/*!
 * satellite-js v5.0.0
 * (c) 2013 Shashwat Kandadai and UCSC
 * https://github.com/shashwatak/satellite-js
 * License: MIT
 */

"use strict";

Object.defineProperty(exports, "__esModule", {
  value: true
});
exports.days2mdhms = days2mdhms;
exports.invjday = invjday;
exports.jday = jday;
/* -----------------------------------------------------------------------------
 *
 *                           procedure days2mdhms
 *
 *  this procedure converts the day of the year, days, to the equivalent month
 *    day, hour, minute and second.
 *
 *  algorithm     : set up array for the number of days per month
 *                  find leap year - use 1900 because 2000 is a leap year
 *                  loop through a temp value while the value is < the days
 *                  perform int conversions to the correct day and month
 *                  convert remainder into h m s using type conversions
 *
 *  author        : david vallado                  719-573-2600    1 mar 2001
 *
 *  inputs          description                    range / units
 *    year        - year                           1900 .. 2100
 *    days        - julian day of the year         0.0  .. 366.0
 *
 *  outputs       :
 *    mon         - month                          1 .. 12
 *    day         - day                            1 .. 28,29,30,31
 *    hr          - hour                           0 .. 23
 *    min         - minute                         0 .. 59
 *    sec         - second                         0.0 .. 59.999
 *
 *  locals        :
 *    dayofyr     - day of year
 *    temp        - temporary extended values
 *    inttemp     - temporary int value
 *    i           - index
 *    lmonth[12]  - int array containing the number of days per month
 *
 *  coupling      :
 *    none.
 * --------------------------------------------------------------------------- */
function days2mdhms(year, days) {
  var lmonth = [31, year % 4 === 0 ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
  var dayofyr = Math.floor(days);

  //  ----------------- find month and day of month ----------------
  var i = 1;
  var inttemp = 0;
  while (dayofyr > inttemp + lmonth[i - 1] && i < 12) {
    inttemp += lmonth[i - 1];
    i += 1;
  }
  var mon = i;
  var day = dayofyr - inttemp;

  //  ----------------- find hours minutes and seconds -------------
  var temp = (days - dayofyr) * 24.0;
  var hr = Math.floor(temp);
  temp = (temp - hr) * 60.0;
  var minute = Math.floor(temp);
  var sec = (temp - minute) * 60.0;
  return {
    mon: mon,
    day: day,
    hr: hr,
    minute: minute,
    sec: sec
  };
}

/* -----------------------------------------------------------------------------
 *
 *                           procedure jday
 *
 *  this procedure finds the julian date given the year, month, day, and time.
 *    the julian date is defined by each elapsed day since noon, jan 1, 4713 bc.
 *
 *  algorithm     : calculate the answer in one step for efficiency
 *
 *  author        : david vallado                  719-573-2600    1 mar 2001
 *
 *  inputs          description                    range / units
 *    year        - year                           1900 .. 2100
 *    mon         - month                          1 .. 12
 *    day         - day                            1 .. 28,29,30,31
 *    hr          - universal time hour            0 .. 23
 *    min         - universal time min             0 .. 59
 *    sec         - universal time sec             0.0 .. 59.999
 *
 *  outputs       :
 *    jd          - julian date                    days from 4713 bc
 *
 *  locals        :
 *    none.
 *
 *  coupling      :
 *    none.
 *
 *  references    :
 *    vallado       2007, 189, alg 14, ex 3-14
 *
 * --------------------------------------------------------------------------- */
function jdayInternal(year, mon, day, hr, minute, sec) {
  var msec = arguments.length > 6 && arguments[6] !== undefined ? arguments[6] : 0;
  return 367.0 * year - Math.floor(7 * (year + Math.floor((mon + 9) / 12.0)) * 0.25) + Math.floor(275 * mon / 9.0) + day + 1721013.5 + ((msec / 60000 + sec / 60.0 + minute) / 60.0 + hr) / 24.0 // ut in days
  // # - 0.5*sgn(100.0*year + mon - 190002.5) + 0.5;
  ;
}

function jday(year, mon, day, hr, minute, sec, msec) {
  if (year instanceof Date) {
    var date = year;
    return jdayInternal(date.getUTCFullYear(), date.getUTCMonth() + 1,
    // Note, this function requires months in range 1-12.
    date.getUTCDate(), date.getUTCHours(), date.getUTCMinutes(), date.getUTCSeconds(), date.getUTCMilliseconds());
  }
  return jdayInternal(year, mon, day, hr, minute, sec, msec);
}

/* -----------------------------------------------------------------------------
 *
 *                           procedure invjday
 *
 *  this procedure finds the year, month, day, hour, minute and second
 *  given the julian date. tu can be ut1, tdt, tdb, etc.
 *
 *  algorithm     : set up starting values
 *                  find leap year - use 1900 because 2000 is a leap year
 *                  find the elapsed days through the year in a loop
 *                  call routine to find each individual value
 *
 *  author        : david vallado                  719-573-2600    1 mar 2001
 *
 *  inputs          description                    range / units
 *    jd          - julian date                    days from 4713 bc
 *
 *  outputs       :
 *    year        - year                           1900 .. 2100
 *    mon         - month                          1 .. 12
 *    day         - day                            1 .. 28,29,30,31
 *    hr          - hour                           0 .. 23
 *    min         - minute                         0 .. 59
 *    sec         - second                         0.0 .. 59.999
 *
 *  locals        :
 *    days        - day of year plus fractional
 *                  portion of a day               days
 *    tu          - julian centuries from 0 h
 *                  jan 0, 1900
 *    temp        - temporary double values
 *    leapyrs     - number of leap years from 1900
 *
 *  coupling      :
 *    days2mdhms  - finds month, day, hour, minute and second given days and year
 *
 *  references    :
 *    vallado       2007, 208, alg 22, ex 3-13
 * --------------------------------------------------------------------------- */
function invjday(jd, asArray) {
  // --------------- find year and days of the year -
  var temp = jd - 2415019.5;
  var tu = temp / 365.25;
  var year = 1900 + Math.floor(tu);
  var leapyrs = Math.floor((year - 1901) * 0.25);

  // optional nudge by 8.64x10-7 sec to get even outputs
  var days = temp - ((year - 1900) * 365.0 + leapyrs) + 0.00000000001;

  // ------------ check for case of beginning of a year -----------
  if (days < 1.0) {
    year -= 1;
    leapyrs = Math.floor((year - 1901) * 0.25);
    days = temp - ((year - 1900) * 365.0 + leapyrs);
  }

  // ----------------- find remaing data  -------------------------
  var mdhms = days2mdhms(year, days);
  var mon = mdhms.mon,
    day = mdhms.day,
    hr = mdhms.hr,
    minute = mdhms.minute;
  var sec = mdhms.sec - 0.00000086400;
  if (asArray) {
    return [year, mon, day, hr, minute, Math.floor(sec)];
  }
  return new Date(Date.UTC(year, mon - 1, day, hr, minute, Math.floor(sec)));
}