Find the last Sunday of each month

From Rosetta Code
Task
Find the last Sunday of each month
You are encouraged to solve this task according to the task description, using any language you may know.

Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc.).

Example of an expected output:

./last_sundays 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Cf.

Ada[edit]

The program from [[1]] solves this task, as well.

Output:
>./last_weekday_in_month sunday 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

AutoHotkey[edit]

InputBox, Year, , Enter a year., , 300, 135
Date := Year . "0101"
 
while SubStr(Date, 1, 4) = Year {
FormatTime, WD, % Date, WDay
if (WD = 1)
MM := LTrim(SubStr(Date, 5, 2), "0"), Day%MM% := SubStr(Date, 7, 2)
Date += 1, Days
}
 
Gui, Font, S10, Courier New
Gui, Add, Text, , % "Last Sundays of " Year ":`n---------------------"
 
Loop, 12 {
FormatTime, Month, % Year (A_Index > 9 ? "" : "0") A_Index, MMMM
Gui, Add, Text, y+1, % Month (StrLen(Month) > 7 ? "" : "`t") "`t" Day%A_Index%
}
 
Gui, Show
return
Output:
Last Sundays of 2013:
---------------------
January		27
February	24
March		31
April		28
May		26
June		30
July		28
August		25
September	29
October		27
November	24
December	29

Batch File[edit]

Uses day of week, last day of month and leapyear routines

 
@echo off
setlocal enabledelayedexpansion
set /p yr= Enter year:
echo.
call:monthdays %yr% list
set mm=1
for %%i in (!list!) do (
call:calcdow !yr! !mm! %%i dow
set/a lsu=%%i-dow
set mf=0!mm!
echo !yr!-!mf:~-2!-!lsu!
set /a mm+=1
)
pause
exit /b
 
:monthdays yr &list
setlocal
call:isleap %1 ly
for /L %%i in (1,1,12) do (
set /a "nn = 30 + ^!(((%%i & 9) + 6) %% 7) + ^!(%%i ^^ 2) * (ly - 2)
set list=!list! !nn!
)
endlocal & set %2=%list%
exit /b
 
:calcdow yr mt dy &dow  :: 0=sunday
setlocal
set/a a=(14-%2)/12,yr=%1-a,m=%2+12*a-2,"dow=(%3+yr+yr/4-yr/100+yr/400+31*m/12)%%7"
endlocal & set %~4=%dow%
exit /b
 
:isleap yr &leap  :: remove ^ if not delayed expansion
set /a "%2=^!(%1%%4)+(^!^!(%1%%100)-^!^!(%1%%400))"
exit /b
 
Output:
Enter year: 2016

2016-01-31
2016-02-28
2016-03-27
2016-04-24
2016-05-29
2016-06-26
2016-07-31
2016-08-28
2016-09-25
2016-10-30
2016-11-27
2016-12-25

BBC BASIC[edit]

 
INSTALL @lib$+"DATELIB"
 
INPUT "What year to calculate (YYYY)? " Year%
 
PRINT '"Last Sundays in ";Year%;" are on:"
FOR Month%=1 TO 12
PRINT Year% "-" RIGHT$("0"+STR$Month%,2) "-";FN_dim(Month%,Year%)-FN_dow(FN_mjd(FN_dim(Month%,Year%),Month%,Year%))
NEXT
END
 
Output:
What year to calculate (YYYY)? 2013

Last Sundays in 2013 are on:
      2013-01-27
      2013-02-24
      2013-03-31
      2013-04-28
      2013-05-26
      2013-06-30
      2013-07-28
      2013-08-25
      2013-09-29
      2013-10-27
      2013-11-24
      2013-12-29

Befunge[edit]

This is essentially identical to Last Friday of each month except for the initial day offset.

":raeY",,,,,&>55+,:::45*:*%\"d"%!*\4%+!3v
v2++6**"I"5\+/*:*54\-/"d"\/4::-1::p53+g5<
>:00p5g4-+7%\:0\v>,"-",5g+:55+/68*+,55+%v
^<<_$$vv*86%+55:<^+*86%+55,+*86/+55:-1:<6
>$$^@$<>+\55+/:#^_$>:#,_$"-",\:04-\-00g^8
^<# #"#"##"#"##!` +76:+1g00,+55,+*<
Output:
Year:2013

2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

C[edit]

Identical to Last Friday of each month except for the initial day offset.

 
#include <stdio.h>
#include <stdlib.h>
 
int main(int argc, char *argv[])
{
int days[] = {31,29,31,30,31,30,31,31,30,31,30,31};
int m, y, w;
 
if (argc < 2 || (y = atoi(argv[1])) <= 1752) return 1;
days[1] -= (y % 4) || (!(y % 100) && (y % 400));
w = y * 365 + 97 * (y - 1) / 400 + 4;
 
for(m = 0; m < 12; m++) {
w = (w + days[m]) % 7;
printf("%d-%02d-%d\n", y, m + 1,
days[m] + (w < 5 ? -2 : 5) - w);
}
 
return 0;
}
 

C++[edit]

 
#include <windows.h>
#include <iostream>
#include <string>
 
//--------------------------------------------------------------------------------------------------
using namespace std;
 
//--------------------------------------------------------------------------------------------------
class lastSunday
{
public:
lastSunday()
{
m[0] = "JANUARY: "; m[1] = "FEBRUARY: "; m[2] = "MARCH: "; m[3] = "APRIL: ";
m[4] = "MAY: "; m[5] = "JUNE: "; m[6] = "JULY: "; m[7] = "AUGUST: ";
m[8] = "SEPTEMBER: "; m[9] = "OCTOBER: "; m[10] = "NOVEMBER: "; m[11] = "DECEMBER: ";
}
 
void findLastSunday( int y )
{
year = y;
isleapyear();
 
int days[] = { 31, isleap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 },
d;
for( int i = 0; i < 12; i++ )
{
d = days[i];
while( true )
{
if( !getWeekDay( i, d ) ) break;
d--;
}
lastDay[i] = d;
}
 
display();
}
 
private:
void isleapyear()
{
isleap = false;
if( !( year % 4 ) )
{
if( year % 100 ) isleap = true;
else if( !( year % 400 ) ) isleap = true;
}
}
 
void display()
{
system( "cls" );
cout << " YEAR " << year << endl << "=============" << endl;
for( int x = 0; x < 12; x++ )
cout << m[x] << lastDay[x] << endl;
 
cout << endl << endl;
}
 
int getWeekDay( int m, int d )
{
int y = year;
 
int f = y + d + 3 * m - 1;
m++;
if( m < 3 ) y--;
else f -= int( .4 * m + 2.3 );
 
f += int( y / 4 ) - int( ( y / 100 + 1 ) * 0.75 );
f %= 7;
 
return f;
}
 
int lastDay[12], year;
string m[12];
bool isleap;
};
//--------------------------------------------------------------------------------------------------
int main( int argc, char* argv[] )
{
int y;
lastSunday ls;
 
while( true )
{
system( "cls" );
cout << "Enter the year( yyyy ) --- ( 0 to quit ): ";
cin >> y;
if( !y ) return 0;
 
ls.findLastSunday( y );
 
system( "pause" );
}
return 0;
}
//--------------------------------------------------------------------------------------------------
 
Output:
  YEAR 2013
=============
JANUARY:   27
FEBRUARY:  24
MARCH:     31
APRIL:     28
MAY:       26
JUNE:      30
JULY:      28
AUGUST:    25
SEPTEMBER: 29
OCTOBER:   27
NOVEMBER:  24
DECEMBER:  29

Other solution, based on the Boost DateTime library:

#include <iostream>
#include <boost/date_time/gregorian/gregorian.hpp>
#include <cstdlib>
 
int main( int argc , char* argv[ ] ) {
using namespace boost::gregorian ;
 
int year = std::atoi( argv[ 1 ] ) ;
for ( int i = 1 ; i < 13 ; i++ ) {
try {
date d( year , i , 1 ) ;
d = d.end_of_month( ) ;
day_iterator d_itr ( d ) ;
while ( d_itr->day_of_week( ) != Sunday ) {
--d_itr ;
}
std::cout << to_simple_string ( *d_itr ) << std::endl ;
} catch ( bad_year by ) {
std::cout << "Terminated because of " << by.what( ) << "\n" ;
}
}
return 0 ;
}
Output:
2013-Jan-27
2013-Feb-24
2013-Mar-31
2013-Apr-28
2013-May-26
2013-Jun-30
2013-Jul-28
2013-Aug-25
2013-Sep-29
2013-Oct-27
2013-Nov-24
2013-Dec-29

