C Program to find day of week given date

Solution 1:

As reported also by Wikipedia, in 1990 Michael Keith and Tom Craver published an expression to minimise the number of keystrokes needed to enter a self-contained function for converting a Gregorian date into a numerical day of the week.

The expression does preserve neither y nor d, and returns a zero-based index representing the day, starting with Sunday, i.e. if the day is Monday the expression returns 1.

A code example which uses the expression follows:

int d    = 15   ; //Day     1-31
int m    = 5    ; //Month   1-12`
int y    = 2013 ; //Year    2013` 

int weekday  = (d += m < 3 ? y-- : y - 2, 23*m/9 + d + 4 + y/4- y/100 + y/400)%7;  

The expression uses the comma operator, as discussed in this answer.

Enjoy! ;-)

Solution 2:

A one-liner is unlikely, but the strptime function can be used to parse your date format and the struct tm argument can be queried for its tm_wday member on systems that modify those fields automatically (e.g. some glibc implementations).

int get_weekday(char * str) {
  struct tm tm;
  memset((void *) &tm, 0, sizeof(tm));
  if (strptime(str, "%d-%m-%Y", &tm) != NULL) {
    time_t t = mktime(&tm);
    if (t >= 0) {
      return localtime(&t)->tm_wday; // Sunday=0, Monday=1, etc.
    }
  }
  return -1;
}

Or you could encode these rules to do some arithmetic in a really long single line:

• 1 Jan 1900 was a Monday.
• Thirty days has September, April, June and November; all the rest have thirty-one, saving February alone, which has twenty-eight, rain or shine, and on leap years, twenty-nine.
• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

EDIT: note that this solution only works for dates after the UNIX epoch (1970-01-01T00:00:00Z).

Solution 3:

Here's a C99 version based on wikipedia's article about Julian Day

#include <stdio.h>

const char *wd(int year, int month, int day) {
  /* using C99 compound literals in a single line: notice the splicing */
  return ((const char *[])                                         \
          {"Monday", "Tuesday", "Wednesday",                       \
           "Thursday", "Friday", "Saturday", "Sunday"})[           \
      (                                                            \
          day                                                      \
        + ((153 * (month + 12 * ((14 - month) / 12) - 3) + 2) / 5) \
        + (365 * (year + 4800 - ((14 - month) / 12)))              \
        + ((year + 4800 - ((14 - month) / 12)) / 4)                \
        - ((year + 4800 - ((14 - month) / 12)) / 100)              \
        + ((year + 4800 - ((14 - month) / 12)) / 400)              \
        - 32045                                                    \
      ) % 7];
}

int main(void) {
  printf("%d-%02d-%02d: %s\n", 2011, 5, 19, wd(2011, 5, 19));
  printf("%d-%02d-%02d: %s\n", 2038, 1, 19, wd(2038, 1, 19));
  return 0;
}

By removing the splicing and spaces from the return line in the wd() function, it can be compacted to a 286 character single line :)

Solution 4:

This is my implementation. It's very short and includes error checking. If you want dates before 01-01-1900, you could easily change the anchor to the starting date of the Gregorian calendar.

#include <stdio.h>

int main(int argv, char** arv) {
    int month[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    char* day[] = {"Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"};   
    int d, m, y, i; 

    printf("Fill in a date after 01-01-1900 as dd-mm-yyyy: ");
    scanf("%d-%d-%d",  &d, &m, &y);

    // correction for leap year
    if (y % 4 == 0 && (y % 100 != 0 || y % 400 == 0))
        month[1] = 29;

    if (y < 1900 || m < 1 || m > 12 || d < 1 || d > month[m - 1]) {
        printf("This is an invalid date.\n");
        return 1;
    }

    for (i = 1900; i < y; i++)
        if (i % 4 == 0 && (i % 100 != 0 || i % 400 == 0))
            d += 366;
        else
            d += 365;

    for (i = 0; i < m - 1; i++) 
        d += month[i];

    printf("This is a %s.\n", day[d % 7]);
    return 0;
}