Solving this problem is again relatively straightforward if you use the date library. However, this time, you have to use the following types:

• date::year_month_day, a structure that represents a day with fields for year, month (1 to 12), and day (1 to 31).
• date::iso_week::year_weeknum_weekday, from the iso_week.h header, is a structure that has fields for year, number of weeks in a year, and number of days in a week (1 to 7). This class is implicitly convertible to and from date::sys_days, which makes it explicitly convertible to any other calendar system that is implicitly convertible to and from date::sys_days, such as date::year_month_day.

With that being said, the problem resolves to creating a year_month_day object to represent the desired date and then a year_weeknum_weekday object from it, and retrieving the day of the week with weekday():

unsigned int week_day(int const y, unsigned int const m, 
                      unsigned int const d)
{
   using namespace date;

   if(m < 1 || m > 12 || d < 1 || d > 31) return 0;

   auto const dt = date::year_month_day{year{ y }, month{ m }, day{ d }};
   auto const tiso = iso_week::year_weeknum_weekday{ dt };

   return (unsigned int)tiso.weekday();
}

int main()
{
   auto wday = week_day(2018, 5, 9);
}