C#[edit]

using System;
 
namespace LastSundayOfEachMonth
{
class Program
{
static void Main()
{
Console.Write("Year to calculate: ");
 
string strYear = Console.ReadLine();
int year = Convert.ToInt32(strYear);
 
DateTime date;
for (int i = 1; i <= 12; i++)
{
date = new DateTime(year, i, DateTime.DaysInMonth(year, i), System.Globalization.CultureInfo.CurrentCulture.Calendar);
while (date.DayOfWeek != DayOfWeek.Sunday)
{
date = date.AddDays(-1);
}
Console.WriteLine(date.ToString("yyyy-MM-dd"));
}
}
}
}
 
Output:
Year to calculate: 2013
2013-Jan-27
2013-Feb-24
2013-Mar-31
2013-Apr-28
2013-May-26
2013-Jun-30
2013-Jul-28
2013-Aug-25
2013-Sep-29
2013-Oct-27
2013-Nov-24
2013-Dec-29

Clojure[edit]

(ns last-sundays.core
(:require [clj-time.core :as time]
[clj-time.periodic :refer [periodic-seq]]
[clj-time.format :as fmt])
(:import (org.joda.time DateTime DateTimeConstants))
(:gen-class))
 
(defn sunday? [t]
(= (.getDayOfWeek t) (DateTimeConstants/SUNDAY)))
 
(defn sundays [year]
(take-while #(= (time/year %) year)
(filter sunday? (periodic-seq (time/date-time year 1 1) (time/days 1)))))
 
(defn last-sundays-of-months [year]
(->> (sundays year)
(group-by time/month)
(vals)
(map (comp first #(sort-by time/day > %)))
(map #(fmt/unparse (fmt/formatters :year-month-day) %))
(interpose "\n")
(apply str)))
 
(defn -main [& args]
(println (last-sundays-of-months (Integer. (first args)))))
 
Output:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

COBOL[edit]

 
program-id. last-sun.
data division.
working-storage section.
1 wk-date.
2 yr pic 9999.
2 mo pic 99 value 1.
2 da pic 99 value 1.
1 rd-date redefines wk-date pic 9(8).
1 binary.
2 int-date pic 9(8).
2 dow pic 9(4).
2 sunday pic 9(4) value 7.
procedure division.
display "Enter a calendar year (1601 thru 9999): "
with no advancing
accept yr
if yr >= 1601 and <= 9999
continue
else
display "Invalid year"
stop run
end-if
perform 12 times
move 1 to da
add 1 to mo
if mo > 12 *> to avoid y10k in 9999
move 12 to mo
move 31 to da
end-if
compute int-date = function
integer-of-date (rd-date)
if mo =12 and da = 31 *> to avoid y10k in 9999
continue
else
subtract 1 from int-date
end-if
compute rd-date = function
date-of-integer (int-date)
compute dow = function mod
((int-date - 1) 7) + 1
compute dow = function mod ((dow - sunday) 7)
subtract dow from da
display yr "-" mo "-" da
add 1 to mo
end-perform
stop run
.
end program last-sun.
 
Output:
2016-01-31
2016-02-28
2016-03-27
2016-04-24
2016-05-29
2016-06-26
2016-07-31
2016-08-28
2016-09-25
2016-10-30
2016-11-27
2016-12-25

D[edit]

void lastSundays(in uint year) {
import std.stdio, std.datetime;
 
foreach (immutable month; 1 .. 13) {
auto date = Date(year, month, 1);
date.day(date.daysInMonth);
date.roll!"days"(-(date.dayOfWeek + 7) % 7);
date.writeln;
}
}
 
void main() {
lastSundays(2013);
}
Output:
2013-Jan-27
2013-Feb-24
2013-Mar-31
2013-Apr-28
2013-May-26
2013-Jun-30
2013-Jul-28
2013-Aug-25
2013-Sep-29
2013-Oct-27
2013-Nov-24
2013-Dec-29

Elixir[edit]

defmodule RC do
def lastSunday(year) do
Enum.map(1..12, fn month ->
lastday = :calendar.last_day_of_the_month(year, month)
daynum = :calendar.day_of_the_week(year, month, lastday)
sunday = lastday - rem(daynum, 7)
{year, month, sunday}
end)
end
end
 
y = String.to_integer(hd(System.argv))
Enum.each(RC.lastSunday(y), fn {year, month, day} ->
 :io.format "~4b-~2..0w-~2..0w~n", [year, month, day]
end)
Output:
C:\Elixir>elixir lastSunday.exs 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Erlang[edit]

 
-module( last_date_each_month ).
 
-export( [monday/1, tuesday/1, wednesday/1, thursday/1, friday/1, saturday/1, sunday/1] ).
 
monday( Year ) -> last( Year, 1 ).
tuesday( Year ) -> last( Year, 2 ).
wednesday( Year ) -> last( Year, 3 ).
thursday( Year ) -> last( Year, 4 ).
friday( Year ) -> last( Year, 5 ).
saturday( Year ) -> last( Year, 6 ).
sunday( Year ) -> last( Year, 7 ).
 
 
 
last( Year, Week_day ) ->
Months = lists:seq( 1, 12 ),
Months_days = [{X, Y} || X <- Months, Y <- lists:seq(calendar:last_day_of_the_month(Year, X), calendar:last_day_of_the_month(Year, X) - 7, -1), calendar:valid_date(Year, X, Y), calendar:day_of_the_week(Year, X, Y) =:= Week_day],
[{Year, X, proplists:get_value(X, Months_days)} || X <- Months].
 
Output:
30> [io:fwrite("~B-~2.10.0B-~B~n", [Y,M,D]) || {Y,M,D} <- last_date_each_month:sunday(2013)].
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

FBSL[edit]

#APPTYPE CONSOLE
 
DIM date AS INTEGER, dayname AS STRING
FOR DIM i = 1 TO 12
FOR DIM j = 31 DOWNTO 1
date = 20130000 + (i * 100) + j
IF CHECKDATE(i, j, 2013) THEN
dayname = DATECONV(date, "dddd")
IF dayname = "Sunday" THEN
PRINT 2013, " ", i, " ", j
EXIT FOR
END IF
END IF
NEXT
NEXT
 
PAUSE
 
Output:
2013-1-27
2013-2-24
2013-3-31
2013-4-28
2013-5-26
2013-6-30
2013-7-28
2013-8-25
2013-9-29
2013-10-27
2013-11-24
2013-12-29

Press any key to continue...

Fortran[edit]

Having already functions that calculate a daynumber from a date and the reverse, the notorious routines of H. F. Fliegel and T. C. van Flandern, the problem becomes simple. Their routine calculates the Julian day number but my version adjusts so that day zero is the thirty-first of December 1899, which is a Sunday. Thus, the day of the week is given by MOD(Daynumber,7) with zero for Sunday up to six for Saturday, as per Genesis.

Their routine is not only fast, but flexible, in particular allowing Month + 1 so that DAYNUM(Y,M + 1,0) gives the last day of month M for M = 1 to 12, with no agony over which month has what last day and leap years or not. If W is the day of week desired, all that remains is to use MOD(D - W,7) to determine the offset back to the desired day of the week and make that shift. There is a problem with the MOD function when negative numbers are involved: it may or may not (depending on computer, compiler, language) return a negative result; by adding 7 and taking a MOD again, this variation is suppressed.

The source uses the MODULE protocol for convenience in its usage in larger projects (which contain a calculation for Easter, daylight savings changeover dates, solar azimuth and altitude, etc. which have been removed here), but requires only F77 features, except for the annoyance of the MUNYAD function returning a TYPE(DATEBAG) result to get the three parts. This is an example where a palindromic programming protocol would be so much better. One would write
      D = DAYNUM(Y,M,D)    !Daynumber from date.
DAYNUM(Y,M,D) = D !Date parts from a day number.

But alas, only pl/i offers palindromic functions, and only for SUBSTR.

