C Program to find day of week given date

Question

Is there a way to find out day of the week given date in just one line of C code?

For example

Given 19-05-2011(dd-mm-yyyy) gives me Thursday

Answer 1

Not in one line of code, there's nothing for dealing with dates in the C standard library. It would be fairly simple to write a function based on the Doomsday algorithm, or similar, though.

Answer 2

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 :)

Answer 3

#include<stdio.h>
#include<math.h>
#include<conio.h>
int fm(int date, int month, int year) {
int fmonth, leap;
if ((year % 100 == 0) && (year % 400 != 0))
leap = 0;
   else if (year % 4 == 0)
  leap = 1;
else
  leap = 0;

  fmonth = 3 + (2 - leap) * ((month + 2) / (2 * month))+ (5 * month + month / 9) / 2;

 fmonth = fmonth % 7;

  return fmonth;
}
int day_of_week(int date, int month, int year) {

   int dayOfWeek;
   int YY = year % 100;
   int century = year / 100;

   printf("\nDate: %d/%d/%d \n", date, month, year);

   dayOfWeek = 1.25 * YY + fm(date, month, year) + date - 2 * (century % 4);

   //remainder on division by 7
   dayOfWeek = dayOfWeek % 7;

   switch (dayOfWeek) {
      case 0:
         printf("weekday = Saturday");
         break;
      case 1:
         printf("weekday = Sunday");
         break;
      case 2:
         printf("weekday = Monday");
         break;
      case 3:
         printf("weekday = Tuesday");
         break;
      case 4:
         printf("weekday = Wednesday");
         break;
      case 5:
         printf("weekday = Thursday");
         break;
      case 6:
         printf("weekday = Friday");
         break;
      default:
         printf("Incorrect data");
   }
   return 0;
}
int main() {
   int date, month, year;

   printf("\nEnter the year ");
   scanf("%d", &year);

   printf("\nEnter the month ");
   scanf("%d", &month);

   printf("\nEnter the date ");
   scanf("%d", &date);

   day_of_week(date, month, year);

   return 0;
}

OUTPUT: Enter the year 2012

Enter the month 02

Enter the date 29

Date: 29/2/2012

weekday = Wednesday

Answer 4

/*
Program to calculate the day on a given date by User
*/

#include<stdio.h>
#include<conio.h>
#include<process.h>
void main()
{
int dd=0,mm=0,i=0,yy=0,odd1=0,todd=0;//variable declaration for inputing the date
int remyr=0,remyr1=0,lyrs=0,oyrs=0,cyr=0,upyr=0,leap=0;//variable declaration for calculation of odd days
int montharr[12]={31,28,31,30,31,30,31,31,30,31,30,31};//array of month days
clrscr();
printf("Enter the date as DD-MM-YY for which you want to know the day\t:");
scanf("%d%d%d",&dd,&mm,&yy);  //input the date
/*
check out correct date or not?
*/
if(yy%100==0)
    {
    if(yy%400==0)
        {
        //its the leap year
        leap=1;
        if(dd>29&&mm==2)
            {

            printf("You have entered wrong date");
            getch();
            exit(0);
            }
        }
    else if(dd>28&&mm==2)
        {
        //not the leap year
        printf("You have entered wrong date");
        getch();
        exit(0);

        }
    }
else if(yy%4==0)
    {
    //again leap year
    leap=1;
    if(dd>29&mm==2)
        {
        printf("You have entered wrong date");
        getch();
        exit(0);

        }
    }
else if(dd>28&&mm==2)
    {
    //not the leap year
    printf("You have entered wrong date");
    getch();
    exit(0);

    }
//if the leap year feb month contains 29 days
if(leap==1)
{
montharr[1]=29;

}
//check date,month,year should not be beyond the limits

if((mm>12)||(dd>31)|| (yy>5000))
    {
    printf("Your date is wrong");
    getch();
    exit(0);

    }
//odd months should not contain more than 31 days
if((dd>31 && (mm == 1||mm==3||mm==5||mm==7||mm==8||mm==10||mm==12)))
    {
    printf("Your date is wrong");
    getch();
    exit(0);

    }
//even months should not contains more than 30 days

if((dd>30 && (mm == 4||mm==6||mm==9||mm==11)))
    {
    printf("Your date is wrong");
    getch();
    exit(0);

    }

//logic to calculate odd days.....
printf("\nYou have entered date: %d-%d-%d ",dd,mm,yy);
remyr1=yy-1;
remyr=remyr1%400;

cyr=remyr/100;
if(remyr==0)
    {
    oyrs=0;
    }
else if(cyr==0 && remyr>0)
    {
    oyrs=0;
    }


else if(cyr==1)
    {
    oyrs=5;
    }
else if(cyr==2)
    {
    oyrs=3;
    }
else if(cyr==3)
    {
    oyrs=1;
    }

    upyr=remyr%100;
    lyrs=upyr/4;
    odd1=lyrs+upyr;
    odd1=odd1%7;
    odd1=odd1+oyrs;
    for(i=0;i<mm-1;i++)
        {
        odd1=odd1+montharr[i];
        }
    todd=odd1+dd;
    if(todd>7)
        todd=todd%7;    //total odd days gives the re quired day....
    printf("\n\nThe day on %d-%d-%d :",dd,mm,yy);

    if(todd==0)
        printf("Sunday");
    if(todd==1)
        printf("Monday");
    if(todd==2)
        printf("Tuesday");
    if(todd==3)
        printf("Wednesday");
    if(todd==4)
        printf("Thrusday");
    if(todd==5)
        printf("Friday");
    if(todd==6)
        printf("Saturday");
getch();
}

Answer 5

The answer I came up with:

const int16_t TM_MON_DAYS_ACCU[12] = {
    0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334
};

int tm_is_leap_year(unsigned year) {
    return ((year & 3) == 0) && ((year % 400 == 0) || (year % 100 != 0));
}

// The "Doomsday" the the day of the week of March 0th,
// i.e the last day of February.
// In common years January 3rd has the same day of the week,
// and on leap years it's January 4th.
int tm_doomsday(int year) {
    int result;
    result  = TM_WDAY_TUE;
    result += year;       // I optimized the calculation a bit:
    result += year >>= 2; // result += year / 4
    result -= year /= 25; // result += year / 100
    result += year >>= 2; // result += year / 400
    return result;
}

void tm_get_wyday(int year, int mon, int mday, int *wday, int *yday) {
    int is_leap_year = tm_is_leap_year(year);
    // How many days passed since Jan 1st?
    *yday = TM_MON_DAYS_ACCU[mon] + mday + (mon <= TM_MON_FEB ? 0 : is_leap_year) - 1;
    // Which day of the week was Jan 1st of the given year?
    int jan1 = tm_doomsday(year) - 2 - is_leap_year;
    // Now just add these two values.
    *wday = (jan1 + *yday) % 7;
}

with these defines (matching struct tm of time.h):

#define TM_WDAY_SUN 0
#define TM_WDAY_MON 1
#define TM_WDAY_TUE 2
#define TM_WDAY_WED 3
#define TM_WDAY_THU 4
#define TM_WDAY_FRI 5
#define TM_WDAY_SAT 6

#define TM_MON_JAN  0
#define TM_MON_FEB  1
#define TM_MON_MAR  2
#define TM_MON_APR  3
#define TM_MON_MAY  4
#define TM_MON_JUN  5
#define TM_MON_JUL  6
#define TM_MON_AUG  7
#define TM_MON_SEP  8
#define TM_MON_OCT  9
#define TM_MON_NOV 10
#define TM_MON_DEC 11

Answer 6

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).