Find the difference between the dates and group it under some category - c++

Question

I have a start date and an end date. I need to find the difference between these dates and group it under the following categories.
< 1 year, < 2 year and so on till X years.
I'm trying to write a unix C++ program for this problem.
I can easily find the unix time difference between start and end date and compare with the 1 year's time stamp (12 * 30 * 20 * 60 * 60) and so on.
Is there any C++ function that returns the difference in years given the start and end date? Also let's say, the difference is 8 years, I suppose I have to write conditions like this,
if((end_date - start_date) < 12 * 30 * 24 * 60 * 60)
group = " less than 1 year"
...
...
Until what point do I stop at, as I won't know what the maximum difference is between the dates?
Is there any easy way to compute this?
I know i'm confusing here, but i ve put all my efforts to explain the problem here. Thanks in advance.
Also note, this is not a assignment or anything.

Answer 1

Assuming "precise years" (in other words, all years are 365 days long) is not an issue, I would do something like this (counting the number of times each year happens in this case - since the original question doesn't really say WHAT to do with each year)
const int MAX_YEARS = 10;
const int YEAR_IN_SECONDS = 365 * 24 * 60 * 60;
std::array<int, MAX_YEARS+1> bins;
int years = static_cast<int>(difftime(end_date - start_date) / YEAR_IN_SECONDS);
// Outside of range, put it at the end of range...
// We could discard or do something else in this case.
if (years > MAX_YEARS)
{
years = MAX_YEARS;
}
bins[years]++; // Seen one more of "this year".
Obviously, what you do with "bins", and what/how you store data there really depends on what you actually are trying to achieve.
An alternative solution would be to use const double YEAR_IN_SECONDS = 365.25 * 24 * 60 * 60;, which would slightly better cover for leap-years. If you want to be precise about it, you'd have to find out if you are before or after each of the leapday in a particular year that is divisible by 4 (and keep in mind that there are special cases for years divisible by 100 and other rules at 400).

Answer 2

#include <chrono>
using years = std::chrono::duration<std::chrono::system_clock::rep, std::ratio<365 * 24 * 60 * 60, 1>>;
std::chrono::system_clock::time_point end_date = std::chrono::system_clock::now();
std::chrono::system_clock::time_point start_date = end_date - years(2);
years how_many = std::chrono::duration_cast<years>(end_date - start_date);
int how_many_as_int = how_many.count();
std::cout << how_many_as_int << std::endl;
std::unordered_map<int, std::list<whatever>> m;
m[how_many_as_int].push_back(...);