 
MODULE DATEGNASH
C Calculate conforming to complex calendarical contortions.
C Astronomer Simon Newcomb determined that the tropical year of 1900
c contained 31556925.9747 seconds, or 365.24219879 days.
c Subsequent definitions involve "no measurable differences",
c whereas in 45BC (when the Julian calendar was adopted), the year was
c 365.24232 days long, going by modern calculations.
c The "tropical" year is the time between the same equinoxes, and thus
c contains the effect of the precession of the Earth's axis, which would
c otherwise cause the seasons to likewise precess around the "fixed star"
c year, that being the time for midnight to point in the same direction
c amongst the "fixed" stars. Specifically, the vernal equinox
c (the northern hemisphere's spring equinox: cultural colonialism)
c is meant to hover around the 21'st of March, although it may fall
c within the 19'th or 20'th, which last has been the most popular in
c the 20'th century, until the leap year of 2000 resynchronised
c the civil calendar.
C By contrast, the computations of astrologers are still based on the
c constellations as oriented in Babylonian times...
 
c So, to add .24219879 to 365 days...
c Adjustment per year Nett Discrepancy remaining.
c +1/4 +.25 +.25 -.00780121 5h 48m 45.98s
c -1/100 -.01 .24 +.00219879 -11m 14.02s
c +1/400 +.0025 .2425 -.00030121 -26.02s
c -1/4000 -.00025 .24225 -.00005121 -4.42s
c
c The remnant of -.00005121 (meaning that the calendar year is too long)
c amounts to needing to drop one day on 19,527 years, and while this could be
c accommodated nicely enough by -1/20000 to give a calendar year of
c 365.24420 days with a remaining discrepancy of -.00000121 or -.01sec/year,
c there is a problem. The Earth's spin is slowing by a similar amount.
c Similar to leap years and leap days are the leap seconds that since the
c development of clocks based on atomic oscillations, have been added or
c sometimes removed to keep clock time aligned with astronomical observations.
c This additional confusion is not further considered.
 
TYPE DateBag !Pack three parts into one.
INTEGER DAY,MONTH,YEAR !The usual suspects.
END TYPE DateBag !Simple enough.
 
CHARACTER*9 MONTHNAME(12),DAYNAME(0:6) !Re-interpretations.
PARAMETER (MONTHNAME = (/"January","February","March","April",
1 "May","June","July","August","September","October","November",
2 "December"/))
PARAMETER (DAYNAME = (/"Sunday","Monday","Tuesday","Wednesday",
1 "Thursday","Friday","Saturday"/)) !Index this array with DayNum mod 7.
CHARACTER*3 MTHNAME(12) !The standard abbreviations.
PARAMETER (MTHNAME = (/"JAN","FEB","MAR","APR","MAY","JUN",
1 "JUL","AUG","SEP","OCT","NOV","DEC"/))
 
INTEGER*4 JDAYSHIFT !INTEGER*2 just isn't enough.
PARAMETER (JDAYSHIFT = 2415020) !Thus shall 31/12/1899 give 0, a Sunday, via DAYNUM.
DOUBLE PRECISION DAYSINYEAR !A real ache.
PARAMETER (DAYSINYEAR = 365.24219879D0) !The "D0" demands DOUBLE PRECISION precision.
INTEGER*4 SECONDSINDAY !This has its uses.
PARAMETER (SECONDSINDAY = 24*60*60) !86400. Disregarding "leap" seconds.
INTEGER NOTADAYNUMBER !Might as well settle on one value.
PARAMETER (NOTADAYNUMBER = -2147483648) !And everyone use it.
 
PARAMETER NZBASEPLACE = "Mt. Cook Trig, Wellington." !Name the place.
DOUBLE PRECISION NZBASELAT,NZBASELONG !Alas, DATA statements do not allow arithmetic expressions.
PARAMETER (NZBASELAT = -((59.3 D0/60 + 17)/60 + 41)) !Degrees South, thus negative.
PARAMETER (NZBASELONG = ((34.65D0/60 + 46)/60 + 174)) !Degrees East are oriented as positive: left to right with N up.
C Seconds Minutes Degrees.
C This is the location of the Mt. Cook trigonometrical base point for New Zealand.
C (It's in the foyer of what used to be the Dominion Museum, Wellington)
C Determined by Henry Jackson, chief surveyor, in 1870.
 
TYPE Terroir !Collate the attributes of location, as so far needed.
CHARACTER*28 PLACENAME !Name the location.
DOUBLE PRECISION LATITUDE,LONGITUDE !Locate the location.
DOUBLE PRECISION ZONETIME !The time zone of its civil clock, not necessarily a whole hour.
END TYPE Terroir !The nature of the climate, soil, etc. is not yet involved.
TYPE(Terroir) BASE !Righto, let's have one of them.
DATA BASE/Terroir("Mt. Cook Trig, Wellington.", !The compiler bungles if NZBASEPLACE is used here.
1 NZBASELAT,NZBASELONG,+12.0)/ !Where it's at. +12 hours ahead = +180 degrees Eastward of Greenwhich.
Careful! New Zealand is not centred on longitude 180, but its civil clock's time zone is.
DOUBLE PRECISION SINBASELAT,COSBASELAT !Calculated in SOLARDIRECTION and ZAPME.
 
CONTAINS !Let the madness begin.
 
INTEGER*4 FUNCTION DAYNUM(YY,M,D) !Computes (JDayN - JDayShift), not JDayN.
C Conversion from a Gregorian calendar date to a Julian day number, JDayN.
C Valid for any Gregorian calendar date producing a Julian day number
C greater than zero, though remember that the Gregorian calendar
C was not used before y1582m10d15 and often, not after that either.
C thus in England (et al) when Wednesday 2'nd September 1752 (Julian style)
C was followed by Thursday the 14'th, occasioning the Eleven Day riots
C because creditors demanded a full month's payment instead of 19/30'ths.
C The zero of the Julian day number corresponds to the first of January
C 4713BC on the *Julian* calendar's naming scheme, as extended backwards
C with current usage into epochs when it did not exist: the proleptic Julian calendar.
c This function employs the naming scheme of the *Gregorian* calendar,
c and if extended backwards into epochs when it did not exist (thus the
c proleptic Gregorian calendar) it would compute a zero for y-4713m11d24 *if*
c it is supposed there was a year zero between 1BC and 1AD (as is convenient
c for modern mathematics and astronomers and their simple calculations), *but*
c 1BC immediately precedes 1AD without any year zero in between (and is a leap year)
c thus the adjustment below so that the date is y-4714m11d24 or 4714BCm11d24,
c not that this name was in use at the time...
c Although the Julian calendar (introduced by himself in what we would call 45BC,
c which was what the Romans occasionally called 709AUC) was provoked by the
c "years of confusion" resulting from arbitrary application of the rules
c for the existing Roman calendar, other confusions remain unresolved,
c so precise dating remains uncertain despite apparently precise specifications
c (and much later, Dennis the Short chose wrongly for the birth of Christ)
c and the Roman practice of inclusive reckoning meant that every four years
c was interpreted as every third (by our exclusive reckoning) so that the
c leap years were not as we now interpret them. This was resolved by Augustus
c but exactly when (and what date name is assigned) and whose writings used
c which system at the time of writing is a matter of more confusion,
c and this has continued for centuries.
C Accordingly, although an algorithm may give a regular sequence of date names,
c that does not mean that those date names were used at the time even if the
c calendar existed then, because the interpretation of the algorithm varied.
c This in turn means that a date given as being on the Julian calendar
c prior to about 10AD is not as definite as it may appear and its alignment
c with the astronomical day number is uncertain even though the calculation
c is quite definite.
c
C Computationally, year 1 is preceded by year 0, in a smooth progression.
C But there was never a year zero despite what astronomers like to say,
C so the formula's year 0 corresponds to 1BC, year -1 to 2BC, and so on back.
C Thus y-4713 in this counting would be 4714BC on the Gregorian calendar,
C were it to have existed then which it didn't.
C To conform to the civil usage, the incoming YY, presumed a proper BC (negative)
C and AD (positive) year is converted into the computational counting sequence, Y,
C and used in the formula. If a YY = 0 is (improperly) offered, it will manifest
C as 1AD. Thus YY = -4714 will lead to calculations with Y = -4713.
C Thus, 1BC is a leap year on the proleptic Gregorian calendar.
C For their convenience, astronomers decreed that a day starts at noon, so that
C in Europe, observations through the night all have the same day number.
C The current Western civil calendar however has the day starting just after midnight
C and that day's number lasts until the following midnight.
C
C There is no constraint on the values of D, which is just added as it stands.
C This means that if D = 0, the daynumber will be that of the last day of the
C previous month. Likewise, M = 0 or M = 13 will wrap around so that Y,M + 1,0
C will give the last day of month M (whatever its length) as one day before
C the first day of the next month.
C
C Example: Y = 1970, M = 1, D = 1; JDAYN = 2440588, a Thursday but MOD(2440588,7) = 3.
C and with the adjustment JDAYSHIFT, DAYNUM = 25568; mod 7 = 4 and DAYNAME(4) = "Thursday".
C The Julian Day number 2440588.0 is for NOON that Thursday, 2440588.5 is twelve hours later.
C And Julian Day number 2440587.625 is for three a.m. Thursday.
C
C DAYNUM and MUNYAD are the infamous routines of H. F. Fliegel and T.C. van Flandern,
C presented in Communications of the ACM, Vol. 11, No. 10 (October, 1968).
Carefully typed in again by R.N.McLean (whom God preserve) December XXMMIIX.
C Though I remain puzzled as to why they used I,J,K for Y,M,D,
C given that the variables were named in the INTEGER statement anyway.
INTEGER*4 JDAYN !Without rebasing, this won't fit in INTEGER*2.
INTEGER YY,Y,M,MM,D !NB! Full year number, so 1970, not 70.
Caution: integer division in Fortran does not produce fractional results.
C The fractional part is discarded so that 4/3 gives 1 and -4/3 gives -1.
C Thus 4/3 might be Trunc(4/3) or 4 div 3 in other languages. Beware of negative numbers!
Y = YY !I can fiddle this copy without damaging the original's value.
IF (Y.LT.1) Y = Y + 1 !Thus YY = -2=2BC, -1=1BC, +1=1AD, ... becomes Y = -1, 0, 1, ...
MM = (M - 14)/12 !Calculate once. Note that this is integer division, truncating.
JDAYN = D - 32075 !This is the proper astronomer's Julian Day Number.
a + 1461*(Y + 4800 + MM)/4
b + 367*(M - 2 - MM*12)/12
c - 3*((Y + 4900 + MM)/100)/4
DAYNUM = JDAYN - JDAYSHIFT !Thus, *NOT* the actual *Julian* Day Number.
END FUNCTION DAYNUM !But one such that Mod(n,7) gives day names.
 
