C++ parse datetime string with "last"-day notation

Question

std::chrono::year_month_day_last provides a very convenient way to express last days of year/month pairs.

Is there a possibility using std::chrono::parse, Howard Hinnant's free, open-source, C++11, header-only datetime library or boost to parse a date-string where the "last" part is modeled as the value zero or the absence of a value, f.e. parse "2023-01-00" or "2023-01" as year_month_day_last{2023y / January / last}?

Answer 1

How about the last part being modeled with the string "last"?

#include "date/date.h"
#include <chrono>
#include <iostream>
#include <sstream>

date::year_month_day
my_parse(std::istream& is)
{
    using namespace date;
    using namespace std;
    using namespace chrono;

    year_month_day result{};
    year_month ym;
    is >> parse(" %Y-%m-", ym);
    if (!is.fail())
    {
        if (isdigit(is.peek()))
        {
            day d;
            is >> parse("%d", d);
            if (!is.fail())
                result = ym/d;
        }
        else
        {
            string word;
            is >> word;
            if (word != "last")
                is.setstate(ios::failbit);
            else
                result = ym/last;
        }
    }
    return result;
}

For C++20, drop #include "date/date.h" and using namespace date;.

Just read in a year, month and the trailing -, and then test to see if you have a number or not following. If number it is a day number, else it better be the string "last".

The beginning space in the format string matches zero or more white space characters.

This could be exercised with:

int
main()
{
    using namespace std;

    istringstream in{"2023-01-30 2023-01-last"};
    in.exceptions(ios::failbit);
    auto date1 = my_parse(in);
    cout << date1 << '\n';
    auto date2 = my_parse(in);
    cout << date2 << '\n';
}

which outputs:

2023-01-30
2023-01-31