C++ Int to Any Date Without External Library

Question

I need to convert some integers to date. First off I do know of the Boost::Gregorian library, but I can't use it because it won't compile with Clang, which is where my application gets it best performance.

I am parsing the raw database files and so performance will be important since the conversions will happen hundreds of thousands of times to represent birthdates, timestamps, appointment times and so on.

I have a few different origin dates depending on which database I am parsing. The origin dates I am using are:

System 1: 1706-02-24
System 2: 1840-01-01

I tried it this way, but I get an error that timeinfo2 is null when I try printing it out:

time_t rawtime;
struct tm* timeinfo;
time(&rawtime);
timeinfo = localtime(&rawtime);
timeinfo->tm_year = 1706 - 1900;
timeinfo->tm_mon = 2 - 1;
timeinfo->tm_mday = 24;
timeinfo->tm_mday += 98040; // days since origin

time_t newtime;

struct tm* timeinfo2;
newtime = mktime(timeinfo);
timeinfo2 = localtime(&newtime);

Result should be: 1968-08-12

Answer 1

I have a lot of respect for what Howard has put together, but I needed something that would execute as fast as possible and worried about including the entire date library when all I needed was an int to date.

Here is what I came up with:

string GetDateFromDaysSincePointInTime(int days)
{
    int a, b, c, d, e, m, dd, mm, yyyy;
    a = days + 2374475;

    b = (4 * a + 3) / 146097;
    c = -b * 146097 / 4 + a;
    d = (4 * c + 3) / 1461;
    e = -1461 * d / 4 + c;
    m = (5 * e + 2) / 153;
    dd = -(153 * m + 2) / 5 + e + 1;
    mm = -m / 10 * 12 + m + 3;
    yyyy = b * 100 + d - 4800 + m / 10;
    return  to_string(yyyy) + "-" + to_string(mm) + '-' + to_string(dd);
}

To use this, simply call it with an int GetDateFromDaysSincePointInTime(113908). That will give you a date. Assuming my starting point is different from your starting point, go to the site https://www.timeanddate.com/date/dateadd.html and add/subtract your desired date from the date that was output. Then alter the int value on variable a by that amount and run again to get your corrected date.

From there, it should be able to be easily altered to have leading zeros if desired:

std::ostringstream month;
month << std::setw(2) << std::setfill('0') << mm;
std::ostringstream day;
day << std::setw(2) << std::setfill('0') << dd;
return  to_string(yyyy) + "-" + month.str() + '-' + day.str()

Alternate way

Here is another way that is more readable, and there doesn't seem to be any real performance hit on 30,000 records:

 string GetDateFromInt(int days)
 {
    int startYear = 1600;
    int year = days / 365.2421875;
    float peryear = 365.2421875;
    int remainder = fmod(days, peryear); // here you could add what day of the year to get a date in the middle of the year
    bool leapyear= ((year & 3) == 0  && (year % 100 != 0));
    int leapYearIndex = leapyear ? 1 : 0;
    int daysInYear = leapYearIndex ? 366 : 365;
    const unsigned short int __mon_yday[2][13] =
    {
        /* Normal years.  */
        { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365 },
        /* Leap years.  */
        { 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366 }
    };
    int dayOfYear = remainder;

    if (dayOfYear >= 1 && dayOfYear <= daysInYear) {
        for (int mon = 0; mon < 12; mon++) {
            if (dayOfYear <= __mon_yday[leapYearIndex][mon + 1]) {
                int month = mon + 1;
                int dayOfMonth = dayOfYear - __mon_yday[leapYearIndex][mon];
                std::ostringstream months;
                months << std::setw(2) << std::setfill('0') << month;
                std::ostringstream day;
                day << std::setw(2) << std::setfill('0') << dayOfMonth;

                return  to_string(startYear + year) + "-" + months.str() + '-' + day.str();
            }
        }
    }
}