocks|key|1413786|text|在您的daysInMonth函数中要注意闰年。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1413787|如果您想要非常高的性能，您可以预计算成对，以便在一步中达到month%2Byear，然后计算天/小时/分钟/秒。|1413788|一个好的解决方案是gmtime+source+code中的解决方案。|offset|length|1413789|/*
+*+gmtime+-+convert+the+calendar+time+into+broken+down+time
+*/
/*+$Header:+gmtime.c,v+1.4+91/04/22+13:20:27+ceriel+Exp+$+*/

#include++++++++<time.h>
#include++++++++<limits.h>
#include++++++++"loc_time.h"

struct+tm+*
gmtime(register+const+time_t+*timer)
{
++++++++static+struct+tm+br_time;
++++++++register+struct+tm+*timep+=+&br_time;
++++++++time_t+time+=+*timer;
++++++++register+unsigned+long+dayclock,+dayno;
++++++++int+year+=+EPOCH_YR;

++++++++dayclock+=+(unsigned+long)time+%25+SECS_DAY;
++++++++dayno+=+(unsigned+long)time+/+SECS_DAY;

++++++++timep->tm_sec+=+dayclock+%25+60;
++++++++timep->tm_min+=+(dayclock+%25+3600)+/+60;
++++++++timep->tm_hour+=+dayclock+/+3600;
++++++++timep->tm_wday+=+(dayno+%2B+4)+%25+7;+++++++/*+day+0+was+a+thursday+*/
++++++++while+(dayno+>=+YEARSIZE(year))+{
++++++++++++++++dayno+-=+YEARSIZE(year);
++++++++++++++++year%2B%2B;
++++++++}
++++++++timep->tm_year+=+year+-+YEAR0;
++++++++timep->tm_yday+=+dayno;
++++++++timep->tm_mon+=+0;
++++++++while+(dayno+>=+_ytab[LEAPYEAR(year)][timep->tm_mon])+{
++++++++++++++++dayno+-=+_ytab[LEAPYEAR(year)][timep->tm_mon];
++++++++++++++++timep->tm_mon%2B%2B;
++++++++}
++++++++timep->tm_mday+=+dayno+%2B+1;
++++++++timep->tm_isdst+=+0;

++++++++return+timep;
}|code-block|syntax|javascript|1413790|entityMap|0|LINK|mutability|MUTABLE|url|http://www.google.com/codesearch/p?hl=en&sa=N&cd=2&ct=rc#4kyKlxPG9x4/current/l4minix-03-03-10/src/lib/l4mxc/ansi/gmtime.c&q=gmtime^0|0|0|9|I|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@$F|X|G|Y|1|Z]]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Be careful about leap years in your daysInMonth function.

If you want very high performance, you can precompute the pair to get to month+year in one step, and then calculate the day/hour/min/sec.

A good solution is the one in the <a href="http://www.google.com/codesearch/p?hl=en&amp;sa=N&amp;cd=2&amp;ct=rc#4kyKlxPG9x4/current/l4minix-03-03-10/src/lib/l4mxc/ansi/gmtime.c&amp;q=gmtime" rel="noreferrer">gmtime source code</a>:

<pre><code>/*
 * gmtime - convert the calendar time into broken down time
 */
/* $Header: gmtime.c,v 1.4 91/04/22 13:20:27 ceriel Exp $ */

#include &lt;time.h&gt;
#include &lt;limits.h&gt;
#include "loc_time.h"

struct tm *
gmtime(register const time_t *timer)
{
 static struct tm br_time;
 register struct tm *timep = &amp;br_time;
 time_t time = *timer;
 register unsigned long dayclock, dayno;
 int year = EPOCH_YR;

 dayclock = (unsigned long)time % SECS_DAY;
 dayno = (unsigned long)time / SECS_DAY;

 timep-&gt;tm_sec = dayclock % 60;
 timep-&gt;tm_min = (dayclock % 3600) / 60;
 timep-&gt;tm_hour = dayclock / 3600;
 timep-&gt;tm_wday = (dayno + 4) % 7; /* day 0 was a thursday */
 while (dayno &gt;= YEARSIZE(year)) {
 dayno -= YEARSIZE(year);
 year++;
 }
 timep-&gt;tm_year = year - YEAR0;
 timep-&gt;tm_yday = dayno;
 timep-&gt;tm_mon = 0;
 while (dayno &gt;= _ytab[LEAPYEAR(year)][timep-&gt;tm_mon]) {
 dayno -= _ytab[LEAPYEAR(year)][timep-&gt;tm_mon];
 timep-&gt;tm_mon++;
 }
 timep-&gt;tm_mday = dayno + 1;
 timep-&gt;tm_isdst = 0;