Could compute the day of the year somewhat as follows...
c DN:=D + (61*Month + (Month div 8)) div 2 - 30
c + if Month > 2 then FebLength - 30 else 0;
 
TYPE(DATEBAG) FUNCTION MUNYAD(DAYNUM) !Oh for palindromic programming!
Conversion from a Julian day number to a Gregorian calendar date. See JDAYN/DAYNUM.
INTEGER*4 DAYNUM,JDAYN !Without rebasing, this won't fit in INTEGER*2.
INTEGER Y,M,D,L,N !Y will be a full year number: 1950 not 50.
JDAYN = DAYNUM + JDAYSHIFT !Revert to a proper Julian day number.
L = JDAYN + 68569 !Further machinations of H. F. Fliegel and T.C. van Flandern.
N = 4*L/146097
L = L - (146097*N + 3)/4
Y = 4000*(L + 1)/1461001
L = L - 1461*Y/4 + 31
M = 80*L/2447
D = L - 2447*M/80
L = M/11
M = M + 2 - 12*L
Y = 100*(N - 49) + Y + L
IF (Y.LT.1) Y = Y - 1 !The other side of conformity to BC/AD, as in DAYNUM.
MUNYAD%YEAR = Y !Now place for the world to see.
MUNYAD%MONTH = M
MUNYAD%DAY = D
END FUNCTION MUNYAD !A year has 365.2421988 days...
 
CHARACTER*10 FUNCTION SLASHDATE(DAYNUM) !This is relatively innocent.
Caution! The Gregorian calendar did not exist prior to 15/10/1582!
Confine expected operation to four-digit years, since fixed-field sizes are in mind.
Can use this function in WRITE statements with FORMAT, since this function does not use them.
Compilers of lesser merit can concoct code that bungles such double usage otherwise.
INTEGER*4 DAYNUM !-32768 to 32767 is just not adequate.
TYPE(DATEBAG) D !Though these numbers are more restrained.
INTEGER N,L !Workers.
IF (DAYNUM.EQ.NOTADAYNUMBER) THEN !Perhaps some work can be dodged.
SLASHDATE = " Undated!!" !No proper day number has been placed.
RETURN !So give up, rather than show odd results.
END IF !So much for confusion.
D = MUNYAD(DAYNUM) !Get the pieces.
IF (D%DAY.GT.9) THEN !Here we go.
SLASHDATE(1:1) = CHAR(D%DAY/10 + ICHAR("0")) !Faster than a table look-up?
ELSE !Even if not,
SLASHDATE(1:1) = " " !This should be quick.
END IF !So much for the tens digit.
SLASHDATE(2:2) = CHAR(MOD(D%DAY,10) + ICHAR("0")) !The units digit.
SLASHDATE(3:3) = "/" !Enough of the day number. The separator.
IF (D%MONTH.GT.9) THEN !Now for the month.
SLASHDATE(4:4) = CHAR(D%MONTH/10 + ICHAR("0")) !The tens digit.
ELSE !Not so often used. A table beckons...
SLASHDATE(4:4) = " " !Some might desire leading zeroes here.
END IF !Enough of October, November and December.
SLASHDATE(5:5) = CHAR(MOD(D%MONTH,10) + ICHAR("0")) !The units digit.
SLASHDATE(6:6) = "/" !Enough of the month number. The separator.
L = 10 !The year value deserves a loop, it having four digits.
N = ABS(D%YEAR) !Should never be zero. 1BC is year -1 and 1AD is year = +1.
1 SLASHDATE(L:L) = CHAR(MOD(N,10) + ICHAR("0")) !But if it is, this will place a zero.
N = N/10 !Drop a power of ten.
L = L - 1 !Step back for the next digit.
IF (L.GT.6) GO TO 1 !Thus always four digits, even if they lead with zero.
IF (N.GT.0) SLASHDATE(7:7) = "?" !Y > 9999? Might as well do something.
IF (D%YEAR.LT.0) SLASHDATE(7:7) = "-" !Years BC? Rather than give no indication.
c WRITE (SLASHDATE,1) D%DAY,D%MONTH,D%YEAR !Some compilers will bungle this.
c 1 FORMAT (I2,"/",I2,"/",I4) !If so, a local variable must be used.
RETURN !Enough. !As when SLASHDATE is invoked in a WRITE statement.
END FUNCTION SLASHDATE !Simple enough.
END MODULE DATEGNASH
 
PROGRAM LASTSUNDAY
USE DATEGNASH
INTEGER D,W,M,Y
WRITE (6,1)
1 FORMAT ("Employs the Gregorian calendar pattern.",/,
1 "You specify a day of the week, then nominate a year.",/,
2 "For each month of that year, this calculates the date ",
3 "of the last such day.",//
4 "So, what day (0 = Sunday, 6 = Saturday):",$)
READ (5,*) W
IF (W.LT.0 .OR. W.GT.6) STOP "Not a good week day number!"
 
10 WRITE (6,11)
11 FORMAT ("What year (non-positive to quit):",$)
READ (5,*) Y
IF (Y.LE.0) STOP
DO M = 1,12
D = DAYNUM(Y,M + 1,0) !Zeroth day = last day of previous month.
D = D - MOD(MOD(D - W,7) + 7,7) !Protect against MOD(D,7) giving negative values for D < 0.
WRITE (6,*) SLASHDATE(D)," : ",DAYNAME(MOD(D,7))
END DO
GO TO 10
END
 

And after all that, results:

Employs the Gregorian calendar pattern.
You specify a day of the week, then nominate a year.
For each month of that year, this calculates the date of the last such day.

