# Find the last Sunday of each month

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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```

Related tasks

## 11l

```F last_sundays(year)
[String] sundays
L(month) 1..12
V last_day_of_month = I month < 12 {Time(year, month + 1)} E Time(year + 1)
L
last_day_of_month -= TimeDelta(days' 1)
I last_day_of_month.strftime(‘%w’) == ‘0’
sundays [+]= year‘-’(‘#02’.format(month))‘-’last_day_of_month.strftime(‘%d’)
L.break
R sundays

print(last_sundays(2013).join("\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
```

## 360 Assembly

The program uses one ASSIST macro (XPRNT) to keep the code as short as possible.

```*        Last Sunday of each month 31/01/2017
LASTSUND CSECT
USING  LASTSUND,R13       base register
B      72(R15)            skip savearea
DC     17F'0'             savearea
STM    R14,R12,12(R13)    prolog
ST     R13,4(R15)         " <-
ST     R15,8(R13)         " ->
LR     R13,R15            " addressability
L      R4,YEAR            year
SRDA   R4,32              .
D      R4,=F'400'         year/400
LTR    R4,R4              if year//400=0
BZ     LEAP
L      R4,YEAR            year
SRDA   R4,32              .
D      R4,=F'4'           year/4
LTR    R4,R4              if year//4=0
BNZ    NOTLEAP
L      R4,YEAR            year
SRDA   R4,32              .
D      R4,=F'100'         year/400
LTR    R4,R4              if year//100=0
BZ     NOTLEAP
LEAP     MVC    DAYS+4(4),=F'29'   days(2)=29
NOTLEAP  L      R8,YEAR            year
BCTR   R8,0               y=year-1
LA     R7,42              42
AR     R7,R8              +y
LR     R3,R8              y
SRA    R3,2               y/4
AR     R7,R3              +y/4
LR     R4,R8              y
SRDA   R4,32              .
D      R4,=F'100'         y/100
LA     R4,0               .
M      R4,=F'6'           *6
AR     R7,R5              +6*(y/100)
LR     R4,R8              y
SRDA   R4,32              .
D      R4,=F'400'         y/100
AR     R7,R5              k=42+y+y/4+6*(y/100)+y/400
LA     R6,1               m=1
LOOPM    C      R6,=F'12'          do m=1 to 12
BH     ELOOPM
LR     R1,R6              m
SLA    R1,2               .
L      R2,DAYS-4(R1)      days(m)
AR     R7,R2              k=k+days(m)
LR     R4,R7              k
SRDA   R4,32              .
D      R4,=F'7'           k/7
SR     R2,R4              days(m)-k//7
LR     R9,R2              d=days(m)-k//7
L      R1,YEAR            year
CVD    R1,DW              year: binary to packed
OI     DW+7,X'0F'           zap sign
UNPK   PG(4),DW             unpack (ZL4)
CVD    R6,DW              m : binary to packed
OI     DW+7,X'0F'           zap sign
UNPK   PG+5(2),DW           unpack (ZL2)
CVD    R9,DW              d: binary to packed
OI     DW+7,X'0F'           zap sign
UNPK   PG+8(2),DW           unpack (ZL2)
XPRNT  PG,L'PG            print buffer
LA     R6,1(R6)           m=m+1
B      LOOPM
ELOOPM   L      R13,4(0,R13)       epilog
LM     R14,R12,12(R13)    " restore
XR     R15,R15            " rc=0
BR     R14                exit
YEAR     DC     F'2013'            <== input year
DAYS     DC     F'31',F'28',F'31',F'30',F'31',F'30'
DC     F'31',F'31',F'30',F'31',F'30',F'31'
PG       DC     CL80'YYYY-MM-DD'   buffer
XDEC     DS     CL12               temp
DW       DS     D                  packed (PL8) 15num
YREGS
END    LASTSUND```
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
```

## Action!

Day of the week is determined using Sakamoto method.

```;https://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week#Sakamoto.27s_methods
BYTE FUNC DayOfWeek(INT y BYTE m,d)	;1<=m<=12, y>1752
BYTE ARRAY t=[0 3 2 5 0 3 5 1 4 6 2 4]
BYTE res

IF m<3 THEN
y==-1
FI
res=(y+y/4-y/100+y/400+t(m-1)+d) MOD 7
RETURN (res)

BYTE FUNC IsLeapYear(INT y)
IF y MOD 100=0 THEN
IF y MOD 400=0 THEN
RETURN (1)
ELSE
RETURN (0)
FI
FI

IF y MOD 4=0 THEN
RETURN (1)
FI
RETURN (0)

INT FUNC GetMaxDay(INT y BYTE m)
BYTE ARRAY MaxDay=[31 28 31 30 31 30 31 31 30 31 30 31]

IF m=2 AND IsLeapYear(y)=1 THEN
RETURN (29)
FI
RETURN (MaxDay(m-1))

PROC PrintB2(BYTE x)
IF x<10 THEN
Put('0)
FI
PrintB(x)
RETURN

PROC Main()
INT MinYear=[1753],MaxYear=[9999],y
BYTE m,d,last,maxD

DO
PrintF("Input year in range %I...%I: ",MinYear,MaxYear)
y=InputI()
UNTIL y>=MinYear AND y<=MaxYear
OD

FOR m=1 TO 12
DO
last=0
maxD=GetMaxDay(y,m)
FOR d=1 TO maxD
DO
IF DayOfWeek(y,m,d)=0 THEN
last=d
FI
OD
PrintI(y) Put('-)
PrintB2(m) Put('-)
PrintB2(last) PutE()
OD
RETURN```
Output:
```Input year in range 1753...9999: 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
```

## Ada

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```

## ALGOL 68

Translation of: ALGOL W
```BEGIN # find the last Sunday in each month of a year             #
# returns true if year is a leap year, false otherwise       #
# assumes year is in the Gregorian Calendar                  #
PROC is leap year = ( INT year )BOOL:
year MOD 400 = 0 OR ( year MOD 4 = 0 AND year MOD 100 /= 0 );
# returns the day of the week of the specified date (d/m/y)  #
#         Sunday = 1                                         #
PROC day of week = ( INT d, m, y )INT:
BEGIN
INT mm := m;
INT yy := y;
IF mm <= 2 THEN
mm := mm + 12;
yy := yy - 1
FI;
INT j = yy OVER 100;
INT k = yy MOD  100;
(d + ( ( mm + 1 ) * 26 ) OVER 10 + k + k OVER 4 + j OVER 4 + 5 * j ) MOD 7
END # day of week # ;
# returns an array of the last Sunday of each month in year  #
PROC last sundays = ( INT year )[]INT:
BEGIN
[ 1 : 12 ]INT last days := ( 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 );
IF is leap year( year ) THEN last days[ 2 ] := 29 FI;
# for each month, determine the day number of the    #
# last Sunday                                        #
[ 1 : 12 ]INT last;
FOR m pos TO 12 DO
INT dow := day of week( last days[ m pos ], m pos, year );
IF dow = 0 # Saturday # THEN dow := 7 FI;
# calculate the offset for the previous Sunday   #
last[ m pos ] := ( last days[ m pos ] + 1 ) - dow
OD;
last
END # last sundays # ;
# test the last sundays procedure                            #
INT   year = 2021;
[]INT last = last sundays( year );
FOR m pos TO 12 DO
print( ( whole( year, 0 )
, IF m pos < 10 THEN "-0" ELSE "-1" FI
, whole( m pos MOD 10, 0 )
, "-"
, whole( last[ m pos ], 0 )
, newline
)
)
OD
END```
Output:
```2021-01-31
2021-02-28
2021-03-28
2021-04-25
2021-05-30
2021-06-27
2021-07-25
2021-08-29
2021-09-26
2021-10-31
2021-11-28
2021-12-26
```

## ALGOL-M

```BEGIN

% CALCULATE P MOD Q %
INTEGER FUNCTION MOD(P, Q);
INTEGER P, Q;
BEGIN
MOD := P - Q * (P / Q);
END;

COMMENT
RETURN DAY OF WEEK (SUN=0, MON=1, ETC.) FOR A GIVEN
GREGORIAN CALENDAR DATE USING ZELLER'S CONGRUENCE;
INTEGER FUNCTION DAYOFWEEK(MO, DA, YR);
INTEGER MO, DA, YR;
BEGIN
INTEGER Y, C, Z;
IF MO < 3 THEN
BEGIN
MO := MO + 10;
YR := YR - 1;
END
ELSE MO := MO - 2;
Y := MOD(YR, 100);
C := YR / 100;
Z := (26 * MO - 2) / 10;
Z := Z + DA + Y + (Y / 4) + (C / 4) - 2 * C + 777;
DAYOFWEEK := MOD(Z, 7);
END;

% RETURN 1 IF Y IS A LEAP YEAR, OTHERWISE 0 %
INTEGER FUNCTION ISLEAPYR(Y);
INTEGER Y;
BEGIN
IF MOD(Y,4) <> 0 THEN  % QUICK EXIT IN MOST COMMON CASE %
ISLEAPYR := 0
ELSE IF MOD(Y,400) = 0 THEN
ISLEAPYR := 1
ELSE IF MOD(Y,100) = 0 THEN
ISLEAPYR := 0
ELSE                   % NON-CENTURY AND DIVISIBLE BY 4 %
ISLEAPYR := 1;
END;

% RETURN THE NUMBER OF DAYS IN THE SPECIFIED MONTH %
INTEGER FUNCTION MONTHDAYS(MO, YR);
INTEGER MO, YR;
BEGIN
IF MO = 2 THEN
BEGIN
IF ISLEAPYR(YR) = 1 THEN
MONTHDAYS := 29
ELSE
MONTHDAYS := 28;
END
ELSE IF (MO = 4) OR (MO = 6) OR (MO = 9) OR (MO = 11) THEN
MONTHDAYS := 30
ELSE
MONTHDAYS := 31;
END;

COMMENT
RETURN THE DAY OF THE MONTH CORRESPONDING TO LAST OCCURRENCE
OF WEEKDAY K (SUN=0, MON=1, ETC.) FOR A GIVEN MONTH AND YEAR;
INTEGER FUNCTION LASTKDAY(K, M, Y);
INTEGER K, M, Y;
BEGIN
INTEGER D, W;
% DETERMINE WEEKDAY FOR THE LAST DAY OF THE MONTH %
D := MONTHDAYS(M, Y);
W := DAYOFWEEK(M, D, Y);
% BACK UP AS NEEDED TO DESIRED WEEKDAY %
IF W >= K THEN
D := D - (W - K)
ELSE
D := D - (7 - K - W);
LASTKDAY := D;
END;

STRING FUNCTION MONTHNAME(N);
INTEGER N;
BEGIN
CASE (N - 1) OF
BEGIN
MONTHNAME := "JAN";
MONTHNAME := "FEB";
MONTHNAME := "MAR";
MONTHNAME := "APR";
MONTHNAME := "MAY";
MONTHNAME := "JUN";
MONTHNAME := "JUL";
MONTHNAME := "AUG";
MONTHNAME := "SEP";
MONTHNAME := "OCT";
MONTHNAME := "NOV";
MONTHNAME := "DEC";
END;
END;

% EXERCISE THE ROUTINE %
INTEGER M, Y, SUNDAY;
SUNDAY := 0;
WRITE("DISPLAY LAST SUNDAYS FOR WHAT YEAR?");
READ(Y);
FOR M:=1 STEP 1 UNTIL 12 DO
WRITE(MONTHNAME(M), LASTKDAY(SUNDAY,M,Y));

END```
Output:
```DISPLAY LAST SUNDAYS FOR WHAT YEAR?
-> 2021
JAN    31
FEB    28
MAR    28
APR    25
MAY    30
JUN    27
JUL    25
AUG    29
SEP    26
OCT    31
NOV    28
DEC    26
```

## ALGOL W

Uses the Day_of_week and isLeapYear procedures from the day-of-the-week and leap-year tasks - included here for convenience.

