将时代时间转换为“实际”约会时间 [英] Converting epoch time to "real" date/time

问题描述

我想做的是将一个时代（从1970年1月1日以来的秒数）转换为真实时间（m / d / yh：m：s）

到目前为止，我有以下算法，对我来说感觉有点丑陋：

  void DateTime :: splitTicks （time_t time）{
秒=时间％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& days）{
 int year; 
（年= 1970;天> daysInYear（年）;年++）{
 days  -  = daysIn年（年）; 
} 
回报年; 
} 
 
 int DateTime :: reduceDaysToMonths（time_t& days，int year）{
 int month; （month = 0; days> daysInMonth（month，year）; month ++）
 days  -  = daysInMonth（month，year）; 
 
返回月
} 
  

你可以假设成员秒，分钟，小时，月 日和年都存在。

使用循环修改原来的时间感觉有点偏离，我想知道是否有更好的解决方案。

解决方案

小心闰年在你的日子中的功能。

如果你想要非常高的性能，你可以预先计算该对可以一个月到达+年，然后计算日/小时/分/秒。

一个很好的解决方案是 * gmtime - 将日历时间转换为分解时间
* /
/ * $ He ader：gmtime.c，v 1.4 91/04/22 13:20:27 ceriel Exp $ * /

#include< time.h>
#include< limits.h>
#includeloc_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;
注册unsigned long dayclock，dayno;
int year = EPOCH_YR;

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

timep-> tm_sec = dayclock％60;
timep-> tm_min =（dayclock％3600）/ 60;
timep-> tm_hour = dayclock / 3600;
timep-> tm_wday =（dayno + 4）％7; / *第0天是星期日* /
while（dayno> = YEARSIZE（year））{
dayno - = YEARSIZE（year）;
年++;
}
timep-> tm_year =年 - 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 ++;
}
timep-> tm_mday = dayno + 1;
timep-> tm_isdst = 0;

返回timep;
}

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:

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 &days) {
    int year;
    for (year=1970;days>daysInYear(year);year++) {
        days -= daysInYear(year);
    }
    return year;
}

int DateTime::reduceDaysToMonths(time_t &days,int year) {
    int month;
    for (month=0;days>daysInMonth(month,year);month++)
        days -= daysInMonth(month,year);
    return month;
}

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

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

解决方案

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 gmtime source 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        <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 % SECS_DAY;
        dayno = (unsigned long)time / SECS_DAY;

        timep->tm_sec = dayclock % 60;
        timep->tm_min = (dayclock % 3600) / 60;
        timep->tm_hour = dayclock / 3600;
        timep->tm_wday = (dayno + 4) % 7;       /* day 0 was a thursday */
        while (dayno >= YEARSIZE(year)) {
                dayno -= YEARSIZE(year);
                year++;
        }
        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++;
        }
        timep->tm_mday = dayno + 1;
        timep->tm_isdst = 0;

        return timep;
}

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