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Difference between two dates C++

  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    I am trying to solve a problem that asks me to give the total days between two dates. I have to take care of the some matters between those two dates such as leap years and the way of inputting the years by the users. (For example, if you input 1 and 17, the code will still give you the difference is 16 years (2017 - 2001 = 16). I am not supposed to change ANYTHING inside the main() function. Here is my code.

    2. Relevant equations
    #include <iostream>
    #include <cmath>

    using namespace std;

    class date
    {
    private:
    int m;
    int d;
    int y;

    public:
    date();
    date(int, int, int);
    int countLeapYears(date&);
    int getDifference(date&, date&);


    };

    int main()
    {
    int day, month, year;
    char c;

    cout << "Enter a start date: " << endl;
    cin >> month >> c >> day >> c >> year;

    date start = date(month, day, year);

    cout << "Enter an end date: " << endl;
    cin >> month >> c >> day >> c >> year;

    date end = date(month, day, year);

    int duration = end - start;

    cout << "The number of days between those two dates are: " << duration << endl;

    return 0;
    }

    date::date()
    {
    m = 0;
    d = 0;
    y = 0;
    }

    date::date(int a, int b, int c)
    {
    m = a;
    d = b;
    y = c;
    }

    const int monthDays[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    int date::countLeapYears(date& d)
    {
    int years = d.y;
    if (d.m <= 2)
    years--;


    return years / 4 - years / 100 + years / 400;
    }

    int date::getDifference(date& start, date& end)
    {

    int n1 = start.y*365 + start.d;

    for (int i=0; i<start.m - 1; i++)
    {
    n1 += monthDays;
    n1 += countLeapYears(start);
    }

    int n2 = end.y*365 + end.d;

    for (int i=0; i<end.m - 1; i++)
    {
    n2 += monthDays;
    n2 += countLeapYears(end);
    }
    int duration = n2 - n1;
    return duration;
    }


    3. The attempt at a solution
    have an issue with my code above, and I need your help please. When I ran it, it said invalid binary operation between "date" and "date". Now, I assume that when I initialized int duration = end - start, I should have got a number. I guess what I am doing wrong here is I failed to convert the (end - start) date type into integer. I thought my function getDifference already took care of that issue. Somehow, it appeared that I did not take care of that issue.
     
  2. jcsd
  3. Nov 12, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Your code has no idea that the minus sign should trigger the getDifference function. With the current functionality you would have to use int duration = date::getDifference(end, start);. To use the minus sign you have to overload the minus operator.
     
  4. Nov 12, 2016 #3
    let me update my code.
    #include <iostream>
    #include <cmath>

    using namespace std;

    class date
    {
    private:
    int m;
    int d;
    int y;

    public:
    date(int, int, int);
    int countLeapYears(date&);
    int getDifference(date&);
    friend int operator-(const date&);

    };



    date::date(int a, int b, int c)
    {
    m = a;
    d = b;
    y = c;
    }

    const int monthDays[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

    int date::countLeapYears(date& d)
    {
    int years = d.y;
    if (d.m <= 2)
    years--;
    return years / 4 - years / 100 + years / 400;
    }

    int date::getDifference(date& start)
    {

    int n1 = start.y*365 + start.d;

    for (int i=0; i<start.m - 1; i++)
    {
    n1 += monthDays;
    n1 += countLeapYears(start);
    }

    return n1;
    }

    int date operator-(const date& d)
    {
    int difference = getDiffernce(d);
    return difference;
    }

    int main()
    {
    int day, month, year;
    char c;

    cout << "Enter a start date: " << endl;
    cin >> month >> c >> day >> c >> year;

    date start = date(month, day, year);

    cout << "Enter an end date: " << endl;
    cin >> month >> c >> day >> c >> year;

    date end = date(month, day, year);
    int duration = end-start;

    cout << "The number of days between those two dates are: " << duration << endl;

    return 0;
    }
     
  5. Nov 12, 2016 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You can write [i] as [plain][i][/plain] to avoid that the forum interprets it as marker to make the text italic. Or put the whole code in [code]...[/code]-tags, then things like indentation stay visible.

    The new code is better, but exactly like that it won't even compile.
     
  6. Nov 12, 2016 #5
    it gave me the wrong total difference days between two input dates.
    It gave me

    727081
    as the difference between 1/1/1992 and 1/1/2016
     
  7. Nov 12, 2016 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Where does your getDifference consider the other date?

    727081/365.25 = 1990.6
     
  8. Nov 12, 2016 #7
    What do you mean by that? Please help me
     
  9. Nov 13, 2016 #8

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Code (Text):
    int date::getDifference(date& start)
    {

      int n1 = start.y*365 + start.d;

      for (int i=0; i<start.m - 1; i++)
      {
        n1 += monthDays[i];
        n1 += countLeapYears(start);
      }

      return n1;
    }
    Does the output change in any way if you have a different end date? Can you calculate the difference between two dates if you don't even consider the end date?

    An unrelated error: you add days for LeapYears every month. 0 times in your example, but up to 11 times if the start date is in December. You should add them exactly once.
     
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