So, what day (0 = Sunday, 6 = Saturday):0
What year (non-positive to quit):2013
 27/ 1/2013 : Sunday
 24/ 2/2013 : Sunday
 31/ 3/2013 : Sunday
 28/ 4/2013 : Sunday
 26/ 5/2013 : Sunday
 30/ 6/2013 : Sunday
 28/ 7/2013 : Sunday
 25/ 8/2013 : Sunday
 29/ 9/2013 : Sunday
 27/10/2013 : Sunday
 24/11/2013 : Sunday
 29/12/2013 : Sunday
What year (non-positive to quit):0

FreeBASIC[edit]

' version 23-06-2015
' compile with: fbc -s console
 
#Ifndef TRUE ' define true and false for older freebasic versions
#Define FALSE 0
#Define TRUE Not FALSE
#EndIf
 
Function leapyear(Year_ As Integer) As Integer
' from the leapyeat entry
If (Year_ Mod 4) <> 0 Then Return FALSE
If (Year_ Mod 100) = 0 AndAlso (Year_ Mod 400) <> 0 Then Return FALSE
Return TRUE
 
End Function
 
Function wd(m As Integer, d As Integer, y As Integer) As Integer
' Zellerish
' 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday
' 4 = Thursday, 5 = Friday, 6 = Saturday
 
If m < 3 Then ' If m = 1 Or m = 2 Then
m += 12
y -= 1
End If
Return (y + (y \ 4) - (y \ 100) + (y \ 400) + d + ((153 * m + 8) \ 5)) Mod 7
End Function
 
' ------=< MAIN >=------
 
Type month_days
m_name As String
days As UByte
End Type
 
Dim As month_days arr(1 To 12)
Data "January", 31, "February", 28, "March", 31, "April", 30
Data "May", 31, "June", 30, "July", 31, "August", 31
Data "September", 30, "October", 31, "November", 30, "December", 31
 
Dim As Integer yr, d, i, x
Dim As String keypress
 
For i = 1 To 12
With arr(i)
Read .m_name
Read .days
End With
Next
 
Do
 
Do
Print "For what year do you want to find the last Sunday of the month"
Input "any number below 1800 stops program, year in YYYY format";yr
' empty input also stops
If yr < 1800 Then
End
Else
Exit Do
End If
Loop
 
Print : Print
Print "Last Sunday of the month for"; yr
 
For i = 1 To 12
d = arr(i).days
If i = 2 AndAlso leapyear(yr) = TRUE Then d = d + 1
x = wd(i, d, yr)
d = d - x ' don't test it just do it
Print d; " "; arr(i).m_name
Next
 
' empty key buffer
While Inkey <> "" : keypress = Inkey : Wend
Print : Print
Print "Find last Sunday for a other year [Y|y], anything else stops"
keypress =""
While keypress = "" : keypress = Inkey : Wend
If LCase(keypress) <> "y" Then Exit Do
Print : Print
 
Loop
End
Output:
For what year do you want to find the last Sunday of the month
any number below 1800 stops program, year in YYYY format? 2015

Last Sunday of the month for 2015
 25 January
 22 February
 29 March
 26 April
 31 May
 28 June
 26 July
 30 August
 27 September
 25 October
 29 November
 27 December

Go[edit]

This is different from the Go code for Last Friday of each month. It uses the fact that time functions in Go correct for dates: if you enter 32nd of october, Go corrects to 1st of November. So that if 29th of February is corrected to 1st of March, chosen year is not a leap year.

package main
 
import (
"fmt"
"time"
)
 
func main() {
 
var year int
var t time.Time
var lastDay = [12]int { 31,29,31,30,31,30,31,31,30,31,30,31 }
 
for {
fmt.Print("Please select a year: ")
_, err := fmt.Scanf("%d", &year)
if err != nil {
fmt.Println(err)
continue
} else {
break
}
}
 
fmt.Println("Last Sundays of each month of", year)
fmt.Println("==================================")
 
for i := 1;i < 13; i++ {
j := lastDay[i-1]
if i == 2 {
if time.Date(int(year), time.Month(i), j, 0, 0, 0, 0, time.UTC).Month() == time.Date(int(year), time.Month(i), j-1, 0, 0, 0, 0, time.UTC).Month() {
j = 29
} else {
j = 28
}
}
for {
t = time.Date(int(year), time.Month(i), j, 0, 0, 0, 0, time.UTC)
if t.Weekday() == 0 {
fmt.Printf("%s: %d\n", time.Month(i), j)
break
}
j = j - 1
}
}
}
 
Please select a year: 2013
Last Sundays of each month of 2013
==================================
January: 27
February: 24
March: 31
April: 28
May: 26
June: 30
July: 28
August: 25
September: 29
October: 27
November: 24
December: 29

Groovy[edit]

Solution:

enum Day {
Sun, Mon, Tue, Wed, Thu, Fri, Sat
static Day valueOf(Date d) { Day.valueOf(d.format('EEE')) }
}
 
def date = Date.&parse.curry('yyyy-MM-dd')
def month = { it.format('MM') }
def days = { year -> (date("${year}-01-01")..<date("${year+1}-01-01")) }
def weekDays = { dayOfWeek, year -> days(year).findAll { Day.valueOf(it) == dayOfWeek } }
 
def lastWeekDays = { dayOfWeek, year ->
weekDays(dayOfWeek, year).reverse().inject([:]) { months, sunday ->
def monthStr = month(sunday)
!months[monthStr] ? months + [(monthStr):sunday]  : months
}.values().sort()
}

Test:

def ymd = { it.format('yyyy-MM-dd') }
def lastSundays = lastWeekDays.curry(Day.Sun)
lastSundays(args[0] as int).each { println (ymd(it)) }

Execution (Cygwin on Windows 7):

[2201] groovy lastSundays.groovy 2013
Output:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Icon and Unicon[edit]

This is a trivial adaptation of the solution to the "Last Friday of each month" task and works in both languages:

procedure main(A)
every write(lastsundays(!A))
end
 
procedure lastsundays(year)
every m := 1 to 12 do {
d := case m of {
2 : if IsLeapYear(year) then 29 else 28
4|6|9|11 : 30
default : 31
} # last day of month
 
z := 0
j := julian(m,d,year) + 1 # first day of next month
until (j-:=1)%7 = 6 do z -:=1 # backup to last sunday (6)
suspend sprintf("%d-%d-%d",year,m,d+z)
}
end
 
link datetime, printf

Sample run:

->lsem 2013 2014
2013-1-27
2013-2-24
2013-3-31
2013-4-28
2013-5-26
2013-6-30
2013-7-28
2013-8-25
2013-9-29
2013-10-27
2013-11-24
2013-12-29
2014-1-26
2014-2-23
2014-3-30
2014-4-27
2014-5-25
2014-6-29
2014-7-27
2014-8-31
2014-9-28
2014-10-26
2014-11-30
2014-12-28
->

J[edit]

Same solution as for Last_Friday_of_each_month#J

require'dates'
last_sundays=: 12 {. [: ({:/.~ }:"1)@(#~ 0 = weekday)@todate (i.366) + todayno@,&1 1

Example use:

   last_sundays 2013
2013 1 27
2013 2 24
2013 3 31
2013 4 28
2013 5 26
2013 6 30
2013 7 28
2013 8 25
2013 9 29
2013 10 27
2013 11 24
2013 12 29

Java[edit]

import java.util.Scanner;
 
public class LastSunday
{
static final String[] months={"January","February","March","April","May","June","July","August","September","October","November","December"};
 
public static int[] findLastSunday(int year)
{
boolean isLeap = isLeapYear(year);
 
int[] days={31,isLeap?29:28,31,30,31,30,31,31,30,31,30,31};
int[] lastDay=new int[12];
 
for(int m=0;i<12;i++)
{
int d;
for(d=days[m]; getWeekDay(year,m,d)!=0; d--)
;
lastDay[m]=d;
}
 
return lastDay;
}
 
private static boolean isLeapYear(int year)
{
if(year%4==0)
{
if(year%100!=0)
return true;
else if (year%400==0)
return true;
}
return false;
}
 
private static int getWeekDay(int y, int m, int d)
{
int f=y+d+3*m-1;
m++;
 
if(m<3)
y--;
else
f-=(int)(0.4*m+2.3);
 
f+=(int)(y/4)-(int)((y/100+1)*0.75);
f%=7;
 
return f;
}
 
private static void display(int year, int[] lastDay)
{
System.out.println("\nYEAR: "+year);
for(int m=0;i<12;i++)
System.out.println(months[m]+": "+lastDay[m]);
}
 
public static void main(String[] args) throws Exception
{
System.out.print("Enter year: ");
Scanner s=new Scanner(System.in);
 
int y=Integer.parseInt(s.next());
 
int[] lastDay = findLastSunday(y);
display(y, lastDay);
 
s.close();
}
}
Output:
Enter year: 2013

YEAR: 2013
January: 27
February: 24
March: 31
April: 28
May: 26
June: 30
July: 28
August: 25
September: 29
October: 27
November: 24
December: 29

