Universal formula to calculate the number of days in a month taking into account leap years

Question

Does someone know a math formula to calculate the number of days in a month like this one

28 + (x + Math.floor(x/8)) % 2 + 2 % x + 2 * Math.floor(1/x);

but which also takes in to account leap years? It should also take into account that the Gregorian calendar omits 3 leap days every 400 years, which is the length of its leap cycle.

Answer 1

It isn't very hard to add a term (m == 2) * leapyear(yyyy) to the expression to determine the correct number of days in February of a leap year. This C code shows a way to do it:

#include <stdio.h>
#include <stdbool.h>

static inline bool leapyear(int yy)
{
    if (yy %   4 != 0) return false;
    if (yy % 100 != 0) return true;
    if (yy % 400 != 0) return false;
    return true;
}

static inline int old_dim(int mm)
{
    return (28 + (mm + (mm/8)) % 2 + 2 % mm + 2 * (1/mm));
}

static inline int new_dim(int mm, int yyyy)
{
    return (28 + (mm + (mm/8)) % 2 + 2 % mm + 2 * (1/mm) + ((mm == 2) * leapyear(yyyy)));
}

int main(void)
{
    /*28 + (x + Math.floor(x/8)) % 2 + 2 % x + 2 * Math.floor(1/x);*/
    for (int mm = 1; mm <= 12; mm++)
        printf("mm = %2d, DIM = %2d\n", mm, old_dim(mm));

    for (int yyyy = 1900; yyyy < 2101; yyyy += 5)
    {
        for (int mm = 1; mm <= 12; mm++)
            printf("yyyy = %4d, mm = %2d: DIM = %2d\n", yyyy, mm, new_dim(mm, yyyy));
    }

    return 0;
}

The output for mm = 2 (filtered from the full output) is:

yyyy = 1900, mm =  2: DIM = 28
yyyy = 1905, mm =  2: DIM = 28
yyyy = 1910, mm =  2: DIM = 28
yyyy = 1915, mm =  2: DIM = 28
yyyy = 1920, mm =  2: DIM = 29
yyyy = 1925, mm =  2: DIM = 28
yyyy = 1930, mm =  2: DIM = 28
yyyy = 1935, mm =  2: DIM = 28
yyyy = 1940, mm =  2: DIM = 29
yyyy = 1945, mm =  2: DIM = 28
yyyy = 1950, mm =  2: DIM = 28
yyyy = 1955, mm =  2: DIM = 28
yyyy = 1960, mm =  2: DIM = 29
yyyy = 1965, mm =  2: DIM = 28
yyyy = 1970, mm =  2: DIM = 28
yyyy = 1975, mm =  2: DIM = 28
yyyy = 1980, mm =  2: DIM = 29
yyyy = 1985, mm =  2: DIM = 28
yyyy = 1990, mm =  2: DIM = 28
yyyy = 1995, mm =  2: DIM = 28
yyyy = 2000, mm =  2: DIM = 29
yyyy = 2005, mm =  2: DIM = 28
yyyy = 2010, mm =  2: DIM = 28
yyyy = 2015, mm =  2: DIM = 28
yyyy = 2020, mm =  2: DIM = 29
yyyy = 2025, mm =  2: DIM = 28
yyyy = 2030, mm =  2: DIM = 28
yyyy = 2035, mm =  2: DIM = 28
yyyy = 2040, mm =  2: DIM = 29
yyyy = 2045, mm =  2: DIM = 28
yyyy = 2050, mm =  2: DIM = 28
yyyy = 2055, mm =  2: DIM = 28
yyyy = 2060, mm =  2: DIM = 29
yyyy = 2065, mm =  2: DIM = 28
yyyy = 2070, mm =  2: DIM = 28
yyyy = 2075, mm =  2: DIM = 28
yyyy = 2080, mm =  2: DIM = 29
yyyy = 2085, mm =  2: DIM = 28
yyyy = 2090, mm =  2: DIM = 28
yyyy = 2095, mm =  2: DIM = 28
yyyy = 2100, mm =  2: DIM = 28

This correctly considers 1900 and 2100 as non-leap years, but 2000 as a leap year.

yyyy = 1900, mm =  1: DIM = 31
yyyy = 1900, mm =  2: DIM = 28
yyyy = 1900, mm =  3: DIM = 31
yyyy = 1900, mm =  4: DIM = 30
yyyy = 1900, mm =  5: DIM = 31
yyyy = 1900, mm =  6: DIM = 30
yyyy = 1900, mm =  7: DIM = 31
yyyy = 1900, mm =  8: DIM = 31
yyyy = 1900, mm =  9: DIM = 30
yyyy = 1900, mm = 10: DIM = 31
yyyy = 1900, mm = 11: DIM = 30
yyyy = 1900, mm = 12: DIM = 31

…

yyyy = 2000, mm =  1: DIM = 31
yyyy = 2000, mm =  2: DIM = 29
yyyy = 2000, mm =  3: DIM = 31
yyyy = 2000, mm =  4: DIM = 30
yyyy = 2000, mm =  5: DIM = 31
yyyy = 2000, mm =  6: DIM = 30
yyyy = 2000, mm =  7: DIM = 31
yyyy = 2000, mm =  8: DIM = 31
yyyy = 2000, mm =  9: DIM = 30
yyyy = 2000, mm = 10: DIM = 31
yyyy = 2000, mm = 11: DIM = 30
yyyy = 2000, mm = 12: DIM = 31