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

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! ;-)

Answer 2

Here is a simple code that I created in c that should fix your problem :

#include <conio.h>

int main()
{
  int y,n,oy,ly,td,a,month,mon_,d,days,down,up; // oy==ordinary year, td=total days, d=date

    printf("Enter the year,month,date: ");
    scanf("%d%d%d",&y,&month,&d);
    n= y-1; //here we subtracted one year because we have to find on a particular day of that year, so we will not count whole year.
    oy= n%4;

    if(oy==0) // for leap year
      {
         mon_= month-1;
         down= mon_/2;  //down means months containing 30 days.
           up= mon_-down; // up means months containing 31 days.
           if(mon_>=2)
             {
               days=(up*31)+((down-1)*30)+29+d; // here in down case one month will be of feb so we subtracted 1 and after that seperately
               td= (oy*365)+(ly*366)+days;      // added 29 days as it is the if block of leap year case.
             }
           if(mon_==1)
             {
               days=(up*31)+d;
               td= (oy*365)+(ly*366)+days;
             } 
           if(mon_==0)
             {
               days= d;
               td= (oy*365)+(ly*366)+days;
             }    
      }
    else
      {
         mon_= month-1;
         down= mon_/2;
           up= mon_-down;
           if(mon_>=2)
             {
               days=(up*31)+((down-1)*30)+28+d;
               td= (oy*365)+(ly*366)+days;
             }
           if(mon_==1)
             {
               days=(up*31)+d;
               td= (oy*365)+(ly*366)+days;
             } 
           if(mon_==0)
             {
               days= d;
               td= (oy*365)+(ly*366)+days;
             }    
      }  

    ly= n/4;
     a= td%7;

     if(a==0)
       printf("\nSunday");
     if(a==1)
       printf("\nMonday");
     if(a==2)
       printf("\nTuesday");
     if(a==3)
       printf("\nWednesday");
     if(a==4)
       printf("\nThursday");
     if(a==5)
       printf("\nFriday");
     if(a==6)
       printf("\nSaturday");
  return 0;   
}