# Last Friday of each month

Last Friday 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 date of the last Fridays 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_fridays 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## 360 Assembly

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

*        Last Friday of each month 17/07/2016
LASTFRI  CSECT
USING  LASTFRI,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)         " ->
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     R7,44              44
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=44+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'2016'            <== 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    LASTFRI
Output:
2016-01-29
2016-02-26
2016-03-25
2016-04-29
2016-05-27
2016-06-24
2016-07-29
2016-08-26
2016-09-30
2016-10-28
2016-11-25
2016-12-30

## Action!

;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)=5 THEN
last=d
FI
OD
PrintI(y) Put('-)
PrintB2(m) Put('-)
PrintB2(last) PutE()
OD
RETURN
Output:
Input year in range 1753...9999: 2021
2021-01-29
2021-02-26
2021-03-26
2021-04-30
2021-05-28
2021-06-25
2021-07-30
2021-08-27
2021-09-24
2021-10-29
2021-11-26
2021-12-31

Uses GNAT. Applicable to any day of the week, cf. [[1]].

procedure Last_Weekday_In_Month is

use GNAT.Calendar.Time_IO;
begin
Ada.Text_IO.Put_Line(Image(Date => T, Picture => ISO_Date));
end Put_Line;

subtype Day_Name is Formatting.Day_Name; use type Formatting.Day_Name;

T, Selected : Time;
Year : Year_Number := Integer'Value (Ada.Command_Line.Argument (2));

begin
for Month in 1 .. 12 loop
T := Time_Of (Year => Year, Month => Month, Day => 01);
if Formatting.Day_Of_Week (T) = Weekday then
Selected := T;
end if;
T := T + Day_Count(1);
end loop;
Put_Line(Selected);
end loop;
end Last_Weekday_In_Month;
Output:
>./last_weekday_in_month friday 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## ALGOL 68

Basically the same as the "Find the Last Sunday Of Each Month" task solution.

Translation of: ALGOL W
BEGIN # find the last Friday 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 Friday of each month in year  #
PROC last fridays = ( 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 Friday                                        #
[ 1 : 12 ]INT last;
FOR m pos TO 12 DO
INT dow := day of week( last days[ m pos ], m pos, year );
# dow     = 1 Sun, 2 Mon, ... , 6 Fri, 0 Sat     #
# change to 2 Sun, 3 Mon, ... , 0 Fri, 1 Sat     #
dow := ( dow + 1 ) MOD 7;
# offset the last day to the last Friday         #
last[ m pos ] := last days[ m pos ] - dow
OD;
last
END # last fridays # ;
# test the last fridays procedure                            #
INT   year = 2021;
[]INT last = last fridays( 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:
2021-01-29
2021-02-26
2021-03-26
2021-04-30
2021-05-28
2021-06-25
2021-07-30
2021-08-27
2021-09-24
2021-10-29
2021-11-26
2021-12-31

## ALGOL W

Basically the same as the "Find the Last Sunday Of Each Month" task solution, uses the Day_of_week and isLeapYear procedures from the the day-of-the-week and leap-year tasks (included here for convenience).

begin % find the last Friday 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, Friday = 6, Saturday = 0               %
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 Friday   %
% of each month in year                                     %
procedure lastFridays ( 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 Friday                                       %
for mPos := 1 until 12 do begin
integer dow;
dow := Day_of_week( lastDays( mPos ), mPos, year );
% dow     = 1 Sun, 2 Mon, ... , 6 Fri, 0 Sat    %
% change to 2 Sun, 3 Mon, ... , 0 Fri, 1 Sat    %
dow := ( dow + 1 ) rem 7;
% offset the last day to the last Friday        %
last( mPos ) := lastDays( mPos ) - dow
end for_mPos
end lastFridays ;
begin
% test the lastFridays procedure                        %
integer array last ( 1 :: 12 );
integer year;
year := 2020;
lastFridays( 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:
2020-01-31
2020-02-28
2020-03-27
2020-04-24
2020-05-29
2020-06-26
2020-07-31
2020-08-28
2020-09-25
2020-10-30
2020-11-27
2020-12-25

