Converting epoch time to "real" date/time

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.

Solution 1:

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;
}

Solution 2:

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

see: http://www.cplusplus.com/reference/clibrary/ctime/

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.

Solution 3:

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

std::time_t result = std::time(nullptr);
std::cout << std::asctime(std::localtime(&result));

Output: Wed Sep 21 10:27:52 2011

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

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

Taken from: http://en.cppreference.com/w/cpp/chrono/c/time

Solution 4:

time_t t = unixTime;
cout << ctime(&t) << endl;

Solution 5:

This code might help you.

#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;
}