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

This one works: I took January 2006 as a reference. (It is a Sunday)

int isLeapYear(int year) {

     if(((year%4==0)&&(year%100!=0))||((year%400==0))) 
        return 1;
     else 
        return 0;

 }

int isDateValid(int dd,int mm,int yyyy) {

    int isValid=-1; 
    if(mm<0||mm>12) {
        isValid=-1;
    }
    else {

    if((mm==1)||(mm==3)||(mm==5)||(mm==7)||(mm==8)||(mm==10)||(mm==12)) {
           if((dd>0)&&(dd<=31))
             isValid=1;
    }  else if((mm==4)||(mm==6)||(mm==9)||(mm==11)) {
         if((dd>0)&&(dd<=30))
             isValid=1;
    }  else {
             if(isLeapYear(yyyy)){
                 if((dd>0)&&dd<30)
                     isValid=1;
             } else {
                   if((dd>0)&&dd<29)
                     isValid=1;
             }
        }  



    }
               return isValid;



 }


int calculateDayOfWeek(int dd,int mm,int yyyy)  {

                if(isDateValid(dd,mm,yyyy)==-1) {
                    return -1;
                }

                 int days=0;
                  int i; 

                   for(i=yyyy-1;i>=2006;i--) {
                         days+=(365+isLeapYear(i));

                     }
                     printf("days after years is %d\n",days);

                   for(i=mm-1;i>0;i--) {

                       if((i==1)||(i==3)||(i==5)||(i==7)||(i==8)||(i==10)) {

                           days+=31;
                       }
                       else if((i==4)||(i==6)||(i==9)||(i==11)) {
                           days+=30;
                       } else {

                           days+= (28+isLeapYear(i));

                       }


                   }
                       printf("days after months is %d\n",days);

                   days+=dd;

                   printf("days after days is %d\n",days);

                   return ((days-1)%7);                


              }

Answer 2

#include<stdio.h>
int main(void) {
int n,y;
int ly=0;
int mon;
printf("enter the date\n");
scanf("%d",&n);
printf("enter the month in integer\n");
scanf("%d",&mon);
mon=mon-1;
printf("enter year\n");
scanf("%d",&y);
int dayT;
dayT=n%7;
if((y%4==0&&y%100!=0)|(y%4==0&&y%100==0&&y%400==0))
{
    ly=y;
    printf("the given year is a leap year\n");
}
char a[12]={6,2,2,5,0,3,5,1,4,6,2,4};
if(ly!=0)
{
    a[0]=5;
    a[1]=1;
}
int m,p;
m=a[mon];
int i,j=0,t=1;
for(i=1600;i<=3000;i++)
{
 i=i+99;
    if(i<y)
    {

        if(t==1)
        {
        p=5;t++;
        }
        else if(t==2)
        {
            p=3;
            t++;
        }
        else if(t==3)
        {
            p=1;
            t++;
        }
        else
        {
            p=0;
            t=1;
        }

}}
int q,r,s;
q=y%100;
r=q%7;
s=q/4;
int yTerm;
yTerm=p+r+s;
int w=dayT+m+yTerm;
w=w%7;
if(w==0)
printf("SUNDAY");
else if(w==1)
printf("MONDAY");
else if(w==2)
printf("TUESDAY");
else if(w==3)
printf("WEDNESDAY");
else if(w==4)
printf("THURSDAY");
else if(w==5)
printf("FRIDAY");
else
printf("SATURDAY");
return 0;

}

Answer 3

I think you can find that in glib: http://developer.gnome.org/glib/unstable/glib-Date-and-Time-Functions.html#g-date-get-day

Regards

Answer 4

#include<stdio.h>
static char day_tab[2][13] = {
        {0,31,28,31,30,31,30,31,31,30,31,30,31},
        {0,31,29,31,30,31,30,31,31,30,31,30,31}
        };
int main()
{
     int year,month;
     scanf("%d%d%d",&year,&month,&day);
     printf("%d\n",day_of_year(year,month,day));
     return 0;
}
int day_of_year(int year ,int month,int day)
{
        int i,leap;
        leap = year%4 == 0 && year%100 != 0 || year%400 == 0;
        if(month < 1 || month >12)
        return -1;
        if (day <1 || day > day_tab[leap][month])
        return -1;
        for(i= 1;i<month ; i++)
        {
                day += day_tab[leap][year];
        }
        return day;
}

Answer 5

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;
}

Answer 6

For Day of Week, years 2000 - 2099.

uint8_t rtc_DayOfWeek(uint8_t year, uint8_t month, uint8_t day)
{
    //static const uint8_t month_offset_table[] = {0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}; // Typical table.

    // Added 1 to Jan, Feb. Subtracted 1 from each instead of adding 6 in calc below.

    static const uint8_t month_offset_table[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};

    // Year is 0 - 99, representing years 2000 - 2099

    // Month starts at 0.

    // Day starts at 1.

    // Subtract 1 in calc for Jan, Feb, only in leap years.
    // Subtracting 1 from year has the effect of subtracting 2 in leap years, subtracting 1 otherwise.
    // Adding 1 for Jan, Feb in Month Table so calc ends up subtracting 1 for Jan, Feb, only in leap years.
    // All of this complication to avoid the check if it is a leap year.
    if (month < 2) {
        year--;
    }

    // Century constant is 6. Subtract 1 from Month Table, so difference is 7.

    // Sunday (0), Monday (1) ...

    return (day + month_offset_table[month] + year + (year >> 2)) % 7;

} /* end rtc_DayOfWeek() */