Java 8[edit]

import java.time.*;
import java.util.stream.*;
import static java.time.temporal.TemporalAdjusters.*;
 
public class FindTheLastSundayOfEachMonth {
public static Stream<LocalDate> lastSundaysOf(int year) {
return IntStream.rangeClosed(1, 12).mapToObj(month ->
LocalDate.of(year, month, 1).with(lastDayOfMonth())
.with(previousOrSame(DayOfWeek.SUNDAY))
);
}
 
public static java.util.List<LocalDate> listLastSundaysOf(int year) {
return lastSundaysOf(year).collect(Collectors.toList());
}
 
public static void main(String[] args) throws Exception {
int year = args.length > 0 ? Integer.parseInt(args[0]) : LocalDate.now().getYear();
 
for (LocalDate d : listLastSundaysOf(year)) {
System.out.println(d);
};
 
String result = lastSundaysOf(2013).map(LocalDate::toString).collect(Collectors.joining("\n"));
String test = "2013-01-27\n2013-02-24\n2013-03-31\n2013-04-28\n2013-05-26\n2013-06-30\n2013-07-28\n2013-08-25\n2013-09-29\n2013-10-27\n2013-11-24\n2013-12-29";
if (!test.equals(result)) throw new AssertionError("test failure");
}
 
}

JavaScript[edit]

function lastSundayOfEachMonths(year) {
var lastDay = [31,28,31,30,31,30,31,31,30,31,30,31]
if (year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)) lastDay[2] = 29
for (var date = new Date(), month=0; month<12; month+=1) {
date.setFullYear(year, month, lastDay[month])
date.setDate(date.getDate()-date.getDay())
document.write(date.toISOString().substring(0,10), '<br>')
}
}
 
lastSundayOfEachMonths(2013)
Output:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

jq[edit]

Works with: jq version 1.4

Foundations

# In case your jq does not have "until" defined:
def until(cond; next):
def _until:
if cond then . else (next|_until) end;
_until;
 
# Zeller's Congruence from [[Day_of_the_week#jq]]
 
# Use Zeller's Congruence to determine the day of the week, given
# year, month and day as integers in the conventional way.
# If iso == "iso" or "ISO", then emit an integer in 1 -- 7 where
# 1 represents Monday, 2 Tuesday, etc;
# otherwise emit 0 for Saturday, 1 for Sunday, etc.
#
def day_of_week(year; month; day; iso):
if month == 1 or month == 2 then
[year - 1, month + 12, day]
else
[year, month, day]
end
| .[2] + (13*(.[1] + 1)/5|floor)
+ (.[0]%100) + ((.[0]%100)/4|floor)
+ (.[0]/400|floor) - 2*(.[0]/100|floor)
| if iso == "iso" or iso == "ISO" then 1 + ((. + 5) % 7)
else . % 7
end ;

findLastSundays

# year and month are numbered conventionally 
def findLastSunday(year; month):
def isLeapYear:
year%4 == 0 and ( year%100!=0 or year%400==0 ) ;
def days:
if month == 2 then (if isLeapYear then 29 else 28 end)
else [31, 28, 31,30,31,30,31,31,30,31,30,31][month-1]
end;
year as $year
| month as $month
| days
| until( day_of_week($year; $month; .; null) == 1 ; .-1);
 
# input: year
def findLastSundays:
def months:
["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"];
. as $year
| "YEAR: \(.)",
(range(0;12) | "\(months[.]) \(findLastSunday($year; .+1))") ;
 
$year|tonumber|findLastSundays
 

Example:

$ jq --arg year 2013 -n -r -f findLastSundays.jq
YEAR: 2013
January 27
February 24
March 31
April 28
May 26
June 30
July 28
August 25
September 29
October 27
November 24
December 29

Julia[edit]

 
isdefined(:Date) || using Dates
 
const wday = Dates.Sun
const lo = 1
const hi = 12
 
print("\nThis script will print the last ", Dates.dayname(wday))
println("s of each month of the year given.")
println("(Leave input empty to quit.)")
 
while true
print("\nYear> ")
y = chomp(readline())
0 < length(y) || break
y = try
parseint(y)
catch
println("Sorry, but that does not compute as a year.")
continue
end
println()
for m in Date(y, lo):Month(1):Date(y, hi)
println(" ", tolast(m, wday))
end
end
 

This code uses the Dates module, which is being incorporated into Julian's standard library with the current development version (0.4). I've used isdefined to make this code good for the current stable version (0.3) as well as for future releases. If Dates is not installed on your instance of Julian try Pkg.add("Dates") from the REPL.

Output:
This script will print the last Sundays of each month of the year given.
(Leave input empty to quit.)

Year> 2013

    2013-01-27
    2013-02-24
    2013-03-31
    2013-04-28
    2013-05-26
    2013-06-30
    2013-07-28
    2013-08-25
    2013-09-29
    2013-10-27
    2013-11-24
    2013-12-29

Year> this year
Sorry, but that does not compute as a year.

Year> 

Lasso[edit]

local(
year = integer(web_request -> param('year') || 2013),
date = date(#year + '-1-1'),
lastsu = array,
lastday
)
 
with month in generateseries(1,12) do {
#date -> day = 1
#date -> month = #month
#lastday = #date -> month(-days)
#date -> day = #lastday
loop(7) => {
if(#date -> dayofweek == 1) => {
#lastsu -> insert(#date -> format(`dd MMMM`))
loop_abort
}
#date -> day--
}
}
#lastsu -> join('<br />')
27 January
24 February
31 March
28 April
26 May
30 June
28 July
25 August
29 September
27 October
24 November
29 December

LiveCode[edit]

Abstracted version of "last friday of each month". LiveCode dateItem item 7 is day of week. It is numbered 1 (Sunday) to 7 (Saturday).

function lastDay yyyy, dayofweek
-- year,month num,day of month,hour in 24-hour time,minute,second,numeric day of week.
convert the long date to dateitems
put 1 into item 2 of it
put 1 into item 3 of it
put yyyy into item 1 of it
put it into startDate
convert startDate to dateItems
repeat with m = 1 to 12
put m into item 2 of startDate
repeat with d = 20 to 31
put d into item 3 of startDate
convert startDate to dateItems
-- 1 is Sunday through to 7 Saturday
if item 7 of startDate is dayofweek and item 1 of startDate is yyyy and item 2 of startDate is m then
put item 3 of startDate into mydays[item 2 of startDate]
end if
end repeat
end repeat
combine mydays using cr and space
sort mydays ascending numeric
return mydays
end lastDay
Example
put lastDay(2013, 1)
Output
1 27

2 24 3 31 4 28 5 26 6 30 7 28 8 25 9 29 10 27 11 24

12 29

Mathematica / Wolfram Language[edit]

LastSundays[year_] := 
Table[Last@
DayRange[{year, i},
DatePlus[{year, i}, {{1, "Month"}, {-1, "Day"}}], Sunday], {i,
12}]
LastSundays[2013]
Output:
{{2013, 1, 27}, {2013, 2, 24}, {2013, 3, 31}, {2013, 4, 28}, {2013, 5,
   26}, {2013, 6, 30}, {2013, 7, 28}, {2013, 8, 25}, {2013, 9, 
  29}, {2013, 10, 27}, {2013, 11, 24}, {2013, 12, 29}}

Nim[edit]

import times, os, strutils
 
var timeinfo = getLocalTime getTime()
timeinfo.year = paramStr(1).parseInt
for month in mJan .. mDec:
timeinfo.month = month
for day in countdown(31, 1):
timeinfo.monthday = day
let t = getLocalTime(timeInfoToTime timeinfo)
if t.month == month and t.weekday == dSun:
echo t.format "yyyy-MM-dd"
break

Sample usage:

./lastsunday 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

OCaml[edit]

 
let is_leap_year y =
(* See OCaml solution on Rosetta Code for
determing if it's a leap year *)

if (y mod 100) = 0
then (y mod 400) = 0
else (y mod 4) = 0;;
 
let get_days y =
if is_leap_year y
then
[31;29;31;30;31;30;31;31;30;31;30;31]
else
[31;28;31;30;31;30;31;31;30;31;30;31];;
 
let print_date = Printf.printf "%d/%d/%d\n";;
 
let get_day_of_week y m d =
let y = if m > 2 then y else y - 1 in
let c = y / 100 in
let y = y mod 100 in
let m_shifted = float_of_int ( ((m + 9) mod 12) + 1) in
let m_factor = int_of_float (2.6 *. m_shifted -. 0.2) in
let leap_factor = 5 * (y mod 4) + 3 * (y mod 7) + 5 * (c mod 4) in
(d + m_factor + leap_factor) mod 7;;
 
let get_shift y m last_day =
get_day_of_week y m last_day;;
 
let print_last_sunday y m =
let days = get_days y in
let last_day = List.nth days (m - 1) in
let last_sunday = last_day - (get_shift y m last_day) in
print_date y m last_sunday;;
 
let print_last_sundays y =
let months = [1;2;3;4;5;6;7;8;9;10;11;12] in
List.iter (print_last_sunday y) months;;
 
match (Array.length Sys.argv ) with
2 -> print_last_sundays( int_of_string (Sys.argv.(1)));
|_ -> invalid_arg "Please enter a year";
 

