You want a program that can take 2 dates represented by (m1, d1, y1) and (m2, d2

Question

 You want a program that can take 2 dates represented by (m1, d1, y1) and (m2, d2, y2) Note that m1 and m2 are month numbers (1-12) , d1 and d2 are days in the month, and  y1 and y2 are year numbers.  You will create a function (deltaDays)  which will compute the number of days between the 2 dates.  The function deltaDays  will call some helper functions to do some of the work: isLeapYear, daysInMonth, isLessThanOrEqual.     To do this, copy the following starting code and replace the comments describing the  algorithm with real code.    #include <iostream> #include <iomanip>  using namespace std;  bool isLeapYear(int y) {     // Write code to      // return true if "y" is a leap year, otherwise return false      // LEAP YEAR DEFINITION:      //If a year is not evenly divisible by 100, then a leap year is      // any year that is evenly divisible by 4, otherwise we don't have a leap year      //If a year is evenly divisible by 100, then it is not a leap year unless     // the year is evenly divisible by 400.      }  int daysInMonth(int m, int y) {     // Write code to      // return the number of days in a month (the month number is 1-12 where 1 represents January)     // Note that April, June, September and November have 30 days     // February has 29 days if "y" is a leap year, otherwise February has 28 days     // All other months have 31 days      }  bool isLessThanOrEqual(int m1, int d1, int y1, int m2, int d2, int y2) {     // Write code to return a true if the date represented by     // (m1, d1, y1) is less than or equal to the date represented by (m2, d2, y2)     // HINT:  if y1 != y2 then you know the answer without going any further     // if y1 == y2 then compare m1 and m2 to determine the answer     // if y1 == y2 and m1 == m2, then you will need to look at d1 and d2      } int deltaDays(int m1, int d1, int y1, int m2, int d2, int y2) {     // Return the number of days to go from the date represented by     // (m1, d1, y1) to the date represented by (m2, d2, y2)     // Note this answer can be negative.      //Recommended approach:      int delta;      if (isLessThanOrEqual(m1,d1, y1, m2, d2, y2))     {         delta = d2 -d1;         // if (m1, y1) is not equal to (m2, y2) then         // add to delta "daysInMonth(m, y)" for every          // month (m) and year(y) needed to go from          // (m1, y1) to (m2, d2) -- This will be some kind of while loop          // As an example, suppose that         // (m1, d1, y1) = (3, 10, 2012)         // (m2, d2, y2) = (5, 25, 2012)          // delta =d2 -d1 = 15;         // there are 2 months that need to be added         // for the months 3, and 4 which have 31 and 30 days respectively         // So the total answer is 76      }     else         delta = -deltaDays (m2, d2, y2, m1, d1, y1); //Note: deltaDays can call deltaDays      return delta; }  // Leave this routine alone void test(int m1, int d1, int y1, int m2, int d2, int y2) {     int delta = deltaDays(m1, d1, y1, m2, d2, y2);     cout << delta << " days between " <<          m1 << "/" << d1 << "/" << y1 << " and " <<         m2 << "/" << d2 << "/" << y2 << endl; }  // Run at least the following tests.  You can add more at the end if you like int main() {         test(11, 20, 2001, 11, 28, 2001);  // In the same month     test(11, 20, 2001, 12, 2, 2001);   // different month, same year     test(11, 20, 2001, 1, 2, 2002);    // different year      test(2, 4, 2001, 2, 28, 2001);      // test leap year logic     test(2 , 4, 2001, 3 , 5, 2001);     test(2 , 4, 2004, 3 , 5, 2004);     test(2 , 4, 2000, 3 , 5, 2000);         test(2 , 4, 1900, 3 , 5, 1900);          // Testing a span that is over a year     test (2, 4, 2012, 4, 4, 2013);     test (4, 4, 2013, 2, 4, 2012);   // Negative answer on this one  }
what is the source code? and the output?

Explanation / Answer

#include <iostream>
#include <iomanip>
using namespace std;
bool isLeapYear(int y)
{
return ((y % 4 == 0) && (y% 100 != 0) || (y % 400 == 0));
}
int daysInMonth(int m, int y)
{
switch(m)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12: return 31; break;
case 4:
case 6:
case 9:
case 11: return 30; break;
case 2: if(isLeapYear(y)) return 29; else return 28; break;
}
return 0;
}
bool isLessThanOrEqual(int m1, int d1, int y1, int m2, int d2, int y2)
{
if(y1!=y2)
{
return (y1<=y2);
}
else
{
if(m1!=m2)
{
return (m1<=m2);
}
else
{
return (d1<=d2);
}
}
return true;
}
int deltaDays(int m1, int d1, int y1, int m2, int d2, int y2)
{
    int delta=0;
    if (isLessThanOrEqual(m1,d1, y1, m2, d2, y2))
    {
        delta = d2 -d1;
        int month = m1;
        int year = y1;
        while((year<=y2))
        {
        if((year==y2) && (month == m2)) break;
        delta = delta + daysInMonth(month,year);
        if(month==12)
        {
        year++;
        month = 0; // initialized to zero here because below statement will increase by 1.
        }
        month++;
         }
       
    }
    else
        delta = -deltaDays (m2, d2, y2, m1, d1, y1); //Note: deltaDays can call deltaDays
    return delta;
}

// Leave this routine alone
void test(int m1, int d1, int y1, int m2, int d2, int y2)
{
    int delta = deltaDays(m1, d1, y1, m2, d2, y2);
    cout << delta << " days between " <<
        m1 << "/" << d1 << "/" << y1 << " and " <<
        m2 << "/" << d2 << "/" << y2 << endl;
}

// Run at least the following tests. You can add more at the end if you like
int main()
{  
     test(11, 20, 2001, 11, 28, 2001); // In the same month
     test(11, 20, 2001, 12, 2, 2001);   // different month, same year
     test(11, 20, 2001, 1, 2, 2002);    // different year

     test(2, 4, 2001, 2, 28, 2001);

    // test leap year logic
     test(2 , 4, 2001, 3 , 5, 2001);
     test(2 , 4, 2004, 3 , 5, 2004);
     test(2 , 4, 2000, 3 , 5, 2000);  
     test(2 , 4, 1900, 3 , 5, 1900);  

    // Testing a span that is over a year
      test (2, 4, 2012, 4, 4, 2013);
      test (4, 4, 2013, 2, 4, 2012);   // Negative answer on this one
return 0;
}

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