```begin % find the last Sunday in each month of a year             %
% returns true if year is a leap year, false otherwise       %
% assumes year is in the Gregorian Calendar                  %
logical procedure isLeapYear ( integer value year ) ;
year rem 400 = 0 or ( year rem 4 = 0 and year rem 100 not = 0 );
% returns the day of the week of the specified date (d/m/y)  %
%         Sunday = 1                                         %
integer procedure Day_of_week ( integer value d, m, y );
begin
integer j, k, mm, yy;
mm := m;
yy := y;
if mm <= 2 then begin
mm := mm + 12;
yy := yy - 1;
end if_m_le_2;
j := yy div 100;
k := yy rem 100;
(d + ( ( mm + 1 ) * 26 ) div 10 + k + k div 4 + j div 4 + 5 * j ) rem 7
end Day_of_week;
% sets the elements of last to the day of the last Sunday   %
% of each month in year                                     %
procedure lastSundays ( integer value year
; integer array last ( * )
) ;
begin
integer array lastDays ( 1 :: 12 );
integer m;
% set ld to the day number od the last day of each  %
% month in year                                     %
m := 1;
for ld := 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 do begin
lastDays( m ) := ld;
m             := m + 1
end for_ld ;
if isLeapYear( year ) then lastDays( 2 ) := 29;
% for each month, determine the day number of the   %
% last Sunday                                       %
for mPos := 1 until 12 do begin
integer dow;
dow := Day_of_week( lastDays( mPos ), mPos, year );
if dow = 0 % Saturday % then dow := 7;
% calculate the offset for the previous Sunday  %
last( mPos ) := ( lastDays( mPos ) + 1 ) - dow
end for_mPos
end lastSundays ;
begin
% test the lastSundays procedure                        %
integer array last ( 1 :: 12 );
integer year;
year := 2020;
lastSundays( year, last );
i_w := 1; s_w := 0; % output formatting                 %
for mPos := 1 until 12 do write( year, if mPos < 10 then "-0" else "-1", mPos rem 10, "-", last( mPos ) )
end
end.```
Output:
```2020-01-26
2020-02-23
2020-03-29
2020-04-26
2020-05-31
2020-06-28
2020-07-26
2020-08-30
2020-09-27
2020-10-25
2020-11-29
2020-12-27
```

## AppleScript

Translation of: JavaScript
```-- LAST SUNDAYS OF YEAR ------------------------------------------------------

--  lastSundaysOfYear :: Int -> [Date]
on lastSundaysOfYear(y)

-- lastWeekDaysOfYear :: Int -> Int -> [Date]
script lastWeekDaysOfYear
on |λ|(intYear, iWeekday)

-- lastWeekDay :: Int -> Int -> Date
script lastWeekDay
on |λ|(iLastDay, iMonth)
set iYear to intYear

calendarDate(iYear, iMonth, iLastDay - ¬
(((weekday of calendarDate(iYear, iMonth, iLastDay)) as integer) + ¬
(7 - (iWeekday))) mod 7)
end |λ|
end script

map(lastWeekDay, lastDaysOfMonths(intYear))
end |λ|

-- isLeapYear :: Int -> Bool
on isLeapYear(y)
(0 = y mod 4) and (0 ≠ y mod 100) or (0 = y mod 400)
end isLeapYear

-- lastDaysOfMonths :: Int -> [Int]
on lastDaysOfMonths(y)
{31, cond(isLeapYear(y), 29, 28), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
end lastDaysOfMonths
end script

lastWeekDaysOfYear's |λ|(y, Sunday as integer)
end lastSundaysOfYear

-- TEST ----------------------------------------------------------------------
on run argv

intercalate(linefeed, ¬
map(isoRow, ¬
transpose(map(lastSundaysOfYear, ¬
apply(cond(class of argv is list and argv ≠ {}, ¬
singleYearOrRange, fiveCurrentYears), argIntegers(argv))))))

end run

-- ARGUMENT HANDLING ---------------------------------------------------------

-- Up to two optional command line arguments: [yearFrom], [yearTo]
-- (Default range in absence of arguments: from two years ago, to two years ahead)

-- ~ \$ osascript ~/Desktop/lastSundays.scpt
-- ~ \$ osascript ~/Desktop/lastSundays.scpt 2013
-- ~ \$ osascript ~/Desktop/lastSundays.scpt 2013 2016

-- singleYearOrRange :: [Int] -> [Int]
on singleYearOrRange(argv)
apply(cond(length of argv > 0, my range, my fiveCurrentYears), argv)
end singleYearOrRange

-- fiveCurrentYears :: () -> [Int]
on fiveCurrentYears(_)
set intThisYear to year of (current date)
enumFromTo(intThisYear - 2, intThisYear + 2)
end fiveCurrentYears

-- argIntegers :: maybe [String] -> [Int]
on argIntegers(argv)
if class of argv is list and argv ≠ {} then
{map(my parseInt, argv)}
else
{}
end if
end argIntegers

-- GENERIC FUNCTIONS ---------------------------------------------------------

-- apply (a -> b) -> a -> b
on apply(f, a)
mReturn(f)'s |λ|(a)
end apply

-- calendarDate :: Int -> Int -> Int -> Date
on calendarDate(intYear, intMonth, intDay)
tell (current date)
set {its year, its month, its day, its time} to ¬
{intYear, intMonth, intDay, 0}
return it
end tell
end calendarDate

-- cond :: Bool -> (a -> b) -> (a -> b) -> (a -> b)
on cond(bool, f, g)
if bool then
f
else
g
end if
end cond

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo

-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate

-- isoDateString :: Date -> String
on isoDateString(dte)
(((year of dte) as string) & ¬
"-" & text items -2 thru -1 of ¬
("0" & ((month of dte) as integer) as string)) & ¬
"-" & text items -2 thru -1 of ¬
("0" & day of dte)
end isoDateString

-- isoRow :: [Date] -> String
on isoRow(lstDate)
intercalate(tab, map(my isoDateString, lstDate))
end isoRow

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn

-- parseInt :: String -> Int
on parseInt(s)
s as integer
end parseInt

-- transpose :: [[a]] -> [[a]]
on transpose(xss)
script column
on |λ|(_, iCol)
script row
on |λ|(xs)
item iCol of xs
end |λ|
end script

map(row, xss)
end |λ|
end script

map(column, item 1 of xss)
end transpose
```
Output:
```2015-01-25    2016-01-31    2017-01-29    2018-01-28    2019-01-27
2015-02-22    2016-02-28    2017-02-26    2018-02-25    2019-02-24
2015-03-29    2016-03-27    2017-03-26    2018-03-25    2019-03-31
2015-04-26    2016-04-24    2017-04-30    2018-04-29    2019-04-28
2015-05-31    2016-05-29    2017-05-28    2018-05-27    2019-05-26
2015-06-28    2016-06-26    2017-06-25    2018-06-24    2019-06-30
2015-07-26    2016-07-31    2017-07-30    2018-07-29    2019-07-28
2015-08-30    2016-08-28    2017-08-27    2018-08-26    2019-08-25
2015-09-27    2016-09-25    2017-09-24    2018-09-30    2019-09-29
2015-10-25    2016-10-30    2017-10-29    2018-10-28    2019-10-27
2015-11-29    2016-11-27    2017-11-26    2018-11-25    2019-11-24
2015-12-27    2016-12-25    2017-12-31    2018-12-30    2019-12-29```

More straightforward solution:

AppleScript's weekday constants can be coerced either to English text or to the integers 1 (for Sunday) to 7 (Saturday).

```on lastSundayOfEachMonthInYear(y)
-- Initialise an AppleScript date to the first day of some month in the specified year.
tell (current date) to set {firstDayOfNextMonth, its day, its year} to {it, 1, y}

-- Get a string representation of y, zero-padded if necessary, and initialise the output string.
set y to text 2 thru 5 of ((10000 + y) as text)
set outputText to "./last_sundays " & y
repeat with nextMonth from 2 to 13 -- Yes!
-- For each month in the year, get the first day of the following month.
set firstDayOfNextMonth's month to nextMonth
-- Calculate the date of the Sunday which occurs in the seven days prior to that
-- by subtracting a day plus the difference between the previous day's weekday and the target weekday.
set lastSundayOfThisMonth to firstDayOfNextMonth - (1 + ((firstDayOfNextMonth's weekday) + 5) mod 7) * days
-- Append the required details to the output text.
set {month:m, day:d} to lastSundayOfThisMonth
tell (10000 + m * 100 + d) as text to ¬
set outputText to outputText & (linefeed & y & "-" & text 2 thru 3 & "-" & text 4 thru 5)
end repeat

return outputText
end lastSundayOfEachMonthInYear

lastSundayOfEachMonthInYear(2020)
```
Output:
```"./last_sundays 2020
2020-01-26
2020-02-23
2020-03-29
2020-04-26
2020-05-31
2020-06-28
2020-07-26
2020-08-30
2020-09-27
2020-10-25
2020-11-29
2020-12-27"```

