how to calculate difference between two dates without using any ready library function?

Question

As the title says I want to calculate the difference between two given dates without using any library function such as difftime. If the answer be an algorithm it will be better...

Answer 1

int getDifferentDay(int d1, int m1, int y1, int d2, int m2,int y2) { int x1,x2;

m1 = (m1 + 9) % 12;
y1 = y1 - m1 / 10;
x1= 365*y1 + y1/4 - y1/100 + y1/400 + (m1*306 + 5)/10 + ( d1 - 1 );

m2 = (m2 + 9) % 12;
y2 = y2 - m2 / 10;
x2= 365*y2 + y2/4 - y2/100 + y2/400 + (m2*306 + 5)/10 + ( d2 - 1 );

return (x2 - x1);

}

Answer 2

I would like to tell you in what way you should proceed

First Write few functions ...

A. int valiDate(char *date_string);

This function should take the date as the input entered in the form of a string. It should check whether its a proper Date or Not. I hope You can do this... Reply Back telling me the details of What are the things you checked to take care if its a valid date.It should return 1 if its a proper date or else it should return 0.

B. int year(char *date_string);

This function should return the Year of the Given date. remember you should validatee the date_string befotre returning the year. If the date is not proper it should return a Negative value.

C. int month(char *date_string);

This function should return the MOnth of the Given date. remember you should validatee the date_string befotre returning the month. If the date is not proper it should return a Negative value.

D. int date(char *date_string);

This function should return the date value of the Given date. remember you should validatee the date_string befotre returning the date. If the date is not proper it should return a Negative value.

E. int number_of_days_in_the_year(int year);

This function should take an integer year in the form of input and then return the number of days in that year. For Eg - if input is 2004, output should be 366
if input is 2005, output should be 365

F. int number_of_days_till_now(char *date_string);

This function should return the number of days over till the given date in the given year. For eg Note in all examples i consider the date input string in the DD_MM_YY format.

                       Input                                            Output
                       23-02-2004  (Feb 23 2004)               54
                       01-03-2004  (Mar 1 2004)                 61

                       15-04-2003  (Apr 15 2003)               105
                       01-03-2003  (Mar 1 2003)                  60

I hope You understood this.

Now Lets look at the algorithm

1. Lets say the dates asr stored in the string 'dmy2' and dmy1'.
2. ValiDate both the dates using function 1.
3. Calculate the difference between the years dmy2 and dmy1; lets say it is 'year_diff'

4. if 'year_diff' > 1 then a. Calculate the number of days in the years between dmy1 and dmy2. For eg if dmy2= 15-03-2005 and dmy1=16-02-2001 then the differenc between the years 'year_diff'=(2005-2001)=4 . Therefore Calculate the total number of days in years 2002, 2003,2004 using function 'number_of_days_in_the_year' (function 5) Add them, lets say its stored in "SUM".

5. Calcualte the value SUM1 = number_of_days_till_now(dmy2); i.e number of days over in the dmy2. In our example it is SUM1 = number_of_days_till_now(15-03-2005) = 74;

6. Calculate the value Y2 = number_of_days_in_the_year (dmy1) . I.e Total Number of days in year 1. In our Example it is Y2 = number_of_days_in_the_year (16-02-2001) = 365;

7. Calculate the value SUM2= Y2 - number_of_days_till_now(dmy1); ie. 365 - 47 = 318 ;

8. Now add SUM + SUM1 + SUM2 , You will get the difference between the dates dmy2 and dmy1;

And here is an sample code :

#include<stdio.h>
#include<math.h>
void main()
{
int day1,mon1,year1,day2,mon2,year2;
int ref,dd1,dd2,i;
clrscr();
printf("Enter first day, month, year");
scanf("%d%d%d",&day1,&mon1,&year1);
scanf("%d%d%d",&day2,&mon2,&year2);
ref = year1;
if(year2<year1)
ref = year2;
dd1=0;
dd1=func1(mon1);
for(i=ref;i<year1;i++)
{
    if(i%4==0)
    dd1+=1;
}
dd1=dd1+day1+(year1-ref)*365;
printf("No. of days of first date fronm the Jan 1 %d= %d",year1,dd1);
/* Count for additional days due to leap years*/
dd2=0;
for(i=ref;i<year2;i++)
{
    if(i%4==0)
    dd2+=1;
}
dd2=func1(mon2)+dd2+day2+((year2-ref)*365);
printf("No. of days from the reference year's first Jan = %d",dd2);
printf("Therefore, diff between the two dates is %d",abs(dd2-dd1));

getch();
}

Happy coding.

Answer 3

It might help you to know about the Doomsday algorithm.

Answer 4

Here is some code I just wrote to calculate the difference in years, months, and days. It is in the public domain.

public sealed class DateDifference {
        int years;

        public int Years {
            get { return years; }
        }
        int months;

        public int Months {
            get { return months; }
        }
        int days;

        public int Days {
            get { return days; }
        }

        public override string ToString()
        {
            return string.Format("[DateDifference Years={0}, Months={1}, Days={2}]", years, months, days);
        }


        public DateDifference(DateTime earlier, DateTime later){
            if(later<earlier)
                throw new ArgumentException("later is earlier than 'earlier'.");
            bool isleapday=(earlier.Month==2 && earlier.Day==29);
            DateTime tmp=isleapday ? new DateTime(earlier.Year,2,28) : earlier;
            while(true){
                try {
                    tmp=tmp.AddYears(1);
                    if(isleapday && DateTime.IsLeapYear(tmp.Year))
                        tmp=new DateTime(tmp.Year,2,29);
                } catch(ArgumentOutOfRangeException){
                    break;
                }
                if(tmp<=later){
                    years++;
                    earlier=tmp;
                } else {
                    break;
                }
            }
            // Add months
            tmp=earlier;
            while(true){
                try {
                    tmp=tmp.AddMonths(1);
                if(isleapday && tmp.Day!=29 && tmp.Month!=2)
                    tmp=new DateTime(tmp.Year,tmp.Month,29);
                } catch(ArgumentOutOfRangeException){
                    break;
                }
                if(tmp<=later){
                    months++;
                    earlier=tmp;
                } else {
                    break;
                }
            }
            tmp=earlier;
            while(true){
                try {
                    tmp=tmp.AddDays(1);
                } catch(ArgumentOutOfRangeException){
                    break;
                }
                if(tmp<=later){
                    days++;
                    earlier=tmp;
                } else {
                    break;
                }
            }
        }
    }