Sample usage:

 ocaml sundays.ml  2013
2013/01/27
2013/02/24
2013/03/31
2013/04/28
2013/05/26
2013/06/30
2013/07/28
2013/08/25
2013/09/29
2013/10/27
2013/11/24
2013/12/29

Oforth[edit]

import: date
 
: lastSunday(y)
| m |
Date.JANUARY Date.DECEMBER for: m [
Date newDate(y, m, Date.DaysInMonth(y, m))
while(dup dayOfWeek Date.SUNDAY <>) [ addDays(-1) ] println
] ;
Output:
2013-01-27 00:00:00,000
2013-02-24 00:00:00,000
2013-03-31 00:00:00,000
2013-04-28 00:00:00,000
2013-05-26 00:00:00,000
2013-06-30 00:00:00,000
2013-07-28 00:00:00,000
2013-08-25 00:00:00,000
2013-09-29 00:00:00,000
2013-10-27 00:00:00,000
2013-11-24 00:00:00,000
2013-12-29 00:00:00,000

Perl[edit]

#!/usr/bin/perl
use strict ;
use warnings ;
use DateTime ;
 
for my $i( 1..12 ) {
my $date = DateTime->last_day_of_month( year => $ARGV[ 0 ] ,
month => $i ) ;
while ( $date->dow != 7 ) {
$date = $date->subtract( days => 1 ) ;
}
my $ymd = $date->ymd ;
print "$ymd\n" ;
}
Output:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Perl 6[edit]

Works with: rakudo version 2015-09-10
sub MAIN ($year = Date.new.year) {
for 1..12 -> $mo {
my $month-end = Date.new($year, $mo, Date.days-in-month($year, $mo));
say $month-end - $month-end.day-of-week % 7;
}
}
Output:
2015-01-25
2015-02-22
2015-03-29
2015-04-26
2015-05-31
2015-06-28
2015-07-26
2015-08-30
2015-09-27
2015-10-25
2015-11-29
2015-12-27

PHP[edit]

<?php
// Created with PHP 7.0
 
function printLastSundayOfAllMonth($year)
{
$months = array(
'January', 'February', 'March', 'April', 'June', 'July',
'August', 'September', 'October', 'November', 'December');
 
foreach ($months as $month) {
echo $month . ': ' . date('Y-m-d', strtotime('last sunday of ' . $month . ' ' . $year)) . "\n";
}
}
 
printLastSundayOfAllMonth($argv[1]);
 
Output:
January: 2013-01-27
February: 2013-02-24
March: 2013-03-31
April: 2013-04-28
June: 2013-06-30
July: 2013-07-28
August: 2013-08-25
September: 2013-09-29
October: 2013-10-27
November: 2013-11-24
December: 2013-12-29

PicoLisp[edit]

(de lastSundays (Y)
(for M 12
(prinl
(dat$
(find '((D) (= "Sunday" (day D)))
(mapcar '((D) (date Y M D)) `(range 31 22)) )
"-" ) ) ) )

Test:

: (lastSundays 2013)
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

PowerShell[edit]

 
function last-dayofweek {
param(
[Int][ValidatePattern("[1-9][0-9][0-9][0-9]")]$year,
[String][validateset('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday')]$dayofweek
)
$date = (Get-Date -Year $year -Month 1 -Day 1)
while($date.DayOfWeek -ne $dayofweek) {$date = $date.AddDays(1)}
while($date.year -eq $year) {
if($date.Month -ne $date.AddDays(7).Month) {$date.ToString("yyyy-dd-MM")}
$date = $date.AddDays(7)
}
}
last-dayofweek 2013 "Sunday"
 

Output:

2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Python[edit]

 
import sys
import calendar
 
year = 2013
if len(sys.argv) > 1:
try:
year = int(sys.argv[-1])
except ValueError:
pass
 
for month in range(1, 13):
last_sunday = max(week[-1] for week in calendar.monthcalendar(year, month))
print('{}-{}-{:2}'.format(year, calendar.month_abbr[month], last_sunday))
 

Output:

 
2013-Jan-27
2013-Feb-24
2013-Mar-31
2013-Apr-28
2013-May-26
2013-Jun-30
2013-Jul-28
2013-Aug-25
2013-Sep-29
2013-Oct-27
2013-Nov-24
2013-Dec-29
 

Racket[edit]

 
#lang racket
(require srfi/19 math)
 
(define (days-in-month m y)
(define lengths #(0 31 #f 31 30 31 30 31 31 30 31 30 31))
(define d (vector-ref lengths m))
(or d (days-in-feb y)))
 
(define (leap-year? y)
(and (divides? 4 y)
(or (not (divides? 100 y))
(divides? 400 y))))
 
(define (days-in-feb y)
(if (leap-year? y) 29 28))
 
(define (last-day-in-month m y)
(make-date 0 0 0 0 (days-in-month m y) m y 0))
 
(define (week-day date)
(define days #(sun mon tue wed thu fri sat))
(vector-ref days (date-week-day date)))
 
(define (last-sundays y)
(for/list ([m (in-range 1 13)])
(prev-sunday (last-day-in-month m y))))
 
(define 24hours (make-time time-duration 0 (* 24 60 60)))
 
(define (prev-day d)
(time-utc->date
(subtract-duration
(date->time-utc d) 24hours)))
 
(define (prev-sunday d)
(if (eq? (week-day d) 'sun)
d
(prev-sunday (prev-day d))))
 
(for ([d (last-sundays 2013)])
(displayln (~a (date->string d "~a ~d ~b ~Y"))))
 
Output:
Sun 27 Jan 2013
Sun 24 Feb 2013
Sun 31 Mar 2013
Sun 28 Apr 2013
Sun 26 May 2013
Sun 30 Jun 2013
Sun 28 Jul 2013
Sun 25 Aug 2013
Sun 29 Sep 2013
Sun 27 Oct 2013
Sun 24 Nov 2013
Sun 29 Dec 2013

REBOL[edit]

#!/usr/bin/env rebol
 
last-sundays-of-year: function [
"Return series of last sunday (date!) for each month of the year"
year [integer!] "which year?"
][
d: to-date reduce [1 1 year] ; start with first day of year
collect [
repeat month 12 [
d/month: month + 1 ; move to start of next month
keep d - d/weekday ; calculate last sunday & keep
]
]
]
 
foreach sunday last-sundays-of-year to-integer system/script/args [print sunday]
 
Output:
./last-sundays.reb 2013
27-Jan-2013
24-Feb-2013
31-Mar-2013
28-Apr-2013
26-May-2013
30-Jun-2013
28-Jul-2013
25-Aug-2013
29-Sep-2013
27-Oct-2013
24-Nov-2013
29-Dec-2013

REXX[edit]

This REXX example is an exact replication of the Rosetta Code

  • find last Fridays of each month for any year

except for the innards of the first DO loop.

The   lastDOW   subroutine can be used for any day-of-the-week for any month for any year.