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

blocks|key|1413836|text|标准库提供了执行此操作的函数。gmtime()或localtime()会将time_t+(从纪元开始的秒数，即-+Jan+1+1970+00:00:00)转换为struct+tm。然后，可以使用strftime()根据您指定的格式将struct+tm转换为字符串(char*)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1413837|请参阅：http://www.cplusplus.com/reference/clibrary/ctime/|1413838|日期/时间计算可能会变得很棘手。你最好使用现有的解决方案，而不是试图推出自己的解决方案，除非你有一个非常好的理由。|1413839|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/clibrary/ctime/^0|F|8|O|B|11|6|28|9|2P|A|38|9|3O|5|0|4|1E|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|16|8|@]|D|@$9|17|A|18|1|19]]|E|$]]|$1|H|3|I|5|6|7|1A|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

The standard library provides functions for doing this. <code>gmtime()</code> or <code>localtime()</code> will convert a <code>time_t</code> (seconds since the epoch, i.e.- Jan 1 1970 00:00:00) into a <code>struct tm</code>. <code>strftime()</code> can then be used to convert a <code>struct tm</code> into a string (<code>char*</code>) based on the format you specify.

see: <a href="http://www.cplusplus.com/reference/clibrary/ctime/" rel="nofollow noreferrer">http://www.cplusplus.com/reference/clibrary/ctime/</a>

Date/time calculations can get tricky. You are much better off using an existing solution rather than trying to roll your own, unless you have a really good reason.

blocks|key|1413903|text|如果你的原始时间类型是time_t，你必须使用来自time.h的函数，例如gmtime等来获得可移植的代码。C/C%2B%2B标准没有为time_t指定内部格式(甚至精确类型)，因此您不能直接转换或操作time_t值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1413904|所知道的是，time_t是“算术类型”，但算术运算的结果并未指定-您甚至不能可靠地加/减。在实践中，许多系统对time_t使用整数类型，内部格式为秒，但这不是标准强制执行。|1413905|简而言之，使用gmtime+(以及一般的time.h功能)。|1413906|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

If your original time type is time_t, you have to use functions from time.h i.e. gmtime etc. to get portable code. The C/C++ standards do not specify internal format (or even exact type) for the time_t, so you cannot directly convert or manipulate time_t values.

All that is known is that time_t is "arithmetic type", but results of arithmetic operations are not specified - you cannot even add/subtract reliably. In practice, many systems use integer type for time_t with internal format of seconds since epoch, but this is not enforced by standards.

In short, use gmtime (and time.h functionality in general).

blocks|key|1413949|text|一种简单的方法(尽管与您想要的格式不同)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1413950|std::time_t+result+=+std::time(nullptr);
std::cout+<<+std::asctime(std::localtime(&result));|code-block|syntax|javascript|1413951|输出:+Wed+Sep+21+10:27:52+2011|1413952|请注意，返回的结果将自动与“\n”连接..您可以使用以下命令将其删除：|1413953|std::string::size_type+i+=+res.find("\n");
if+(i+!=+std::string::npos)
++++res.erase(i,+res.length());|1413954|摘自：http://en.cppreference.com/w/cpp/chrono/c/time|offset|length|1413955|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/chrono/c/time^0|0|0|0|0|0|3|1A|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|13|8|@]|9|@$O|14|P|15|1|16]]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

An easy way (though different than the format you wanted):

<pre><code>std::time_t result = std::time(nullptr);
std::cout &lt;&lt; std::asctime(std::localtime(&amp;result));
</code></pre>

Output:
Wed Sep 21 10:27:52 2011

Notice that the returned result will be automatically concatenated with "\n".. you can remove it using:

<pre><code>std::string::size_type i = res.find("\n");
if (i != std::string::npos)
 res.erase(i, res.length());
</code></pre>