## AppleScript

Translation of: JavaScript
-- LAST FRIDAYS OF YEAR ------------------------------------------------------

--  lastFridaysOfYear :: Int -> [Date]
on lastFridaysOfYear(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, Friday as integer)
end lastFridaysOfYear

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

intercalate(linefeed, ¬
map(isoRow, ¬
transpose(map(lastFridaysOfYear, ¬
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/lastFridays.scpt
-- ~ \$ osascript ~/Desktop/lastFridays.scpt 2013
-- ~ \$ osascript ~/Desktop/lastFridays.scpt 2013 2016

-- singleYearOrRange :: [Int] -> [Int]
on singleYearOrRange(argv)
apply(cond(length of argv > 0, my enumFromTo, 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)
-- parseInt :: String -> Int
script parseInt
on |λ|(s)
s as integer
end |λ|
end script

if class of argv is list and argv  {} then
{map(parseInt, argv)}
else
{}
end if
end argIntegers

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

-- Dates and date strings ----------------------------------------------------

-- 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

-- 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

-- Testing and tabulation ----------------------------------------------------

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

-- 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

-- 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

-- 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:
2014-01-31    2015-01-30    2016-01-29    2017-01-27    2018-01-26
2014-02-28    2015-02-27    2016-02-26    2017-02-24    2018-02-23
2014-03-28    2015-03-27    2016-03-25    2017-03-31    2018-03-30
2014-04-25    2015-04-24    2016-04-29    2017-04-28    2018-04-27
2014-05-30    2015-05-29    2016-05-27    2017-05-26    2018-05-25
2014-06-27    2015-06-26    2016-06-24    2017-06-30    2018-06-29
2014-07-25    2015-07-31    2016-07-29    2017-07-28    2018-07-27
2014-08-29    2015-08-28    2016-08-26    2017-08-25    2018-08-31
2014-09-26    2015-09-25    2016-09-30    2017-09-29    2018-09-28
2014-10-31    2015-10-30    2016-10-28    2017-10-27    2018-10-26
2014-11-28    2015-11-27    2016-11-25    2017-11-24    2018-11-30
2014-12-26    2015-12-25    2016-12-30    2017-12-29    2018-12-28

A 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 lastFridayOfEachMonthInYear(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_fridays " & 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 Friday 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 lastFridayOfThisMonth to firstDayOfNextMonth - (1 + (firstDayOfNextMonth's weekday) mod 7) * days
-- Append the required details to the output text.
set {month:m, day:d} to lastFridayOfThisMonth
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 lastFridayOfEachMonthInYear

lastFridayOfEachMonthInYear(2020)
Output:
"./last_fridays 2020
2020-01-31
2020-02-28
2020-03-27
2020-04-24
2020-05-29
2020-06-26
2020-07-31
2020-08-28
2020-09-25
2020-10-30
2020-11-27
2020-12-25"

The above is of course hard-coded for Fridays. It can be made more flexible by taking an AppleScript weekday constant as a parameter:

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(Friday, 2020)
Output:
"./last_Fridays 2020
2020-01-31
2020-02-28
2020-03-27
2020-04-24
2020-05-29
2020-06-26
2020-07-31
2020-08-28
2020-09-25
2020-10-30
2020-11-27
2020-12-25"

## Arturo

lastFridayForMonth: function [m][
ensure -> in? m 1..12

daysOfMonth: [0 31 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" ~"2012-|m|-|d|"
if dt\Day = "Friday" -> return dt
]
]

loop 1..12 'month [
print to :string.format:"yyyy-MM-dd" lastFridayForMonth month
]
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## AutoHotkey

if 1 = ; no parameter passed
{
InputBox, 1, Last Fridays of year, Enter a year:, , , , , , , , %A_YYYY%
If ErrorLevel
ExitApp
}

YYYY = %1% ; retrieve command line parameter
Stmp = %YYYY%0101000000
count= 0

While count < 12
{
FormatTime, ddd, %stmp%, ddd
FormatTime, M, %stmp%, M
If (ddd = "Fri"){
if (M-1 = count){
t := stmp
stmp += 7, days
}
else
res .= SubStr(t, 1, 4) "-" SubStr(t, 5, 2) "-" SubStr(t, 7, 2) "`n"
,count++
,stmp := YYYY . SubStr("0" M, -1) . "01"
}
else
stmp += 1, days
}
MsgBox % res
Output:
for 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## AutoIt

