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), exc

Answer 1

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/
• 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
#include
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 no. of days between d1, d2 and the reference year
    yref = (d1.y < d2.y)? d1.y: d2.y;               //that reference year
    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 no. of days elapsed past the reference year
}
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!