Date to Day of the week algorithm?

Question

What is the algorithm that, given a day, month and year, returns a day of the week?

Answer 1

This can be done using the std::mktime and std::localtime functions. These functions are not just POSIX, they are mandated by the C++ Standard (C++03 §20.5).

#include <ctime>

std::tm time_in = { 0, 0, 0, // second, minute, hour
        4, 9, 1984 - 1900 }; // 1-based day, 0-based month, year since 1900

std::time_t time_temp = std::mktime( & time_in );

// the return value from localtime is a static global - do not call
// this function from more than one thread!
std::tm const *time_out = std::localtime( & time_temp );

std::cout << "I was born on (Sunday = 0) D.O.W. " << time_out->tm_wday << '\n';

Answer 2

One of the easiest algorithm for this is Tomohiko Sakamoto Algorithm:

static int t[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};
y -= m < 3;
return (y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
}

Check this out: https://iq.opengenus.org/tomohiko-sakamoto-algorithm/

I found Wang's method also interesting

w = (d - d^(m) + y^ - y* + [y^/4 - y*/2] - 2( c mod 4)) mod 7

http://rmm.ludus-opuscula.org/PDF_Files/Wang_Day_5_8(3_2015)_high.pdf This pdf is really helpful too.

Thanks!

Answer 3

You need a starting point. Today is fine. Hard-code it.

Then, you need to represent the number of days in a month. This is 31, 28, 31, 30, 31, 30, ... . So you can start adding and subtracting 365 % 7 to the day of the week for each year, and (sum of days in difference of month) % 7 again. And so on.

The caveat: Leap years occur on every 4th year, but not every 100th, unless that year is also a multiple of 400.