Determining the difference between dates

Question

I'm trying to figure out a way for my program to take a date (like February 2nd, 2003) and show the difference between the two with another date (like April 2nd, 2012), excluding leap years. So far I've only been able to figure it out if the dates are in the same month, just by subtracting the "day". In this program I use 2 sets of "month", "day" and "year" integers. I'm pretty much at a loss from where to go from here. This is a completely optional part of my assignment but I'd like to get an idea on how to get it to work. It seems like a hassle to me, but maybe there's a simple math formula I'm not thinking about?

Sorry, I don't have any pre-existing code for this part because the rest of the assignment just deals with having the user enter dates and then adding and subtracting a single day.

Answer 1

Using just the standard library, you can convert a moderately insane date structure into a count of seconds since an arbitrary zero point; then subtract and convert into days:

#include <ctime>

// Make a tm structure representing this date
std::tm make_tm(int year, int month, int day)
{
    std::tm tm = {0};
    tm.tm_year = year - 1900; // years count from 1900
    tm.tm_mon = month - 1;    // months count from January=0
    tm.tm_mday = day;         // days count from 1
    return tm;
}

// Structures representing the two dates
std::tm tm1 = make_tm(2012,4,2);    // April 2nd, 2012
std::tm tm2 = make_tm(2003,2,2);    // February 2nd, 2003

// Arithmetic time values.
// On a posix system, these are seconds since 1970-01-01 00:00:00 UTC
std::time_t time1 = std::mktime(&tm1);
std::time_t time2 = std::mktime(&tm2);

// Divide by the number of seconds in a day
const int seconds_per_day = 60*60*24;
std::time_t difference = (time1 - time2) / seconds_per_day;    

// To be fully portable, we shouldn't assume that these are Unix time;
// instead, we should use "difftime" to give the difference in seconds:
double portable_difference = std::difftime(time1, time2) / seconds_per_day;

Using Boost.Date_Time is a little less weird:

#include "boost/date_time/gregorian/gregorian_types.hpp"

using namespace boost::gregorian;
date date1(2012, Apr, 2);
date date2(2003, Feb, 2);
long difference = (date1 - date2).days();

It seems like a hassle to me, but maybe there's a simple math formula I'm not thinking about?

It is indeed a hassle, but there is a formula, if you want to do the calculation yourself.

Answer 2

There is another way round...

• Given two dates, take the year of the earlier date as the reference year.
• Then calculate no. of days between each of the two given dates and that 1/1/<that year>
• Keep a separate function that tells the number of days elapsed till a specific month.
• The absolute difference of those two no. of days will give the difference between the two given dates.
• Also, do not forget to consider leap years!

The code:

#‎include‬<stdio.h>
#include<math.h>
typedef struct
{
    int d, m, y;
} Date;
int isLeap (int y)
{
    return (y % 4 == 0) && ( y % 100 != 0) || (y % 400 == 0);
}
int diff (Date d1, Date d2)                         //logic here!
{
    int dd1 = 0, dd2 = 0, y, yref;                  //dd1 and dd2 store the <i>no. of days</i> between d1, d2 and the reference year
    yref = (d1.y < d2.y)? d1.y: d2.y;               //that <b>reference year</b>
    for (y = yref; y < d1.y; y++)
        if (isLeap(y))                              //check if there is any leap year between the reference year and d1's year (exclusive)
            dd1++;
    if (isLeap(d1.y) && d1.m > 2) dd1++;                //add another day if the date is past a leap year's February
    dd1 += daysTill(d1.m) + d1.d + (d1.y - yref) * 365;     //sum up all the tiny bits (days)
    for (y = yref; y < d2.y; y++)                       //repeat for d2
        if(isLeap(y))
            dd2++;
    if (isLeap(y) && d2.m > 2) dd2++;
    dd2 += daysTill(d2.m) + d2.d + (d2.y - yref) * 365;
    return abs(dd2 - dd1);                          //return the absolute difference between the two <i>no. of days elapsed past the reference year</i>
}
int daysTill (int month)                            //some logic here too!!
{
    int days = 0;
    switch (month)
    {
        case 1: days = 0;
        break;
        case 2: days = 31;
        break;
        case 3: days = 59;
        break;
        case 4: days = 90;      //number of days elapsed before April in a non-leap year
        break;
        case 5: days = 120;
        break;
        case 6: days = 151;
        break;
        case 7: days = 181;
        break;
        case 8: days = 212;
        break;
        case 9: days = 243;
        break;
        case 10:days = 273;
        break;
        case 11:days = 304;
        break;
        case 12:days = 334;
        break;
    }
    return days;
}
main()
{
    int t;          //no. of test cases
    Date d1, d2;    //d1 is the first date, d2 is the second one! obvious, duh!?
    scanf ("%d", &t);
    while (t--)
    {
        scanf ("%d %d %d", &d1.d, &d1.m, &d1.y);
        scanf ("%d %d %d", &d2.d, &d2.m, &d2.y);
        printf ("%d\n", diff(d1, d2));
    }
}

Standard Input:

1
23 9 1960
11 3 2015

Standard Output:

19892

Code in action: https://ideone.com/RrADFR

Better algorithms, optimizations and edits are always welcome!