Études for Erlang

Étude 6-3: Accumulating the Sum of a List

Add a function julian/1 to the dates module that you wrote in Étude 5-2. Given a string in ISO format ("yyyy-mm-dd"), it returns the Julian date: the day of the year.

Here is some sample output.

1> c(dates).
{ok,dates}
2> dates:julian("2012-12-31").
366
3> dates:julian("2013-12-31").
365
4> dates:julian("2012-02-05").
36
5> dates:julian("2013-02-05").
36
6> dates:julian("1900-03-01").
60
7> dates:julian("2000-03-01").
61
126> dates:julian("2013-01-01").
1

The julian/1 function defines a 12-item list called DaysPerMonth that contains the number of days in each month, splits the date into the year, month, and day (using the date_parts/1 function you wrote in Étude 5-2, and then calls helper function julian/5 (yes, 5).

The julian/5 function does all of the work. Its arguments are the year, month, day, the list of days per month, and an accumulated total, which starts at zero. julian/5 takes the head of the days per month list and adds it to the accumulator, and then calls julian/5 again with the tail of the days per month list and the accumulator value as its last two arguments.

Let’s take, as an example, the sequence of calls for April 18, 2013:

julian(2013, 4, 18, [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31], 0).
julian(2013, 4, 18, [28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31], 31).
julian(2013, 4, 18, [31, 30, 31, 30, 31, 31, 30, 31, 30, 31], 59).
julian(2013, 4, 18, [30, 31, 30, 31, 31, 30, 31, 30, 31], 90).

At this point, the accumulator ...