Taken from: <a href="http://en.cppreference.com/w/cpp/chrono/c/time" rel="nofollow">http://en.cppreference.com/w/cpp/chrono/c/time</a>

blocks|key|1198779|text|time_t+t+=+unixTime;
cout+<<+ctime(&t)+<<+endl;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1198780|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>time_t t = unixTime;
cout &lt;&lt; ctime(&amp;t) &lt;&lt; endl;
</code></pre>

blocks|key|1413978|text|这段代码可能会对你有所帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1413979|#include+<iostream>
#include+<ctime>

using+namespace+std;

int+main()+{
+++//+current+date/time+based+on+current+system
+++time_t+now+=+time(0);
+++
+++//+convert+now+to+string+form
+++char*+dt+=+ctime(&now);

+++cout+<<+"The+local+date+and+time+is:+"+<<+dt+<<+endl;

+++//+convert+now+to+tm+struct+for+UTC
+++tm+*gmtm+=+gmtime(&now);
+++dt+=+asctime(gmtm);
+++cout+<<+"The+UTC+date+and+time+is:"<<+dt+<<+endl;
}|code-block|syntax|javascript|1413980|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This code might help you.
<pre><code>#include &lt;iostream&gt;
#include &lt;ctime&gt;

using namespace std;

int main() {
 // current date/time based on current system
 time_t now = time(0);
 
 // convert now to string form
 char* dt = ctime(&amp;now);

 cout &lt;&lt; &quot;The local date and time is: &quot; &lt;&lt; dt &lt;&lt; endl;

 // convert now to tm struct for UTC
 tm *gmtm = gmtime(&amp;now);
 dt = asctime(gmtm);
 cout &lt;&lt; &quot;The UTC date and time is:&quot;&lt;&lt; dt &lt;&lt; endl;
}
</code></pre>

blocks|key|1198810|text|将纪元字符串转换为UTC的步骤|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1198811|string+epoch_to_utc(string+epoch)+{
++long+temp+=+stol(epoch);
++const+time_t+old+=+(time_t)temp;
++struct+tm+*oldt+=+gmtime(&old);
++return+asctime(oldt);
}|code-block|syntax|javascript|1198812|然后它可以被调用为|1198813|++string+temp+=+"245446047";
++cout+<<+epoch_to_utc(temp);|1198814|输出：|1198815|Tue+Oct+11+19:27:27+1977|1198816|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

To convert a epoch string to UTC
<pre class="lang-cpp prettyprint-override"><code>string epoch_to_utc(string epoch) {
 long temp = stol(epoch);
 const time_t old = (time_t)temp;
 struct tm *oldt = gmtime(&amp;old);
 return asctime(oldt);
}
</code></pre>
and then it can be called as
<pre class="lang-cpp prettyprint-override"><code> string temp = &quot;245446047&quot;;
 cout &lt;&lt; epoch_to_utc(temp);
</code></pre>
outputs:
<pre><code>Tue Oct 11 19:27:27 1977
</code></pre>

What I want to do is convert an epoch time (seconds since midnight 1/1/1970) to "real" time (m/d/y h:m:s)

So far, I have the following algorithm, which to me feels a bit ugly:

<pre><code>void DateTime::splitTicks(time_t time) {
 seconds = time % 60;
 time /= 60;
 minutes = time % 60;
 time /= 60;
 hours = time % 24;
 time /= 24;

 year = DateTime::reduceDaysToYear(time);
 month = DateTime::reduceDaysToMonths(time,year);
 day = int(time);
}

int DateTime::reduceDaysToYear(time_t &amp;days) {
 int year;
 for (year=1970;days&gt;daysInYear(year);year++) {
 days -= daysInYear(year);
 }
 return year;
}

int DateTime::reduceDaysToMonths(time_t &amp;days,int year) {
 int month;
 for (month=0;days&gt;daysInMonth(month,year);month++)
 days -= daysInMonth(month,year);
 return month;
}
</code></pre>

you can assume that the members <code>seconds</code>, <code>minutes</code>, <code>hours</code>, <code>month</code>, <code>day</code>, and <code>year</code> all exist.

Using the <code>for</code> loops to modify the original time feels a little off, and I was wondering if there is a "better" solution to this.

Converting epoch time to "real" date/time

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想要做的是将一个纪元时间(自1970年1月1日午夜以来的秒数)转换为“实时”时间(m/d/y h:m:s)到目前为止，我有以下算法，我觉得有点难看：void DateTime::splitTicks(time_t time) {    seconds = time % 60;    time /= 60;    mi...

问将纪元时间转换为“真实”日期/时间
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