The above is hard-coded for Sundays. As with the "Last Friday of each month" task [Last Friday of each month], the handler can be made more flexible with a weekday constant parameter, so that it can be used for any last weekday of the months:

```on lastWeekdayWOfEachMonthInYear(w, y) -- Parameters: (AppleScript weekday constant, AD year number)
-- Initialise an AppleScript date to the first day of some month in the specified year.
tell (current date) to set {firstDayOfNextMonth, its day, its year} to {it, 1, y}

-- Get a string representation of y, zero-padded if necessary, and initialise the output string.
set y to text 2 thru 5 of ((10000 + y) as text)
set outputText to "./last_" & w & "s " & y
repeat with nextMonth from 2 to 13 -- Yes!
-- For each month in the year, get the first day of the following month.
set firstDayOfNextMonth's month to nextMonth
-- Calculate the date of the target weekday which occurs in the seven days prior to that.
-- The calculation can be described in various ways, the simplest being the subtraction of a day plus the difference between the previous day's weekday and the target weekday:
--	firstDayOfNextMonth - (1 + (((firstDayOfNextMonth's weekday) - 1) - w + 7) mod 7) * days
-- But they all boil down to:
set lastWOfThisMonth to firstDayOfNextMonth - (1 + ((firstDayOfNextMonth's weekday) - w + 6) mod 7) * days
-- Get the day and month of the calculated date and append the required details to the output text.
set {month:m, day:d} to lastWOfThisMonth
tell (10000 + m * 100 + d) as text to ¬
set outputText to outputText & (linefeed & y & "-" & text 2 thru 3 & "-" & text 4 thru 5)
end repeat

return outputText
end lastWeekdayWOfEachMonthInYear

lastWeekdayWOfEachMonthInYear(Sunday, 2020)
```
Output:
```"./last_Sundays 2020
2020-01-26
2020-02-23
2020-03-29
2020-04-26
2020-05-31
2020-06-28
2020-07-26
2020-08-30
2020-09-27
2020-10-25
2020-11-29
2020-12-27"```

## Arturo

```lastSundayForMonth: function [m,y][
ensure -> in? m 1..12

daysOfMonth: @[0 31 (leap? y)? -> 28 -> 27 31 30 31 30 31 31 30 31 30 31]
loop range get daysOfMonth m 1 [d][
dt: to :date.format:"yyyy-M-dd" ~"|y|-|m|-|d|"
if dt\Day = "Sunday" -> return dt
]
]

getLastSundays: function [year][
loop 1..12 'month [
print to :string.format:"yyyy-MM-dd" lastSundayForMonth month year
]
]

getLastSundays 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```

## AutoHotkey

```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```

## AWK

```# syntax: GAWK -f FIND_THE_LAST_SUNDAY_OF_EACH_MONTH.AWK [year]
BEGIN {
split("31,28,31,30,31,30,31,31,30,31,30,31",daynum_array,",") # days per month in non leap year
year = (ARGV[1] == "") ? strftime("%Y") : ARGV[1]
if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) {
daynum_array[2] = 29
}
for (m=1; m<=12; m++) {
for (d=daynum_array[m]; d>=1; d--) {
if (strftime("%a",mktime(sprintf("%d %d %d 0 0 0",year,m,d))) == "Sun") {
printf("%04d-%02d-%02d\n",year,m,d)
break
}
}
}
exit(0)
}
```

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
```

## Batch File

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

```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

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

```#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void last_sunday(struct tm *res, unsigned year, unsigned mon)
{
time_t sec;

*res = (struct tm){
.tm_year = year + mon / 12,
.tm_mon = mon % 12,
.tm_hour = 12,
.tm_isdst = -1
};
sec = mktime(res);
sec -= res->tm_wday * 86400;
*res = *localtime(&sec);
}

int main(int argc, char *argv[])
{
struct tm date;
char str[12];
unsigned m, y;

if (argc < 2)
return 1;
y = strtoul(argv[1], NULL, 0) - 1900;
for (m = 1; m <= 12; ++m) {
last_sunday(&date, y, m);
strftime(str, sizeof str, "%F", &date);
puts(str);
}

return 0;
}
```

## C#

```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);
/* Modification by Albert Zakhia on 2021-16-02
The below code is very slow due to the loop, we will go twice as fast
while (date.DayOfWeek != DayOfWeek.Sunday)
{
date = date.AddDays(-1);
}
*/
// The updated code
int daysOffset = date.DayOfWeek - dayOfWeek; // take the offset to subtract directly instead of looping
if (daysOffset < 0) daysOffset += 7; // if the code is negative, we need to normalize them
date = date.AddDays(-daysOffset ); // now just add the days offset
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
```

## C++

```#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
```

### Using C++20

Using C++20 this task can be completed without external libraries.

```#include <chrono>
#include <iostream>

int main() {
std::cout << "The dates of the last Sunday in each month of 2023:" << std::endl;

for ( unsigned int m = 1; m <= 12; ++m ) {
std::chrono::days days_in_month = std::chrono::sys_days{std::chrono::year{2023}/m/std::chrono::last}
- std::chrono::sys_days{std::chrono::year{2023}/m/1} + std::chrono::days{1};

const unsigned int last_day = days_in_month / std::chrono::days{1};
std::chrono::year_month_day ymd{std::chrono::year{2023}, std::chrono::month{m}, std::chrono::day{last_day}};

while ( std::chrono::weekday{ymd} != std::chrono::Sunday ) {
ymd = std::chrono::sys_days{ymd} - std::chrono::days{1};
}

std::cout << ymd << std::endl;
}
}
```
Output:
```The dates of the last Sunday in each month of 2023:
2023-01-29
2023-02-26
2023-03-26
2023-04-30
2023-05-28
2023-06-25
2023-07-30
2023-08-27
2023-09-24
2023-10-29
2023-11-26
2023-12-31
```

## Clojure

```(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

```       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
```

## Common Lisp

First, calculate the day of week of the 1st day of next month. Then figure out how many days you have to go back until the last Sunday of the month.

```(defun last-sundays (year)
(loop for month from 1 to 12
for last-month-p = (= month 12)
for next-month = (if last-month-p 1 (1+ month))
for year-of-next-month = (if last-month-p (1+ year) year)
for 1st-day-next-month = (encode-universal-time 0 0 0 1 next-month year-of-next-month 0)
for 1st-day-dow = (nth-value 6 (decode-universal-time 1st-day-next-month 0))
;; 0: monday, 1: tuesday, ... 6: sunday
for diff-to-last-sunday = (1+ 1st-day-dow)
for last-sunday = (- 1st-day-next-month (* diff-to-last-sunday 24 60 60))
do (multiple-value-bind (second minute hour date month year)
(decode-universal-time last-sunday 0)
(declare (ignore second minute hour))
(format t "~D-~2,'0D-~2,'0D~%" year month date))))

(last-sundays 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```

## D

```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```

## Delphi

Translation of: C#
```program Find_the_last_Sunday_of_each_month;

{\$APPTYPE CONSOLE}

uses
System.SysUtils,
System.DateUtils;

// ADayOfWeek -> sunday is the first day and is 1
function LastDayOfWeekOfEachMonth(AYear, ADayOfWeek: Word): TArray<TDateTime>;
var
month: word;
daysOffset: Integer;
date: TDatetime;
begin
if (ADayOfWeek > 7) or (ADayOfWeek < 1) then
raise Exception.CreateFmt('Error on FindAllDaysOfWeek: "%d" must be in [1..7] (sun..sat)',
[ADayOfWeek]);

SetLength(Result, 12);

for month := 1 to 12 do
begin
date := EncodeDate(AYear, month, DaysInAMonth(AYear, month));

daysOffset := DayOfWeek(date) - ADayOfWeek;
if daysOffset < 0 then
inc(daysOffset, 7);

Result[month - 1] := date - daysOffset;
end;
end;

var
strYear: string;
Year: Integer;
date: TDateTime;

begin
write('Year to calculate: ');
Readln(strYear);
if not TryStrToInt(strYear, Year) or (Year < 1900) then
raise Exception.CreateFmt('Error: "%s" is not a valid year', [strYear]);

for date in LastDayOfWeekOfEachMonth(Year, 1) do
writeln(FormatDateTime('yyyy-mmm-dd', date));

readln;
end.
```

## EasyLang

```func wkday year month day .
adj = (14 - month) div 12
mm = month + 12 * adj - 2
yy = year - adj
r = day + (13 * mm - 1) div 5 + yy + yy div 4 - yy div 100 + yy div 400
return r mod 7 + 1
.
proc show y . .
days[] = [ 31 28 31 30 31 30 31 31 30 31 30 31 ]
days[2] += if y mod 4 = 0 and (y mod 100 <> 0 or y mod 400 = 0)
for m = 1 to 12
d = days[m]
while wkday y m d <> 1
d -= 1
.
m\$ = m
if m < 10
m\$ = 0 & m
.
print y & "-" & m\$ & "-" & d
.
.
show 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
```

## Elixir

```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
```

## Emacs Lisp

```(require 'calendar)

(defun last-sunday (year)
"Print the last Sunday in each month of year"
(mapcar (lambda (month)
(let*
((days (number-sequence 1 (calendar-last-day-of-month month year)))
(mdy (mapcar (lambda (x) (list month x year)) days))
(weekdays (mapcar #'calendar-day-of-week mdy))
(lastsunday (1+ (cl-position 0 weekdays :from-end t))))
(insert (format "%i-%02i-%02i \n" year month lastsunday))))
(number-sequence 1 12)))

(last-sunday 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 ```

## Erlang

```-module(last_sundays).

-export([in_year/1]).

% calculate all the last sundays in a particular year
in_year(Year) ->
[lastday(Year, Month, 7) || Month <- lists:seq(1, 12)].

% calculate the date of the last occurrence of a particular weekday
lastday(Year, Month, WeekDay) ->
Ldm = calendar:last_day_of_the_month(Year, Month),
Diff = calendar:day_of_the_week(Year, Month, Ldm) rem WeekDay,
{Year, Month, Ldm - Diff}.
```
Output:
```30> [io:fwrite("~B-~2.10.0B-~B~n", [Y,M,D]) || {Y,M,D} <- last_sundays:in_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
```

## Excel

### LAMBDA

Binding the name lastSundayOfEachMonth to the following lambda expression in the Name Manager of the Excel WorkBook:

```lastSundayOfEachMonth
=LAMBDA(y,
LAMBDA(monthEnd,
1 + monthEnd - WEEKDAY(monthEnd)
)(
EDATE(
DATEVALUE(y & "-01-31"),
SEQUENCE(12, 1, 0, 1)
)
)
)
```
Output:

The formula in cell B2 defines an array which populates the range B2:B13.

(Any Excel date format can be applied to those cells)

 =lastSundayOfEachMonth(A2) fx A B 1 Year Last Sundays 2 2013 2013-01-27 3 2013-02-24 4 2013-03-31 5 2013-04-27 6 2013-05-25 7 2013-06-29 8 2013-07-27 9 2013-08-24 10 2013-09-28 11 2013-10-26 12 2013-11-24 13 2013-12-29

## F#

We use a transformation from and to the Julian Day Number, see also PARI/GP or Fortran.

The formulars used here come from [2] (section "Calculation").

