Arithmetics on calendar dates in C or C++ (add N days to given date)

Question

I have been given a date, Which I am taking as an input like (day, month, year): 12, 03, 87.

Now I need to find out the date after n days.<

Answer 1

It might be easier to do the math using seconds since the epoch instead of manipulating the date fields directly.

For example, this program prints the date 7 days from now:

#include <stdio.h>
#include <time.h>

main()
{
    time_t t;
    struct tm *tmp;

    time(&t);

    /* add a week to today */
    t += 7 * 24 * 60 * 60;

    tmp = localtime(&t);
    printf("%02d/%02d/%02d\n", tmp->tm_mon+1, tmp->tm_mday,
        tmp->tm_year % 100);
}

Answer 2

The standard library mktime function includes a trick to make it easy to add a number of months or days into a given date: you can give it a date such as "45th of February" or "2nd day of the 40th month" and mktime will normalize it into a proper date. Example:

#include <time.h>
#include <stdio.h>

int main() {
    int y = 1980;
    int m = 2;
    int d = 5;
    int skip = 40;

    // Represent the date as struct tm.                                                           
    // The subtractions are necessary for historical reasons.
    struct tm  t = { 0 };
    t.tm_mday = d;
    t.tm_mon = m-1;
    t.tm_year = y-1900;

    // Add 'skip' days to the date.                                                               
    t.tm_mday += skip;
    mktime(&t);

    // Print the date in ISO-8601 format.                                                         
    char buffer[30];
    strftime(buffer, 30, "%Y-%m-%d", &t);
    puts(buffer);
}

Compared to doing the arithmetic in seconds using time_t, this approach has the advantage that daylight savings transitions do not cause any problems.