/*REXX program displays dates of last Sundays of each month for any year*/
parse arg yyyy
do j=1 for 12
_ = lastDOW('Sunday', j, yyyy)
say right(_,4)'-'right(j,2,0)"-"left(word(_,2),2)
end /*j*/
exit /*stick a fork in it, we're done.*/
/*┌────────────────────────────────────────────────────────────────────┐
│ lastDOW: procedure to return the date of the last day-of-week of │
│ any particular month of any particular year. │
│ │
│ The day-of-week must be specified (it can be in any case, │
│ (lower-/mixed-/upper-case) as an English name of the spelled day │
│ of the week, with a minimum length that causes no ambiguity. │
│ I.E.: W for Wednesday, Sa for Saturday, Su for Sunday ... │
│ │
│ The month can be specified as an integer 1 ──► 12 │
│ 1=January 2=February 3=March ... 12=December │
│ or the English name of the month, with a minimum length that │
│ causes no ambiguity. I.E.: Jun for June, D for December. │
│ If omitted [or an asterisk(*)], the current month is used. │
│ │
│ The year is specified as an integer or just the last two digits │
│ (two digit years are assumed to be in the current century, and │
│ there is no windowing for a two-digit year). │
│ If omitted [or an asterisk(*)], the current year is used. │
│ Years < 100 must be specified with (at least 2) leading zeroes.│
│ │
│ Method used: find the "day number" of the 1st of the next month, │
│ then subtract one (this gives the "day number" of the last day of │
│ the month, bypassing the leapday mess). The last day-of-week is │
│ then obtained straightforwardly, or via subtraction. │
└────────────────────────────────────────────────────────────────────┘*/

lastdow: procedure; arg dow .,mm .,yy . /*DOW = day of week*/
parse arg a.1,a.2,a.3 /*orig args, errmsg*/
if mm=='' | mm=='*' then mm=left(date('U'),2) /*use default month*/
if yy=='' | yy=='*' then yy=left(date('S'),4) /*use default year */
if length(yy)==2 then yy=left(date('S'),2)yy /*append century. */
/*Note mandatory leading blank in strings below.*/
$=" Monday TUesday Wednesday THursday Friday SAturday SUnday"
!=" JAnuary February MARch APril MAY JUNe JULy AUgust September",
" October November December"
upper $ ! /*uppercase strings*/
if dow=='' then call .er "wasn't specified",1
if arg()>3 then call .er 'arguments specified',4
 
do j=1 for 3 /*any plural args ?*/
if words(arg(j))>1 then call .er 'is illegal:',j
end
 
dw=pos(' 'dow,$) /*find day-of-week*/
if dw==0 then call .er 'is invalid:',1
if dw\==lastpos(' 'dow,$) then call .er 'is ambigious:',1
 
if datatype(mm,'month') then /*if MM is alpha...*/
do
m=pos(' 'mm,!) /*maybe its good...*/
if m==0 then call .er 'is invalid:',1
if m\==lastpos(' 'mm,!) then call .er 'is ambigious:',2
mm=wordpos(word(substr(!,m),1),!)-1 /*now, use true Mon*/
end
 
if \datatype(mm,'W') then call .er "isn't an integer:",2
if \datatype(yy,'W') then call .er "isn't an integer:",3
if mm<1 | mm>12 then call .er "isn't in range 1──►12:",2
if yy=0 then call .er "can't be 0 (zero):",3
if yy<0 then call .er "can't be negative:",3
if yy>9999 then call .er "can't be > 9999:",3
 
tdow=wordpos(word(substr($,dw),1),$)-1 /*target DOW, 0──►6*/
/*day# of last dom.*/
_=date('B',right(yy+(mm=12),4)right(mm//12+1,2,0)"01",'S')-1
?=_//7 /*calc. DOW, 0──►6*/
if ?\==tdow then _=_-?-7+tdow+7*(?>tdow) /*not DOW? Adjust.*/
return date('weekday',_,"B") date(,_,'B') /*return the answer*/
 
.er: arg ,_;say; say '***error!*** (in LASTDOW)';say /*tell error, and */
say word('day-of-week month year excess',arg(2)) arg(1) a._
say; exit 13 /*... then exit. */
Output:
when using the default input (the current year, 2013):
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Ring[edit]

 
see "What year to calculate (yyyy) : "
give year
see "Last Sundays in " + year + " are on :" + nl
month = list(12)
mo = [4,0,0,3,5,1,3,6,2,4,0,2]
mon = [31,28,31,30,31,30,31,31,30,31,30,31]
if year < 2100 leap = year - 1900 else leap = year - 1904 ok
m = ((year-1900)%7) + floor(leap/4) % 7
for n = 1 to 12
month[n] = (mo[n] + m) % 7
next
for n = 1 to 12
for i = (mon[n] - 6) to mon[n]
x = (month[n] + i) % 7
if n < 10 strn = "0" + string(n) else strn = string(n) ok
if x = 4 see year + "-" + strn + "-" + string(i) + nl ok
next
next
 

Output:

What year to calculate (yyyy) : 2013
Last Sundays in 2013 are on :
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Ruby[edit]

require 'date'
 
def last_sundays_of_year(year = Date.today.year)
(1..12).map do |month|
d = Date.new(year, month, -1) # -1 means "last".
d - d.wday
end
end
 
puts last_sundays_of_year(2013)
Output:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Results before the year 1581 may differ from other languages - the Date library takes the Julian reform into account.

Scala[edit]

object FindTheLastSundayOfEachMonth extends App {
import java.util.Calendar._
val cal = getInstance
 
def lastSundaysOf(year: Int) =
(JANUARY to DECEMBER).map{month =>
cal.set(year, month + 1, 1) // first day of next month
(1 to 7).find{_ => cal.add(DAY_OF_MONTH, -1); cal.get(DAY_OF_WEEK) == SUNDAY}
cal.getTime
}
 
val year = args.headOption.map(_.toInt).getOrElse(cal.get(YEAR))
val fmt = new java.text.SimpleDateFormat("yyyy-MM-dd")
println(lastSundaysOf(year).map(fmt.format) mkString "\n")
}

Java 8[edit]

object FindTheLastSundayOfEachMonth extends App {
def lastSundaysOf(year: Int) = (1 to 12).map{month =>
import java.time._; import java.time.temporal.TemporalAdjusters._
LocalDate.of(year, month, 1).`with`(lastDayOfMonth).`with`(previousOrSame(DayOfWeek.SUNDAY))}
 
val year = args.headOption.map(_.toInt).getOrElse(java.time.LocalDate.now.getYear)
println(lastSundaysOf(year) mkString "\n")
}

Seed7[edit]

Uses the libraries time.s7i and duration.s7i. Applicable to any day of the week, cf. [[2]].

$ include "seed7_05.s7i";
include "time.s7i";
include "duration.s7i";
 
const proc: main is func
local
var integer: weekday is 1; # 1 for monday, 2 for tuesday, and so on up to 7 for sunday.
var integer: year is 0;
var integer: month is 1;
var time: aDate is time.value;
var time: selected is time.value;
begin
if length(argv(PROGRAM)) <> 2 then
writeln("usage: lastWeekdayInMonth weekday year");
writeln(" weekday: 1 for monday, 2 for tuesday, and so on up to 7 for sunday.");
else
weekday := integer parse (argv(PROGRAM)[1]);
year := integer parse (argv(PROGRAM)[2]);
for month range 1 to 12 do
aDate := date(year, month, 1);
while aDate.month = month do
if dayOfWeek(aDate) = weekday then
selected := aDate;
end if;
aDate +:= 1 . DAYS;
end while;
writeln(strDate(selected));
end for;
end if;
end func;
Output:
when called with s7 rosetta/lastWeekdayInMonth 7 2013:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Sidef[edit]

var dt = require('DateTime');
var (year=2015) = @ARGV.map{.to_i};
 
for i in (1 .. 12) {
var date = dt.last_day_of_month(
year => year,
month => i
);
 
while (date.dow != 7) {
date = date.subtract(days => 1);
}
 
say date.ymd;
}
Output:
$ sidef last_sunday.sf 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29

Tcl[edit]

proc lastSundays {{year ""}} {
if {$year eq ""} {
set year [clock format [clock seconds] -gmt 1 -format "%Y"]
}
foreach month {2 3 4 5 6 7 8 9 10 11 12 13} {
set d [clock add [clock scan "$month/1/$year" -gmt 1] -1 day]
while {[clock format $d -gmt 1 -format "%u"] != 7} {
set d [clock add $d -1 day]
}
lappend result [clock format $d -gmt 1 -format "%Y-%m-%d"]
}
return $result
}
puts [join [lastSundays {*}$argv] "\n"]
Output:

When called as: “tclsh lastSundays.tcl 2013” (or with the year argument omitted during 2013)

2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-20
2013-11-24
2013-12-29

VBScript[edit]

Works with: Windows Script Host version *
 
strYear = WScript.StdIn.ReadLine
 
For i = 1 To 12
d = DateSerial(strYear, i + 1, 1) - 1
WScript.Echo d - Weekday(d) + 1
Next
 

zkl[edit]

The program from [[3]] solves this task, as well.

Output:
var [const] D=Time.Date;
lastDay(2013,D.Sunday)
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29