```let jdn (year, month, day) =
let a = (14 - month) / 12
let y = year + 4800 - a
let m = month + 12 * a - 3
day + (153*m+2)/5 + 365*y + y/4 - y/100 + y/400 - 32045

let date_from_jdn jdn =
let j = jdn + 32044
let g = j / 146097
let dg = j % 146097
let c = (dg / 36524 + 1) * 3 / 4
let dc = dg - c * 36524
let b = dc / 1461
let db = dc % 1461
let a = (db / 365 + 1) * 3 / 4
let da = db - a * 365
let y = g * 400 + c * 100 + b * 4 + a
let m = (da * 5 + 308) / 153 - 2
let d = da - (m + 4) * 153 / 5 + 122
(y - 4800 + (m + 2) / 12, (m + 2) % 12 + 1, d + 1)

[<EntryPoint>]
let main argv =
let year = System.Int32.Parse(argv.[0])
[for m in 1..12 do yield jdn (year,m+1,0)]
|> List.map (fun x -> date_from_jdn (x - (x+1)%7))
|> List.iter (printfn "%A")
0
```
Output:
```RosettaCode 2016
(2016, 1, 31)
(2016, 2, 28)
(2016, 3, 27)
(2016, 4, 24)
(2016, 5, 29)
(2016, 6, 26)
(2016, 7, 31)
(2016, 8, 28)
(2016, 9, 25)
(2016, 10, 30)
(2016, 11, 27)
(2016, 12, 25)```

## Factor

This program expects a year passed in via command line argument. In case you are wondering — yes — Factor has a last-sunday-of-month word in its calendar vocabulary. This is par for the course when it comes to Factor.

```USING: calendar calendar.format command-line io kernel math math.parser
sequences ;
IN: rosetta-code.last-sunday

: parse-year     (    -- ts  ) (command-line) second string>number <year> ;
: print-last-sun ( ts --     ) last-sunday-of-month (timestamp>ymd) nl ;
: inc-month      ( ts -- ts' ) 1 months time+ ;
: process-month  ( ts -- ts' ) dup print-last-sun inc-month ;
: main           (    --     ) parse-year 12 [ process-month ] times drop ;

MAIN: main
```
Output:
```>factor last-sunday.factor 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

```#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

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
```

## Free Pascal

```program sundays;

Uses sysutils;

type
MonthLength = Array[1..13] of Integer;

procedure sund(y : Integer);
var
dt : TDateTime;
m,mm : Integer;
len : MonthLength;
begin
len[1] := 31; len[2] := 28; len[3] := 31; len[4] := 30;
len[5] := 31; len[6] := 30; len[7] := 31; len[8] := 31;
len[9] := 30; len[10] := 31; len[11] := 30; len[12] := 31; len[13] := 29;
for m := 1 to 12 do
begin
mm := m;
if (m = 2) and IsLeapYear( y ) then
mm := 13;
dt := EncodeDate( y, mm, len[mm] );
dt := EncodeDate( y, mm, len[mm] - DayOfWeek(dt) + 1 );
WriteLn(FormatDateTime('YYYY-MM-DD', dt ));
end;
end;

var
i : integer;
yy: integer;
begin
for i := 1 to paramCount() do begin
Val( paramStr(1), yy );
sund( yy );
end;
end.
```
Output:
```./sundays 2025
2025-01-26
2025-02-23
2025-03-30
2025-04-27
2025-05-25
2025-06-29
2025-07-27
2025-08-31
2025-09-28
2025-10-26
2025-11-30
2025-12-28
```

## FreeBASIC

```' 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 leapyear 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? 2017

Last Sunday of the month for 2017
29 January
26 February
26 March
30 April
28 May
25 June
30 July
27 August
24 September
29 October
26 November
31 December```

## Frink

```d = parseDate[ARGS@0]
for m = 1 to 12
{
d = beginningOfNextMonth[d]
n = d - parseInt[d -> ### u ###] days
println[n->###yyyy-MM-dd###]
}```
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
```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution

The following function retrieves the last day of each month of a year, being a given weekday:

Test case

## FutureBasic

```Long y, c, d, i

void local fn sunday( m as Short, nd as Short)
if d > nd then d -= 7 // Don't overshoot
print y @"–"m @"–"d
d = d + 35 - nd // Advance 5 weeks minus month length
end fn

void local fn month( m as Short )
select m
case 1
if ( y mod 4 > 0 ) or ( y mod 100 == 0 and y mod 400 > 0 )
fn sunday( m, 31)
fn sunday( m + 1, 28)
else
d += 1 // Extra day for leap year
fn sunday( m, 31 )
fn sunday( m + 1, 29 )
end if
case 3, 5, 7, 8, 10, 12
fn sunday( m, 31 )
case 4, 6, 9, 11
fn sunday( m, 30 )
end select
end fn

y = 2024
// Concentrate from D. Knuth: CACM 1962;5:209
c = y / 100 + 1
d = ( 3 * c / 4 ) - ( 5 * y / 4 )
d = ( d mod 7 ) + 36 // A Sunday in February
for i = 1 to 12
fn month( i )
next

handleevents```

Output:

```2024-1-28
2024-2-25
2024-3-31
2024-4-28
2024-5-26
2024-6-30
2024-7-28
2024-8-25
2024-9-29
2024-10-27
2024-11-24
2024-12-29
```

## Gambas

```Public Sub Form_Open()
Dim sYear As String                                                             'To store the year chosen
Dim siDay, siMonth, siWeekday As Short                                          'Day, Month and Weekday

sYear = InputBox("Input year", "Last Sunday of each month")                     'Get the user to enter a year
Print "Last Sundays in " & sYear                                                'Print a heading

For siMonth = 1 To 12                                                           'Loop for each month
For siDay = 31 DownTo 23                                                      'Count backwards from 31 to 23 (Sunday 23rd February 1930)
siWeekday = 7                                                               'Set the Weekday to Saturday in case of error in the next line
Try siWeekday = WeekDay(Date(Val(sYear), siMonth, siDay))                   'TRY and get the Weekday. If there is an error it will be ignored e.g. 31 February
If siWeekday = 0 Then                                                       'If Weekday is Sunday then..
Print Format(Date(Val(sYear), siMonth, siDay), "dddd dd mmmm yyyy")       'Print the date
Break                                                                     'Jump out of this loop
End If
Next
Next

End
```

Output:

```Last Sundays in 2013
Sunday 27 January 2013
Sunday 24 February 2013
Sunday 31 March 2013
Sunday 28 April 2013
Sunday 26 May 2013
Sunday 30 June 2013
Sunday 28 July 2013
Sunday 25 August 2013
Sunday 29 September 2013
Sunday 27 October 2013
Sunday 24 November 2013
Sunday 29 December 2013
```

## Go

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

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```

## Haskell

```import Data.List (find, intercalate, transpose)
import Data.Maybe (fromJust)
import Data.Time.Calendar
( Day,
addDays,
fromGregorian,
gregorianMonthLength,
showGregorian,
)
import Data.Time.Calendar.WeekDate (toWeekDate)

---------------- LAST SUNDAY OF EACH MONTH ---------------

lastSundayOfEachMonth = lastWeekDayDates 7

--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_
putStrLn
( intercalate "  "
<\$> transpose
(lastSundayOfEachMonth <\$> [2013 .. 2017])
)

------------------- NEAREST DAY OF WEEK ------------------

lastWeekDayDates :: Int -> Integer -> [String]
lastWeekDayDates dayOfWeek year =
(showGregorian . mostRecentWeekday dayOfWeek)
. (fromGregorian year <*> gregorianMonthLength year)
<\$> [1 .. 12]

mostRecentWeekday :: Int -> Day -> Day
mostRecentWeekday dayOfWeek date =
fromJust
(find p ((`addDays` date) <\$> [-6 .. 0]))
where
p x =
let (_, _, day) = toWeekDate x
in dayOfWeek == day
```
Output:
```2018-01-28  2019-01-27  2020-01-26  2021-01-31  2022-01-30  2023-01-29
2018-02-25  2019-02-24  2020-02-23  2021-02-28  2022-02-27  2023-02-26
2018-03-25  2019-03-31  2020-03-29  2021-03-28  2022-03-27  2023-03-26
2018-04-29  2019-04-28  2020-04-26  2021-04-25  2022-04-24  2023-04-30
2018-05-27  2019-05-26  2020-05-31  2021-05-30  2022-05-29  2023-05-28
2018-06-24  2019-06-30  2020-06-28  2021-06-27  2022-06-26  2023-06-25
2018-07-29  2019-07-28  2020-07-26  2021-07-25  2022-07-31  2023-07-30
2018-08-26  2019-08-25  2020-08-30  2021-08-29  2022-08-28  2023-08-27
2018-09-30  2019-09-29  2020-09-27  2021-09-26  2022-09-25  2023-09-24
2018-10-28  2019-10-27  2020-10-25  2021-10-31  2022-10-30  2023-10-29
2018-11-25  2019-11-24  2020-11-29  2021-11-28  2022-11-27  2023-11-26
2018-12-30  2019-12-29  2020-12-27  2021-12-26  2022-12-25  2023-12-31```

## Icon and Unicon

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

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

```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

```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

### ES5

#### Iteration

```function lastSundayOfEachMonths(year) {
var lastDay = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
var sundays = [];
var date, month;
if (year % 4 === 0 && (year % 100 !== 0 || year % 400 === 0)) {
lastDay[2] = 29;
}
for (date = new Date(), month = 0; month < 12; month += 1) {
date.setFullYear(year, month, lastDay[month]);
date.setDate(date.getDate() - date.getDay());
sundays.push(date.toISOString().substring(0, 10));
}
return sundays;
}

console.log(lastSundayOfEachMonths(2013).join('\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```

#### Functional composition

```(function () {
'use strict';

// lastSundaysOfYear :: Int -> [Date]
function lastSundaysOfYear(y) {
return lastWeekDaysOfYear(y, days.sunday);
}

// lastWeekDaysOfYear :: Int -> Int -> [Date]
function lastWeekDaysOfYear(y, iWeekDay) {
return [
31,
0 === y % 4 && 0 !== y % 100 || 0 === y % 400 ? 29 : 28,
31, 30, 31, 30, 31, 31, 30, 31, 30, 31
]
.map(function (d, m) {
var dte = new Date(Date.UTC(y, m, d));

return new Date(Date.UTC(
y, m, d - (
(dte.getDay() + (7 - iWeekDay)) % 7
)
));
});
}

// isoDateString :: Date -> String
function isoDateString(dte) {
return dte.toISOString()
.substr(0, 10);
}

// range :: Int -> Int -> [Int]
function range(m, n) {
return Array.apply(null, Array(n - m + 1))
.map(function (x, i) {
return m + i;
});
}

// transpose :: [[a]] -> [[a]]
function transpose(lst) {
return lst[0].map(function (_, iCol) {
return lst.map(function (row) {
return row[iCol];
});
});
}

var days = {
sunday: 0,
monday: 1,
tuesday: 2,
wednesday: 3,
thursday: 4,
friday: 5,
saturday: 6
}

