Études for Erlang

Solution 6-3

Here is a suggested solution for Étude 6-3.

dates.erl

%% @author J D Eisenberg <[e-mail]>
%% @doc Functions for splitting a date into a list of
%% year-month-day and finding Julian date.
%% @copyright 2013 J D Eisenberg
%% @version 0.1

-module(dates).
-export([date_parts/1, julian/1, is_leap_year/1]).

%% @doc Takes a string in ISO date format (yyyy-mm-dd) and
%% returns a list of integers in form [year, month, day].

-spec(date_parts(list()) -> list()).

date_parts(DateStr) ->
  [YStr, MStr, DStr] = re:split(DateStr, "-", [{return, list}]),
  [element(1, string:to_integer(YStr)),
    element(1, string:to_integer(MStr)),
    element(1, string:to_integer(DStr))].

%% @doc Takes a string in ISO date format (yyyy-mm-dd) and
%% returns the day of the year (Julian date).

-spec(julian(string()) -> integer()).

julian(DateStr) ->
  DaysPerMonth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31],
  [Y, M, D] = date_parts(DateStr),
  julian(Y, M, D, DaysPerMonth, 0).

%% @doc Helper function that recursively accumulates the number of days
%% up to the specified date.

-spec(julian(integer(), integer(), integer(), [integer()], integer) -> integer()).

julian(Y, M, D, MonthList, Total) when M > 13 - length(MonthList) ->
  [ThisMonth|RemainingMonths] = MonthList,
  julian(Y, M, D, RemainingMonths, Total + ThisMonth);

julian(Y, M, D, _MonthList, Total) ->
  case M > 2 andalso is_leap_year(Y) of
    true -> Total + D + 1;
    false -> Total + D
  end.

%% @doc Given a year, return true or false depending on ...