Get The Day

Description

 There are a number of ways to calculate what day a given day will be. Here we are using the Julian method. We have also provided an implementation tutorial for this program using Javascript. The web centered program might have more use and portability but Intermediate C programmers would be able to understand this program with ease.

  The theory behind this is that we consider January 1^st of every year to be Julian Day 1. December 31^st will be Julian day 366 for a leap year and 365 for other years. We first find out the Julian day corresponding to the day in question. We calculate the day number using the equation

( julian_day + year + fours - hundreds + four_hundreds)%7

where (each of these are integers)

year = the year in question
fours=(year-1)/4
hundreds = (year-1)/100
four_hundreds = (year-1)/400

  The result of the equation will be a whole number between 0 and 6. If it is 0, the day in question is a Saturday. if it is 1, Sunday and so on. Identifying this and getting the equivalent day is an easy programming task. The program code is given below.

/* Author : Midhun Harikumar - 21 03 2006
* midhunhk@gmail.com
* http://www.geocities.com/mhkonline2/
*==== Program that calculates the day of a given date ====
* Version 1.1 (C)2006 Centrum inc Software Solutions */

# include <stdio.h>
# include <conio.h>

int getdays(int is_leap,int month,int daynum);
int main(void)
{
clrscr();
int is_leap=0,year,julian_day,fours,hundreds,four_hundreds;
int u_yr,u_m,u_d,da_day;
printf("\nEnter a date and I will tell u what day it is...");

printf("\n\n\nEnter the date (dd - mm - yyyy) ...");
scanf("%d %d %d",&u_d,&u_m,&u_yr);

if((u_yr%4 == 0)&&(u_yr%100 != 0)||(u_yr%400 == 0))
is_leap = 1;

julian_day = getdays(is_leap,u_m,u_d);
year = u_yr;
fours = (year - 1)/4;
hundreds = (year - 1)/100;
four_hundreds = (year - 1)/400;
da_day = (year + julian_day +fours - hundreds + four_hundreds)%7;

printf("\n\n%d/%d/%d is a ",u_d,u_m,u_yr);
switch(da_day)
{
case 0 : printf("Saturday"); break;
case 1 : printf("Sunday "); break;
case 2 : printf("Monday "); break;
case 3 : printf("Tuesday "); break;
case 4 : printf("Wednesday"); break;
case 5 : printf("Thursday"); break;
case 6 : printf("Friday"); break;
}
getch();
return (0);
}
/* Function to calculate the number of days for julian days
3 parameters
leap : 0 or 1 indicates whether the yr is leap or not
mnth : 0 to 12 the month
day : 0 to 31 the day
Returns the julian day number in a given date
NB:<Jan 1 of every yr is Julian Day 1, if leap, Dec 31 is Julian Day 366 else 365 */

int getdays(int leap,int mnth,int day)
{
int i,jul_days = 0;
for(i = 1; i<mnth; i++)
{
if((i==1)||(i==3)||(i==5)||(i==7)||(i==8)||(i==10)||(i==12))
jul_days += 31;
else if((i==2)&&(leap==1))
jul_days += 29;
else if((i==2)&&(leap==0))
jul_days += 28;
else
jul_days += 30;
}
jul_days += day;
return jul_days;
}

This is the program in C to know what day a date is.