// TEST

return transpose(
range(2012, 2016)
.map(lastSundaysOfYear)
)
.map(function (row) {
return row
.map(isoDateString)
.join('\t');
})
.join('\n');

})();
```
Output:
```2013-01-27	2014-01-26	2015-01-25	2016-01-31	2017-01-29
2013-02-24	2014-02-23	2015-02-22	2016-02-28	2017-02-26
2013-03-31	2014-03-30	2015-03-29	2016-03-27	2017-03-26
2013-04-28	2014-04-27	2015-04-26	2016-04-24	2017-04-30
2013-05-26	2014-05-25	2015-05-31	2016-05-29	2017-05-28
2013-06-30	2014-06-29	2015-06-28	2016-06-26	2017-06-25
2013-07-28	2014-07-27	2015-07-26	2016-07-31	2017-07-30
2013-08-25	2014-08-31	2015-08-30	2016-08-28	2017-08-27
2013-09-29	2014-09-28	2015-09-27	2016-09-25	2017-09-24
2013-10-27	2014-10-26	2015-10-25	2016-10-30	2017-10-29
2013-11-24	2014-11-30	2015-11-29	2016-11-27	2017-11-26
2013-12-29	2014-12-28	2015-12-27	2016-12-25	2017-12-31```

### ES6

```(() => {
'use strict'

// MAIN -----------------------------------------------
// main :: IO ()
const main = () =>
console.log(unlines(
map(
compose(
intercalate('\t'),
map(isoDateString)
)
)(
transpose(
map(lastWeekDaysOfYear(days.sunday))(
enumFromTo(2019)(2022)
)
)
)
));

// WEEKDAYS -------------------------------------------

// lastWeekDaysOfYear :: Int -> Int -> [Date]
const lastWeekDaysOfYear = iWeekDay =>
y => map((d, m) =>
new Date(Date.UTC(
y, m, d - ((new Date(Date.UTC(y, m, d))
.getDay() + (7 - iWeekDay)) % 7))))([
31,
0 === y % 4 && 0 !== y % 100 || 0 === y % 400 ? 29 : 28,
31, 30, 31, 30, 31, 31, 30, 31, 30, 31
]);

const days = {
sunday: 0,
monday: 1,
tuesday: 2,
wednesday: 3,
thursday: 4,
friday: 5,
saturday: 6
};

// GENERIC FUNCTIONS-----------------------------------

// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
x => fs.reduceRight((a, f) => f(a), x);

// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m => n =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);

// intercalate :: String -> [String] -> String
const intercalate = s => xs =>
xs.join(s);

// isoDateString :: Date -> String
const isoDateString = dte =>
dte.toISOString()
.substr(0, 10);

// map :: (a -> b) -> [a] -> [b]
const map = f => xs =>
(Array.isArray(xs) ? (
xs
) : xs.split('')).map(f);

// If some of the rows are shorter than the following rows,
// their elements are skipped:
// > transpose [[10,11],[20],[],[30,31,32]] == [[10,20,30],[11,31],[32]]

// transpose :: [[a]] -> [[a]]
const transpose = xss => {
const go = xss =>
0 < xss.length ? (() => {
const
h = xss[0],
t = xss.slice(1);
return 0 < h.length ? (
[
[h[0]].concat(t.reduce(
(a, xs) => a.concat(
0 < xs.length ? (
[xs[0]]
) : []
),
[]
))
].concat(go([h.slice(1)].concat(
t.map(xs => xs.slice(1))
)))
) : go(t);
})() : [];
return go(xss);
};

// unlines :: [String] -> String
const unlines = xs => xs.join('\n');

// MAIN ---
return main();
})();
```
Output:
```2019-01-27    2020-01-26    2021-01-31    2022-01-30
2019-02-24    2020-02-23    2021-02-28    2022-02-27
2019-03-31    2020-03-29    2021-03-28    2022-03-27
2019-04-28    2020-04-26    2021-04-25    2022-04-24
2019-05-26    2020-05-31    2021-05-30    2022-05-29
2019-06-30    2020-06-28    2021-06-27    2022-06-26
2019-07-28    2020-07-26    2021-07-25    2022-07-31
2019-08-25    2020-08-30    2021-08-29    2022-08-28
2019-09-29    2020-09-27    2021-09-26    2022-09-25
2019-10-27    2020-10-25    2021-10-31    2022-10-30
2019-11-24    2020-11-29    2021-11-28    2022-11-27
2019-12-29    2020-12-27    2021-12-26    2022-12-25```

## jq

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

```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>
```

## K

```/ List the dates of last Sundays of each month of
/ a given year
/ lastsundt.k

isleap: {(+/~x!' 4 100 400)!2}
wd: {(_jd x)!7}
dom: (31;28;31;30;31;30;31;31;30;31;30;31)
init: {:[isleap x;dom[1]::29;dom[1]::28]}
wdme: {[m;y]; init y; dt:(10000*y)+(100*m)+dom[m-1];jd::(_jd dt);mewd::(wd dt)}
lsd: {[m;y]; wdme[m;y];:[mewd>5;jd::jd+(6-mewd);jd::jd-(1+mewd)];dt:_dj(jd);yy:\$(yr:dt%10000);dd:\$(d:dt!100);mm:\$(mo:((dt-yr*10000)%100));arr::arr,\$(yy,"-",(2\$mm),"-",(2\$dd))}
lsd1: {[y];arr::(); m:1; do[12;lsd[m;y];m+:1]}
main: {[y]; lsd1[y];`0: ,"Dates of last Sundays of ",(\$y); 12 10#arr}
```

The output of a session with the above script is given below:

Output:
```K Console - Enter \ for help

\l lastsundt
main 2013
Dates of last Sundays of 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")

```

## Kotlin

```// version 1.0.6

import java.util.*

fun main(args: Array<String>) {
print("Enter a year : ")
val year = readLine()!!.toInt()

println("The last Sundays of each month in \$year are as follows:")
val calendar = GregorianCalendar(year, 0, 31)
for (month in 1..12) {
val daysInMonth = calendar.getActualMaximum(Calendar.DAY_OF_MONTH)
val lastSunday = daysInMonth - (calendar[Calendar.DAY_OF_WEEK] - Calendar.SUNDAY)
println("\$year-" + "%02d-".format(month) + "%02d".format(lastSunday))
if (month < 12) {
calendar.add(Calendar.DAY_OF_MONTH, 1)
calendar.add(Calendar.MONTH, 1)
calendar.add(Calendar.DAY_OF_MONTH, -1)
}
}
}
```

Sample input/output:

Output:
```Enter a year : 2013
The last Sundays of each month in 2013 are as follows:
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
```

## Lasso

```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```

## Liberty BASIC

This program goes a step further and provides a function definition, NthDayOfMonth(), that returns the date of any occurrence of a day in any given month and year. For example: The third Monday in February, the last Monday in May, the fourth Thursday in November, etc.

```yyyy=2013: if yyyy<1901 or yyyy>2099 then end
nda\$="Lsu"
print "The last Sundays of "; yyyy
for mm=1 to 12
x=NthDayOfMonth(yyyy, mm, nda\$)
select case mm
case  1: print "   January ";   x
case  2: print "   February ";  x
case  3: print "   March ";     x
case  4: print "   April ";     x
case  5: print "   May ";       x
case  6: print "   June ";      x
case  7: print "   July ";      x
case  8: print "   August ";    x
case  9: print "   September "; x
case 10: print "   October ";   x
case 11: print "   November ";  x
case 12: print "   December ";  x
end select
next mm
end

function NthDayOfMonth(yyyy, mm, nda\$)
' nda\$ is a two-part code. The first character, n, denotes
' first, second, third, fourth, and last by 1, 2, 3, 4, or L.
' The last two characters, da, denote the day of the week by
' mo, tu, we, th, fr, sa, or su. For example:
' the nda\$ for the second Monday of a month is "2mo";
' the nda\$ for the last Thursday of a month is "Lth".
if yyyy<1900 or yyyy>2099 or mm<1 or mm>12 then
NthDayOfMonth=0: exit function
end if
nda\$=lower\$(trim\$(nda\$))
if len(nda\$)<>3 then NthDayOfMonth=0: exit function
n\$=left\$(nda\$,1): nC\$="1234l"
da\$=right\$(nda\$,2): daC\$="tuwethfrsasumotuwethfrsasumo"
if not(instr(nC\$,n\$)) or not(instr(daC\$,da\$)) then
NthDayOfMonth=0: exit function
end if
NthDayOfMonth=1
mm\$=str\$(mm): if mm<10 then mm\$="0"+mm\$
db\$=DayOfDate\$(str\$(yyyy)+mm\$+"01")
if da\$<>db\$ then
x=instr(daC\$,db\$): y=instr(daC\$,da\$,x): NthDayOfMonth=1+(y-x)/2
end if
dim MD(12)
MD(1)=31: MD(2)=28: MD(3)=31: MD(4)=30: MD(5)=31: MD(6)=30
MD(7)=31: MD(8)=31: MD(9)=30: MD(10)=31: MD(11)=30: MD(12)=31
if yyyy mod 4 = 0 then MD(2)=29
if n\$<>"1" then
if n\$<>"l" then
NthDayOfMonth=NthDayOfMonth+((val(n\$)-1)*7)
else
if NthDayOfMonth+27<MD(mm) then
NthDayOfMonth=NthDayOfMonth+28
else
NthDayOfMonth=NthDayOfMonth+21
end if
end if
end if
end function

function DayOfDate\$(ObjectDate\$) 'yyyymmdd format
if ObjectDate\$="" then 'today
DaysSince1900 = date\$("days")
else
DaysSince1900 = date\$(mid\$(ObjectDate\$,5,2)+"/"+right\$(ObjectDate\$,2)_
+"/"+left\$(ObjectDate\$,4))
end if
DayOfWeek = DaysSince1900 mod 7
select case DayOfWeek
case 0: DayOfDate\$="tu"
case 1: DayOfDate\$="we"
case 2: DayOfDate\$="th"
case 3: DayOfDate\$="fr"
case 4: DayOfDate\$="sa"
case 5: DayOfDate\$="su"
case 6: DayOfDate\$="mo"
end select
end function```
Output:
```The 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
```

## LiveCode

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```

## Lua

```function isLeapYear (y)
return (y % 4 == 0 and y % 100 ~=0) or y % 400 == 0
end

function dayOfWeek (y, m, d)
local t = os.time({year = y, month = m, day = d})
return os.date("%A", t)
end

function lastWeekdays (wday, year)
local monthLength, day = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
if isLeapYear(year) then monthLength[2] = 29 end
for month = 1, 12 do
day = monthLength[month]
while dayOfWeek(year, month, day) ~= wday do day = day - 1 end
print(year .. "-" .. month .. "-" .. day)
end
end

lastWeekdays("Sunday", tonumber(arg[1]))
```

Command line session:

```>lua lastSundays.lua 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

>```

## Maple

```sundays := proc(year)
local i, dt, change, last_days;
last_days := [31,28,31,30,31,30,31,31,30,31,30,31];
if (Calendar:-IsLeapYear(year)) then
last_days[2] := 28;
end if;
for i to 12 do
dt := Date(year, i, last_days[i]);
change := 0;
if not(Calendar:-DayOfWeek(dt) = 1) then
change := -Calendar:-DayOfWeek(dt) + 1;
end if;
dt := Calendar:-AdjustDateField(dt, "date", change);
printf("%d-%d-%d\n", year, Month(dt), DayOfMonth(dt));
end do;
end proc;

sundays(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
```

## Mathematica / Wolfram Language

```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

```import os, strutils, times

const
DaysInMonth: array[Month, int] = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
DayDiffs:  array[WeekDay, int] = [1, 2, 3, 4, 5, 6, 0]

let year = paramStr(1).parseInt

for month in mJan..mDec:
var lastDay = DaysInMonth[month]
if month == mFeb and year.isLeapYear: lastDay = 29
var date = initDateTime(lastDay, month, year, 0, 0, 0)
date = date - days(DayDiffs[date.weekday])
echo date.format("yyyy-MM-dd")
```

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```

## Nu

Works with: Nushell version 0.96.1
```#!/usr/bin/nu

def last_sunday [y] {
\$'(\$in // 12 + \$y)-(\$in mod 12 + 1)-01T12:00'
| into datetime
| \$in - (\$in | format date '%uday' | into duration)
}

def main [year: int] {
for mon in 1..12 {
\$mon | last_sunday \$year | format date '%F' | print
}
}```

## Oberon-07

First, extend the minimal Dates module from the Long year#Oberon-07 and Leap year#Oberon-07 sample with exported dayOfWeek and OutDate methods. Only the new parts are shown:

```MODULE Dates;
IMPORT Out;

(* returns the day of the week of d/m/y *)
PROCEDURE dayOfWeek* ( d, m, y : INTEGER ) : INTEGER ;
VAR     j, k, mm, yy, w1 : INTEGER;
BEGIN
mm := m;
yy := y;
IF mm <= 2 THEN
INC( mm, 12 );
DEC( yy )
END;
j := yy DIV 100;
k := yy MOD 100;
(* w1 is the day of the week, Saturday = 0, Sunday = 1, ... *)
w1 := ( d + ( ( mm + 1 ) * 26 ) DIV 10 + k + k DIV 4 + j DIV 4 + 5 * j ) MOD 7;
(* convert w1 to Sunday = 0, Monday = 1, ... *)
IF w1 = 0 THEN
w1 := SATURDAY
ELSE
w1 := w1 - 1
END
RETURN w1
END dayOfWeek ;

(* Writes the date dd/mm/yyyy to standard output, in yyyy-mm-dd format *)
PROCEDURE OutDate* ( dd, mm, yyyy : INTEGER ) ;
BEGIN
Out.Int( yyyy, 4 );
IF mm < 10 THEN Out.String( "-0" ) ELSE Out.String( "-1" ) END;
Out.Int( mm MOD 10, 1 );
IF dd < 10 THEN Out.String( "-0" ) ELSE Out.String( "-" ) END;
Out.Int( dd, 1 )
END OutDate ;

END Dates.
```

Next, create a LastDays module that can determine the last particular weekday of each month of a year:

```MODULE LastDays;
IMPORT Dates;

VAR    daysInMonth : ARRAY 12 OF INTEGER;

(* sets the elements of lastWD to the day of the last weekDay of each month in year *)
PROCEDURE DayOfLastWeekDay* ( year, weekDay : INTEGER; VAR lastWD : ARRAY OF INTEGER ) ;
VAR   endOfMonth, dow, mPos, offset : INTEGER ;
BEGIN
FOR mPos := 0 TO 11 DO
endOfMonth := daysInMonth[ mPos ];
IF mPos = 1 THEN
(* February *)
IF Dates.isLeapYear( year ) THEN INC( endOfMonth ) END
END;
dow := Dates.dayOfWeek( endOfMonth, mPos + 1, year );
(* offset the last day to the required day of the week *)
IF dow >= weekDay THEN
offset := dow - weekDay
ELSE
offset := 7 - ( weekDay - dow )
END;
lastWD[ mPos ] := endOfMonth - offset
END
END DayOfLastWeekDay ;

BEGIN
daysInMonth[ 0 ] := 31; daysInMonth[  1 ] := 28; daysInMonth[  2 ] := 31;
daysInMonth[ 3 ] := 30; daysInMonth[  4 ] := 31; daysInMonth[  5 ] := 30;
daysInMonth[ 6 ] := 31; daysInMonth[  7 ] := 31; daysInMonth[  8 ] := 30;
daysInMonth[ 9 ] := 31; daysInMonth[ 10 ] := 30; daysInMonth[ 11 ] := 31
END LastDays.
```

Then, use the Dates and LastDays module to handle the task:

```MODULE LastSundays;
IMPORT LastDays, Dates, Out;
VAR    mPos, year : INTEGER;
last       : ARRAY 12 OF INTEGER;
BEGIN
year := 2024;
LastDays.DayOfLastWeekDay( year, Dates.SUNDAY, last );
FOR mPos := 0 TO 11 DO
Dates.OutDate( last[ mPos ], mPos + 1, year );
Out.Ln
END
END LastSundays.
```
Output:
```2024-01-28
2024-02-25
2024-03-31
2024-04-28
2024-05-26
2024-06-30
2024-07-28
2024-08-25
2024-09-29
2024-10-27
2024-11-24
2024-12-29
```

## OCaml

```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

```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
```

## PARI/GP

```\\ Normalized Julian Day Number from date
njd(D) =
{
my (m = D[2], y = D[1]);
if (D[2] > 2, m++, y--; m += 13);
(1461 * y) \ 4 + (306001 * m) \ 10000 + D[3] - 694024 + 2 - y \ 100 + y \ 400
}

\\ Date from Normalized Julian Day Number
njdate(J) =
{
my (a = J + 2415019, b = (4 * a - 7468865) \ 146097, c, d, m, y);

a += 1 + b - b \ 4 + 1524;
b = (20 * a - 2442) \ 7305;
c = (1461 * b) \ 4;
d = ((a - c) * 10000) \ 306001;
m = d - 1 - 12 * (d > 13);
y = b - 4715 - (m > 2);
d = a - c - (306001 * d) \ 10000;

[y, m, d]
}

for (m=1, 12, a=njd([2013,m+1,0]); print(njdate(a-(a+6)%7)))```

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]
```

## PascalABC.NET

```function LastSunday(month,year: integer): DateTime;
begin
var date := new DateTime(year,month,DateTime.DaysInMonth(year, month));
while date.DayOfWeek <> System.DayOfWeek.Sunday do
date := date.AddDays(-1);
Result := Date;
end;

begin
for var month:=1 to 12 do
Println(LastSunday(month,2024).ToString('yyyy-MM-dd'));
end.
```
Output:
```2024-01-28
2024-02-25
2024-03-31
2024-04-28
2024-05-26
2024-06-30
2024-07-28
2024-08-25
2024-09-29
2024-10-27
2024-11-24
2024-12-29

```

## Perl

```#!/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
```

## Phix

Requires 0.8.1 (day_of_week() is now ISO 8601 compliant)

```include timedate.e

constant SUNDAY=7

procedure showlast(integer dow, integer doy, timedate td)
td = adjust_timedate(td,timedelta(days:=doy-1))
integer {year,month,day} = td
while day_of_week(year,month,day)!=dow do day-=1 end while
printf(1,"%4d-%02d-%02d\n",{year,month,day})
end procedure

procedure last_day_of_month(integer year, integer dow)
integer doy
timedate first = {year,1,1,0,0,0,0,0}
-- start by finding the 1st of the next month, less 1
for i=1 to 11 do
doy = day_of_year(year,i+1,1)-1
showlast(dow,doy,first)
end for
-- do December separately, as 1st would be next year
doy = day_of_year(year,12,31)
showlast(dow,doy,first)
end procedure
last_day_of_month(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
```

## PHP

```<?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

```(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

```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
```

### Alternate Version

This script finds the first and/or last or all dates of any of the days of week; accepts `[Int32]` and `[DateTime]` values for Month and Year parameters; outputs `[DateTime]` objects by default but has an option to output time strings in various formats. This script also allows for pipeline input based mainly upon the Month parameter. This script has a syntax as complex as any PowerShell Cmdlet because it attempts to do everything.

```function Get-Date0fDayOfWeek
{
[CmdletBinding(DefaultParameterSetName="None")]
[OutputType([datetime])]
Param
(
[Parameter(Mandatory=\$false,
ValueFromPipeline=\$true,
ValueFromPipelineByPropertyName=\$true,
Position=0)]
[ValidateRange(1,12)]
[int]
\$Month = (Get-Date).Month,

[Parameter(Mandatory=\$false,
ValueFromPipelineByPropertyName=\$true,
Position=1)]
[ValidateRange(1,9999)]
[int]
\$Year = (Get-Date).Year,

[Parameter(Mandatory=\$true, ParameterSetName="Sunday")]
[switch]
\$Sunday,

[Parameter(Mandatory=\$true, ParameterSetName="Monday")]
[switch]
\$Monday,

[Parameter(Mandatory=\$true, ParameterSetName="Tuesday")]
[switch]
\$Tuesday,

[Parameter(Mandatory=\$true, ParameterSetName="Wednesday")]
[switch]
\$Wednesday,

[Parameter(Mandatory=\$true, ParameterSetName="Thursday")]
[switch]
\$Thursday,

[Parameter(Mandatory=\$true, ParameterSetName="Friday")]
[switch]
\$Friday,

[Parameter(Mandatory=\$true, ParameterSetName="Saturday")]
[switch]
\$Saturday,

[switch]
\$First,

[switch]
\$Last,

[switch]
\$AsString,

[Parameter(Mandatory=\$false)]
[ValidateNotNullOrEmpty()]
[string]
\$Format = "dd-MMM-yyyy"
)

Process
{
[datetime[]]\$dates = 1..[DateTime]::DaysInMonth(\$Year,\$Month) | ForEach-Object {
Get-Date -Year \$Year -Month \$Month -Day \$_ -Hour 0 -Minute 0 -Second 0 |
Where-Object -Property DayOfWeek -Match \$PSCmdlet.ParameterSetName
}

if (\$First -or \$Last)
{
if (\$AsString)
{
if (\$First) {\$dates[0].ToString(\$Format)}
if (\$Last)  {\$dates[-1].ToString(\$Format)}
}
else
{
if (\$First) {\$dates[0]}
if (\$Last)  {\$dates[-1]}
}
}
else
{
if (\$AsString)
{
\$dates | ForEach-Object {\$_.ToString(\$Format)}
}
else
{
\$dates
}
}
}
}
```

The default is to return `[DateTime]` objects:

```1..12 | Get-Date0fDayOfWeek -Year 2013 -Last -Sunday
```
Output:
```Sunday, January 27, 2013 12:00:00 AM
Sunday, February 24, 2013 12:00:00 AM
Sunday, March 31, 2013 12:00:00 AM
Sunday, April 28, 2013 12:00:00 AM
Sunday, May 26, 2013 12:00:00 AM
Sunday, June 30, 2013 12:00:00 AM
Sunday, July 28, 2013 12:00:00 AM
Sunday, August 25, 2013 12:00:00 AM
Sunday, September 29, 2013 12:00:00 AM
Sunday, October 27, 2013 12:00:00 AM
Sunday, November 24, 2013 12:00:00 AM
Sunday, December 29, 2013 12:00:00 AM
```

Return the `[DateTime]` objects as strings (using the default string format):

```1..12 | Get-Date0fDayOfWeek -Year 2013 -Last -Sunday -AsString
```
Output:
```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
```

Return the `[DateTime]` objects as strings (specifying the string format):

```1..12 | Get-Date0fDayOfWeek -Year 2013 -Last -Sunday -AsString -Format yyyy-MM-dd
```
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
```

## PureBasic

```Procedure LastSundayOfEachMonth(yyyy.i,List lfem.i())
Define dv.i=ParseDate("%yyyy",Str(yyyy)), mv.i=1
NewList d.i()
For d=1 To 365
dv=AddDate(dv,#PB_Date_Day,1)
If DayOfWeek(dv)=0
AddElement(d()) : d()=dv
EndIf
Next
dv=0
For mv=1 To 12
ForEach d()
If dv<d() And Month(d())=mv
dv=d()
EndIf
Next
AddElement(lfem()) : lfem()=dv
Next
EndProcedure

NewList lf.i()
Define y.i
OpenConsole("Last Sunday of each month")
Print("Input Year [ 1971 < y < 2038 ]: ")
y=Val(Input())
If y>1971 And y<2038
PrintN("Last Sunday of each month...")
LastSundayOfEachMonth(y,lf())
ForEach lf()
PrintN(FormatDate("%dd.%mm.%yyyy",lf()))
Next
EndIf
Print("...End")
Input()
```
Output:
```Input Year [ 1971 < y < 2038 ]: 2013
Last Sunday of each month...
27.01.2013
24.02.2013
31.03.2013
28.04.2013
26.05.2013
30.06.2013
28.07.2013
25.08.2013
29.09.2013
27.10.2013
24.11.2013
29.12.2013
...End```

## Python

```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
```

## QuickBASIC 4.5

QuickBASIC 4.5 doesn't have a way to manage dates easily. Following you'll see my solution for this task in QuickBASIC/QBASIC

```' PROGRAM Last Sundays in Quick BASIC 4.5 (LASTSQB1)
' This program will calculate the last Sundays of each month in a given year.
' It works assigning the year in the command prompt
' Usage: LASTSQB1 Year
' In the IDE, be sure to assign the COMMAND\$ value in the Run menu.

' Var
DIM iWY AS INTEGER
DIM A AS STRING
DIM iWM AS INTEGER
DIM iWD AS INTEGER
DIM strF AS STRING

' SUBs and FUNCTIONs
DECLARE FUNCTION LastSundayOfMonth% (WhichMonth AS INTEGER, WhichYear AS INTEGER)
DECLARE FUNCTION FirstDayOfWeekOnMonth% (WhichMonth AS INTEGER, WhichYear AS INTEGER)

' Gets the year from the command line
iWY = VAL(COMMAND\$)

' Verifies if given year is in the range.
' If so, runs the program. If not, shows an error message.
IF iWY >= 1753 AND iWY <= 2078 THEN
PRINT "Last Sundays of each month on year:"; iWY
strF = "####_-0#_-##"
FOR iWM = 1 TO 12
IF iWM > 9 THEN strF = "####_-##_-##"
iWD = LastSundayOfMonth(iWM, iWY)
PRINT USING strF; iWY; iWM; iWD
NEXT iWM
ELSE
PRINT "Incorrect year."
PRINT "Usage: LASTSQB1 Year"
PRINT
PRINT "Where Year is a value between 1753 and 2078."
END IF
PRINT
PRINT "End of program"
END

FUNCTION FirstDayOfWeekOnMonth% (WhichMonth AS INTEGER, WhichYear AS INTEGER)
' Adapted from Ray Thomas' SUB GetDay
'  taken from his program Calandar.BAS
'  available in his site http://brisray.com

' Var
DIM iDay AS INTEGER
DIM iMonth AS INTEGER
DIM iYear AS INTEGER
DIM iCentury AS INTEGER
DIM iDoW AS INTEGER
DIM strNewYear AS STRING

' Get the first day of the month
iDay = 1
iMonth = WhichMonth
iYear = WhichYear

' For any date in Jan or Feb add 12 to the month and
'  subtract 1 from the year
IF iMonth < 3 THEN
iMonth = iMonth + 12
iYear = iYear - 1
END IF

' Add 1 to the month and multiply by 2.61
' Drop the fraction (not round) afterwards
iMonth = iMonth + 1
iMonth = FIX(iMonth * 2.61)

' Add Day, Month and the last two digits of the year
strNewYear = LTRIM\$(STR\$(iYear))
iYear = VAL(RIGHT\$(strNewYear, 2))
iDoW = iDay + iMonth + iYear
iCentury = VAL(LEFT\$(strNewYear, 2))

' Add a quarter of the last two digits of the year
'  (truncated not rounded)
iYear = FIX(iYear / 4)
iDoW = iDoW + iYear

' Add the following factors for the year
IF iCentury = 18 THEN iCentury = 2
IF iCentury = 19 THEN iCentury = 0
IF iCentury = 20 THEN iCentury = 6
IF iCentury = 21 THEN iCentury = 4
iDoW = iDoW + iCentury

' The day of the week is the modulus of iDoY divided by 7
iDoW = (iDoW MOD 7) + 1

' The returned value will be between 1 and 7. 1 is Sunday.
FirstDayOfWeekOnMonth = iDoW
END FUNCTION

FUNCTION LastSundayOfMonth% (WhichMonth AS INTEGER, WhichYear AS INTEGER)
' Var
DIM iLDoM AS INTEGER
DIM iFDoWoM AS INTEGER

SELECT CASE WhichMonth
CASE 1, 3, 5, 7, 8, 10, 12
iLDoM = 31
CASE 2
iLDoM = 28 + ABS((WhichYear MOD 4 = 0 AND WhichYear MOD 100 OR WhichYear MOD 400 = 0) <> 0)
CASE ELSE
iLDoM = 30
END SELECT

' Get first Sunday
' All I have to do is to sum the days if the first of the given month
'  is not Sunday.
iFDoWoM = 1 + ((ABS(FirstDayOfWeekOnMonth(WhichMonth, WhichYear) - 7) + 1) MOD 7)

' Get the last Sunday
' I add as many days in multiples of 7 to get the last Sunday
'  in the given amount of days in month
iFDoWoM = iFDoWoM + (7 * ((iLDoM - iFDoWoM) \ 7))

' Returns the last sunday of the given month and year
LastSundayOfMonth = iFDoWoM

END FUNCTION
```
Output:
```C:\>LASTSQB1 2013
Last Sundays of each month on 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
```

### VB-DOS 1.0

As VB-DOS uses serial numbers for dates and includes some useful functions to work with dates, it is easier to do this code in VB-DOS.

```OPTION EXPLICIT

' PROGRAM Last Sundays in VB-DOS 1.0 (LASTVD1)
' This program will calculate the last Sundays of each month in a given year.
' It works assigning the year in the command prompt.
' Usage: LASTSVD1 Year
' In the IDE, be sure to assign the COMMAND\$ value in the Run menu.

' Var
DIM iWY AS INTEGER
DIM A AS STRING
DIM iWM AS INTEGER
DIM iWD AS INTEGER

' SUBs and FUNCTIONs
DECLARE FUNCTION LastSundayOfMonth (WhichMonth AS INTEGER, WhichYear AS INTEGER) AS INTEGER

' Initialize
iWY = VAL(COMMAND\$)

IF iWY >= 1753 AND iWY <= 2078 THEN
PRINT "Last Sundays of each month on year:"; iWY
FOR iWM = 1 TO 12
iWD = LastSundayOfMonth(iWM, iWY)
PRINT FORMAT\$(DATESERIAL(iWY, iWM, iWD), "yyyy-mm-dd")
NEXT iWM
ELSE
PRINT "Incorrect year. Please, specify a value between 1753 and 2078."
END IF

PRINT
PRINT "End of program"
END

FUNCTION LastSundayOfMonth (WhichMonth AS INTEGER, WhichYear AS INTEGER) AS INTEGER
' Var
DIM iLSoM AS INTEGER

SELECT CASE WhichMonth
CASE 1, 3, 5, 7, 8, 10, 12
iLSoM = 31
CASE 2
iLSoM = 28 + ABS((WhichYear MOD 4 = 0 AND WhichYear MOD 100 OR WhichYear MOD 400 = 0) <> 0)
CASE ELSE
iLSoM = 30
END SELECT

' Get last Sunday
iLSoM = iLSoM - (WEEKDAY(DATESERIAL(WhichYear, WhichMonth, iLSoM)) - 1)

LastSundayOfMonth = iLSoM
END FUNCTION
```
Output:
```C:\>LASTSVD1 2013
Last Sundays of each month on 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
```

## Quackery

```  [ over 3 < if [ 1 - ]
dup 4 / over +
over 100 / -
swap 400 / +
swap 1 -
[ table
0 3 2 5 0 3
5 1 4 6 2 4 ]
+ + 7 mod ]                     is dayofweek  ( day month year --> weekday )

[ dup 400 mod 0 = iff
[ drop true ]  done
dup 100 mod 0 = iff
[ drop false ] done
4 mod 0 = ]                     is leap       (           year --> b       )

[ swap 1 -
[ table
31 [ dup leap 28 + ]
31 30 31 30 31 31 30
31 30 31 ]
do nip ]                        is monthdays  (     month year --> n       )

[ number\$
2 times
[ char - join
over 10 < if
[ char 0 join ]
swap number\$ join ]
echo\$ ]                       is echoymd    ( day month year -->         )

[ dip
[ 2dup monthdays
dup temp put
unrot dayofweek ]
- dup 0 < if [ 7 + ]
temp take swap - ]              is lastwkday  ( month year wkd --> n       )

