Converting Gregorian Date to number of days in the year

[fourstar]

Hi all, I have been working on this code all day and can't seem to really get a grasp on it. Essentially I need to calculate the number of days in the year when the user inputs a month (by name), day, and year

My output would look something like this:

<< Enter the (Month Date, Year) i.e. March 23, 1999:
>> january 1, 2009
<< the Julian date is 1

<< Enter the (Month Date, Year) i.e. March 23, 1999
>> december 31, 2000
<< the Julian date is 366

It has to take into account leap years.

This is the code that I have so far. Any direction would be very much appreciated. I'm having trouble trying to visualize it.

#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;

int main()
{
  int daysInMonth[12] = {0, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  int leapYear[13] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
  char month[10];
  int day;
  int year;
  int i, x, mm, sum;

  cout << "Enter the (Month Date, Year) i.e. March 23, 1999: ";
  cin >> month >> day >> year;
  strlwr(month);

       if( (year % 4 == 0 && year % 100 != 0) ||  year % 400 == 0) // leap year
		   daysInMonth[1] = 29;
	   else
		   daysInMonth[1] = 28;

  x = strcmp(month, "january");
    if(x == 0)
      mm = 1;
    else if(strcmp(month, "february") == 0)
      mm = 2;
    else if(strcmp(month, "march") == 0)
      mm = 3;
    else if(strcmp(month, "april") == 0)
      mm = 4;
    else if(strcmp(month, "may") == 0)
      mm = 5;
    else if(strcmp(month, "june") == 0)
      mm = 6;
    else if(strcmp(month, "july") == 0)
      mm = 7;
    else if(strcmp(month, "august") == 0)
      mm = 8;
    else if(strcmp(month, "september") == 0)
      mm = 9;
    else if(strcmp(month, "october") == 0)
      mm = 10;
    else if(strcmp(month, "november") == 0)
      mm = 11;
    else if(strcmp(month, "december") == 0)
      mm = 12;

	sum = 0;

	// counter to add days in the month for the date
    for(i=0; i < mm; i++)
      sum = day + daysInMonth[i];


    cout << endl << sum << endl;
  return 0;
}

I keep getting hung up on the leap year and the counter to add to the input day the number of days prior to what I input.

I'd really appreciate any help at all.

[Dave Sinkula]

Can you use the standard library?

#include <iostream>
#include <ctime>

int main(void)
{
    using namespace std; // I'm being lazy
    struct tm calendar = {0};
    calendar.tm_year = 2009 - 1900;
    calendar.tm_mon  = 6 - 1;
    calendar.tm_mday = 30;
    time_t date = mktime ( &calendar );
    if ( date != (time_t)(-1) )
    {
        cout << ctime(&date) << "tm_yday = " << calendar.tm_yday + 1 << '\n';
    }
    return 0;
}

/* my output
Tue Jun 30 01:00:00 2009
tm_yday = 181
*/

Or something along that line?

[wildgoose]

Use stricmp()
Same as strcmp except it's caseless!

Though I'd recommend strlwr or strupr to convert to lower and upper case but do a strncmp
so that you're only comparing the first 3 letters. That way abbreviation can be used and a match will still work!

To save a running count how about using a base count
for each month and subtract next month from prevous month to get day count for the month. Then Look up November directly instead of adding Jan through Nov.

uint MonthBase[12+1] = {
            0, 31, 59, 90, 120, 151,
             181, 212, 243, 273, 304, 334,       365};
};

days elapsed = MonthBase[ Month {0...11} ] + leapday() if >= March

days in month = MonthBase[ Month{0...11} + 1] - MonthBase[ Month ] + leapday() if February;

Congratulations on knowning about the, "Every four years, except every 100 years, except every 400 years."
Most people don't know those last two!

You should also pick a base year. Something not too long ago but will allow you to handle recent old dates!

[fourstar]

Hi Dave,

Thanks for the quick response. The criteria is to use a 1D array for it.

int daysInMonth[12] = {0, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

Is basically supposed to be the amount of days for each month the user inputs. If the user inputs january 1, 2000 it'll output 1 (which is why the array subscript [0] is holding a '0'.

Then I have to determine whether or not it's a leap year. If it IS a leap year, then the accumulated days after February start to change (+1).

I'm just confused trying to visualize this. I have been writing it out on paper for the last hour and nothing.

So far I have:

1) Set up a 1D array holding the number of days in each month
2) ask for the date from the user
3) see if it's a leap year or not according to the year the user entered
4) if it IS a leap year, count 29 days, otherwise keep it at 28
5) sum the elements of the array
6) have a loop to scroll through the months and days to see if we have chosen the correct day. if so add the days of the month to the sum.