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

New answer for an old question:

chrono-Compatible Low-Level Date Algorithms

has formulas for converting a {year, month, day} triple to a serial count of days and back. You can use it to calculate the number of days between two dates like this:

std::cout << days_from_civil(2012, 4, 2) - days_from_civil(2003, 2, 2) << '\n';

which outputs:

3347

The paper is a how-to manual, not a library. It uses C++14 to demonstrate the formulas. Each formula comes with a detailed description and derivation, that you only have to read if you care about knowing how the formula works.

The formulas are very efficient, and valid over an extremely large range. For example using 32 bit arithmetic, +/- 5 million years (more than enough).

The serial day count is a count of days since (or prior to for negative values) New Years 1970, making the formulas compatible with Unix Time and all known implementations of std::chrono::system_clock.

The days_from_civil algorithm is not novel, and it should look very similar to other algorithms for doing the same thing. But going the other way, from a count of days back to a {year, month, day} triple is trickier. This is the formula documented by civil_from_days and I have not seen other formulations that are as compact as this one.

The paper includes example uses showing typical computations, std::chrono interoperability, and extensive unit tests demonstrating the correctness over +/- 1 million years (using a proleptic Gregorian calendar).

All of the formulas and software are in the public domain.

Answer 2

If you need to do it yourself, then one way to do this pretty easy is by converting dates into a Julian Day. You get formulas at that link, and from conversion on, you only work with floats, where each day is 1 unit.

Answer 3

Here is a complete code to calculating date difference in y/m/d.

Assuming that to and from are date types, and that months and days start from 1 (similar to Qt):

static int increment[12] = { 1, -2, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1 };

int daysInc = 0;
if (to.day() - from.day() < 0)
{
    int month = to.month() - 2; // -1 from zero, -1 previous month.
    if (month < 0)
        month = 11; // Previous month is December.
    daysInc = increment[month];
    if ( (month == 1) && (to.year()%4 == 0) )
        daysInc++; // Increment days for leap year.
}

int total1 = from.year()*360 + from.month()*30 + from.day();
int total2 = to.year()*360 + to.month()*30 + to.day();
int diff = total2 - total1;
int years = diff/360;
int months = (diff - years*360)/30;
int days = diff - years*360 - months*30 + daysInc;

// Extra calculation when we can pass one month instead of 30 days.
if (from.day() == 1 && to.day() == 31) {
    months--;
    days = 30;
}

I tried this algorithm and it is working okay. Let me know if you have troubles using/understanding it.

Answer 4

You should look at the DateTime class.

Also the msdn reference for C++ syntax.

Answer 5

Since you are looking for mathematical formula , it will help you to find a solution to your problem. Let Y be the year,M be the month and D be the day. Do this calculation for both the dates.

Total = Y* 365 + M*30 + D ,then find the difference between 2 totals of the corresponding dates.

While multiplying 30 with M value ,you have to give the number of days in that month. You can do it with #define value or if loop. Similarly you can do for leap year too by multiplying 366 with Y .

Hope this will help u....

Answer 6

I'm not sure what platform are you on? Windows, Linux? But let us pretend that you would like to have a platform independent solution and the langugage is standard C++.

If you can use libraries you can use the Boost::Date_Time library (http://www.boost.org/doc/libs/1_49_0/doc/html/date_time.html)

If you cannot use libraries to solve your assignment, you will need to find a common simple ground. Maybe you could convert all the dates to seconds, or days substract them and then convert that back to the data again. Substracting days or months as integers will not help as it will lead to incorrect results unless you do not take into account the rest. Hope that helps.

Like dbrank0 pointed it out. :)