[ temp put
12 times
[ i^ 1+ over
2dup temp share lastwkday
unrot echoymd cr ]
drop temp release ]            is lastwkdays  (       year wkd -->        )

[ 0 lastwkdays ]                 is lastsundays (           year -->        )

2013 lastsundays```
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```

## R

```last_sundays <- function(year) {
for (month in 1:12) {
if (month == 12) {
date <- as.Date(paste0(year,"-",12,"-",31))
} else {
date <- as.Date(paste0(year,"-",month+1,"-",1))-1
}
while (weekdays(date) != "Sunday") {
date <- date - 1
}
print(date)
}
}
last_sundays(2004)
```
Output:
```[1] "2004-01-25"
[1] "2004-02-29"
[1] "2004-03-28"
[1] "2004-04-25"
[1] "2004-05-30"
[1] "2004-06-27"
[1] "2004-07-25"
[1] "2004-08-29"
[1] "2004-09-26"
[1] "2004-10-31"
[1] "2004-11-28"
[1] "2004-12-26"
```

## Racket

```#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
```

## Raku

(formerly Perl 6)

Works with: rakudo version 2018.03
```sub MAIN (\$year = Date.today.year) {
for 1..12 -> \$month {
my \$month-end = Date.new(\$year, \$month, Date.new(\$year,\$month,1).days-in-month);
say \$month-end - \$month-end.day-of-week % 7;
}
}
```
Output  —  for input `2018`:
```2018-01-28
2018-02-25
2018-03-25
2018-04-29
2018-05-27
2018-06-24
2018-07-29
2018-08-26
2018-09-30
2018-10-28
2018-11-25
2018-12-30```

## REBOL

```#!/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

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

```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
```

## RPL

`WKDAY` is defined at Day of the week

Works with: HP version 48
```≪ → year
≪ { }
.02 .13 FOR month
1 month .13 == DUP .01 month IFTE SWAP year + 1000000 / + +    @ Generate 1st day of the following month
DUP WKDAY NEG DATE+ +
.01 STEP
2 FIX      @ Display October Sunday as ##.10 and not as ##.1
≫ ≫ 'LSTSU' STO
```
```2013 LSTSU
```
Output:
```1: { 27.01 24.02 24.03 28.04 26.05 30.06 28.07 25.08 29.09 27.10 24.11 29.12 }
```

## Ruby

```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.

## Run BASIC

```input "What year to calculate (yyyy) : ";year
print "Last Sundays in ";year;" are:"
dim month(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 then leap = year - 1900 else leap = year - 1904
m = ((year-1900) mod 7) + int(leap/4) mod 7
for n = 1 to 12
month(n) = (val(word\$(mo\$,n)) + m) mod 7
month(n) = (val(word\$(mo\$,n)) + m) mod 7
next
for n = 1 to 12
for i = (val(word\$(mon\$,n)) - 6) to val(word\$(mon\$,n))
x = (month(n) + i) mod 7
if x = 4 then print year ; "-";right\$("0"+str\$(n),2);"-" ; i
next
next```
```What year to calculate (yyyy) : ?2013
Last Sundays in 2013 are:
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```

## Rust

```use std::env::args;
use time::{Date, Duration};

fn main() {
let year = args().nth(1).unwrap().parse::<i32>().unwrap();
(1..=12)
.map(|month| Date::try_from_ymd(year + month / 12, ((month % 12) + 1) as u8, 1))
.filter_map(|date| date.ok())
.for_each(|date| {
let days_back =
Duration::days(((date.weekday().number_from_sunday() as i64 + 5) % 7) + 1);
println!("{}", date - days_back);
});
}
```
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
```

## S-BASIC

```rem - return p mod q
function mod(p, q = integer) = integer
end = p - q * (p/q)

comment
return day of week (Sun = 0, Mon = 1, etc.) for a
given Gregorian calendar date using Zeller's congruence
end
function dayofweek (mo, da, yr = integer) = integer
var y, c, z = integer
if mo < 3 then
begin
mo = mo + 10
yr = yr - 1
end
else mo = mo - 2
y = mod(yr,100)
c = int(yr / 100)
z = int((26 * mo - 2) / 10)
z = z + da + y + int(y/4) + int(c/4) - 2 * c + 777
z = mod(z,7)
end = z

rem - return true if y is a leap year
function isleap(y = integer) = integer
end = mod(y,4)=0 and mod(y,100)<>0 or mod(y,400)=0

rem - return number of days in specified month
function monthdays(m, y = integer) = integer
var n = integer
if m = 2 then
if isleap(y) then
n = 29
else
n = 28
else if (m = 4) or (m = 6) or (m = 9) or (m = 11) then
n = 30
else
n = 31
end = n

comment
return the day of the month corresponding to the last
occurrence of weekday k (Sun=0, Mon=1, etc.) in the given
month and year
end
function lastkday(k, m, y = integer) = integer
var d, w = integer
rem - determine weekday for last day of the month
d = monthdays(m, y)
w = dayofweek(m, d, y)
rem - back up as needed to desired weekday
if w >= k then
d = d - (w - k)
else
d = d - (7 - (k - w))
end = d

rem - return abbreviated month name
function shortmonth (m = integer) = string
end = mid("JanFebMarAprMayJunJulAugSepOctNovDec", m*3-2, 3)

rem - main program starts here

\$constant SUNDAY = 0
var m, y = integer
input "Display last Sundays in what year"; y
for m = 1 to 12
print shortmonth(m);" ";lastkday(SUNDAY, m, y)
next m
end
```
Output:
```Display last Sundays in what year? 2021
Jan 21
Feb 28
Mar 28
Apr 25
May 30
Jun 27
Jul 25
Aug 29
Sep 26
Oct 31
Nov 28
Dec 26```

## Scala

```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

```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

Uses the libraries time.s7i and duration.s7i. Applicable to any day of the week, cf. [[3]].

```\$ 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

```var dt = require('DateTime');
var (year=2016) = 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
```

## Smalltalk

```Pharo Smalltalk

[ :yr | | firstDay firstSunday |
firstDay := Date year: yr month: 1 day: 1.
firstSunday := firstDay addDays: (1 - firstDay dayOfWeek).
(0 to: 53)
collect: [ :each | firstSunday addDays: (each * 7) ]
thenSelect: [ :each |
(((Date daysInMonth: each monthIndex forYear: yr) - each dayOfMonth) <= 6) and: [ each year = yr ] ] ]
```
Output:
```Send value: 2013 to the above block to return an array:
(27 January 2013 24 February 2013 31 March 2013 28 April 2013 26 May 2013 30 June 2013 28 July 2013 25 August 2013 29 September 2013 27 October 2013 24 November 2013 29 December 2013)
```

## Stata

```program last_sundays
args year
clear
qui set obs 12
gen day=dofm(mofd(mdy(_n,1,`year'))+1)-1
qui replace day=day-mod(dow(day),7)
format %td day
list, noobs noheader sep(6)
end

last_sundays 2013

+-----------+
| 27jan2013 |
| 24feb2013 |
| 31mar2013 |
| 28apr2013 |
| 26may2013 |
| 30jun2013 |
|-----------|
| 28jul2013 |
| 25aug2013 |
| 29sep2013 |
| 27oct2013 |
| 24nov2013 |
| 29dec2013 |
+-----------+
```

## Swift

```import Foundation

func lastSundays(of year: Int) -> [Date] {

let calendar = Calendar.current
var dates = [Date]()

for month in 1...12 {

var dateComponents = DateComponents(calendar: calendar,
year: year,
month: month + 1,
day: 0,
hour: 12)

let date = calendar.date(from: dateComponents)!
let weekday = calendar.component(.weekday, from: date)

if weekday != 1 {
dateComponents.day! -= weekday - 1
}

dates.append(calendar.date(from: dateComponents)!)
}
return dates
}

var dateFormatter = DateFormatter()
dateFormatter.dateStyle = .short

print(lastSundays(of: 2013).map(dateFormatter.string).joined(separator: "\n"))
```
Output:
```1/27/13
2/24/13
3/31/13
4/28/13
5/26/13
6/30/13
7/28/13
8/25/13
9/29/13
10/27/13
11/24/13
12/29/13
```

## Tcl

```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
```

## UNIX Shell

Works with: Bourne Again SHell

This uses the programs awk and cal, which are usually installed on any POSIXlike system.

```last_sundays() {
local y=\$1
for (( m=1; m<=12; ++m )); do
cal \$m \$y | awk -vy=\$y -vm=\$m '/^.[0-9]/ {d=\$1} END {print y"-"m"-"d}'
done
}
```
Output:
```\$ 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```

## VBScript

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
```

## Wren

Library: Wren-date
```import "os" for Process
import "./date" for Date

var args = Process.arguments
if (args.count != 1) {
Fiber.abort("Please pass just the year to be processed.")
}

var year = Num.fromString(args[0])
System.print("The dates of the last Sundays in the month for %(year) are:")
Date.default = Date.isoDate
for (m in 1..12) {
var d = Date.monthLength(year, m)
var dt = Date.new(year, m, d)
var wd = dt.dayOfWeek
if (wd == 7) {
System.print(dt)
} else {
System.print(dt.addDays(-wd))
}
}
```
Output:
```\$ wren Find_the_last_Sunday_of_each_month.wren 2013
The dates of the last Sundays in the month for 2013 are:
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

\$ wren Find_the_last_Sunday_of_each_month.wren 2020
The dates of the last Sundays in the month for 2020 are:
2020-01-26
2020-02-23
2020-03-29
2020-04-26
2020-05-31
2020-06-28
2020-07-26
2020-08-30
2020-09-27
2020-10-25
2020-11-29
2020-12-27
```

## XPL0

```func WeekDay(Year, Month, Day);   \Return day of week (0=Sun, 1=Mon ... 6=Sat)
int  Year, Month, Day;            \works for years from 1583 onward
[if Month<=2 then [Month:= Month+12;  Year:= Year-1];
return rem((Day-1 + (Month+1)*26/10 + Year + Year/4 + Year/100*6 + Year/400)/7);
];

int Year, Month, LastDay, WD;
[Year:= IntIn(8);               \from command line
for Month:= 1 to 12 do
[LastDay:= WeekDay(Year, Month+1, 1) - WeekDay(Year, Month, 28);
if LastDay < 0 then LastDay:= LastDay + 7;
LastDay:= LastDay + 27;     \ = number of days in Month
WD:= WeekDay(Year, Month, LastDay);
LastDay:= LastDay - WD;
IntOut(0, Year);  ChOut(0, ^-);
if Month < 10 then ChOut(0, ^0);  IntOut(0, Month);  ChOut(0, ^-);
IntOut(0, LastDay);  CrLf(0);
];
]```
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
```

## zkl

The program from [[4]] 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```