#include <Date.au3>

\$iYear = InputBox('Last Friday in each month', 'Please input the year:')

_GetLastFridays(\$iYear)

Func _GetLastFridays(\$_iYear)
Local \$sResult = 'last fridays in ' & \$_iYear & @LF, \$iDay
If _DateIsLeapYear(\$_iYear) Then \$aDaysInMonth[1] = 29
For \$i = 1 To 12
While 1
If _DateToDayOfWeekISO(\$_iYear, \$i, \$iDay) = 5 Then
\$sResult &= StringFormat('%4d-%02d-%02d', \$_iYear, \$i, \$iDay) & @LF
ExitLoop
EndIf
\$iDay -= 1
WEnd
Next
ConsoleWrite(\$sResult)
EndFunc  ;==>_GetFridays
Output:
last fridays in 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

--BugFix (talk) 13:27, 15 November 2013 (UTC)

## AWK

# syntax: GAWK -f LAST_FRIDAY_OF_EACH_MONTH.AWK year
# converted from Fortran
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]
if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) {
daynum_array[2] = 29
}
y = year - 1
k = 44 + y + int(y/4) + int(6*(y/100)) + int(y/400)
for (m=1; m<=12; m++) {
k += daynum_array[m]
d = daynum_array[m] - (k%7)
printf("%04d-%02d-%02d\n",year,m,d)
}
exit(0)
}
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Befunge

Translation of: C

The algorithm has been slightly simplified to avoid the additional day adjustment inside the loop, and the year is obtained from stdin rather than via the command line.

":raeY",,,,,&>55+,:::45*:*%\"d"%!*\4%+!3v
v2++1**"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:2012

2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## C

Doesn't work with Julian calendar (then again, you probably don't need to plan your weekends for middle ages).

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

int main(int c, char *v[])
{
int days[] = {31,29,31,30,31,30,31,31,30,31,30,31};
int m, y, w;

if (c < 2 || (y = atoi(v[1])) <= 1700) return 1;
days[1] -= (y % 4) || (!(y % 100) && (y % 400));
w = y * 365 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + 6;

for(m = 0; m < 12; m++) {
w = (w + days[m]) % 7;
printf("%d-%02d-%d\n", y, m + 1,
days[m] + (w < 5 ? -2 : 5) - w);
}

return 0;
}

## C#

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;

namespace RosettaCode.LastFridaysOfYear
{
internal static class Program
{
private static IEnumerable<DateTime> LastFridaysOfYear(int year)
{
for (var month = 1; month <= 12; month++)
{
while (date.DayOfWeek != DayOfWeek.Friday)
{
}
yield return date;
}
}

private static void Main(string[] arguments)
{
int year;
var argument = arguments.FirstOrDefault();
if (string.IsNullOrEmpty(argument) || !int.TryParse(argument, out year))
{
year = DateTime.Today.Year;
}

foreach (var date in LastFridaysOfYear(year))
{
Console.WriteLine(date.ToString("d", CultureInfo.InvariantCulture));
}
}
}
}
Output:
01/27/2012
02/24/2012
03/30/2012
04/27/2012
05/25/2012
06/29/2012
07/27/2012
08/31/2012
09/28/2012
10/26/2012
11/30/2012
12/28/2012

## C++

Library: Boost

called with ./last_fridays 2012

#include <boost/date_time/gregorian/gregorian.hpp>
#include <iostream>
#include <cstdlib>

int main( int argc , char* argv[ ] ) {
using namespace boost::gregorian ;

greg_month months[ ] = { Jan , Feb , Mar , Apr , May , Jun , Jul ,
Aug , Sep , Oct , Nov , Dec } ;
greg_year gy = atoi( argv[ 1 ] ) ;
for ( int i = 0 ; i < 12 ; i++ ) {
last_day_of_the_week_in_month lwdm ( Friday , months[ i ] ) ;
date d = lwdm.get_date( gy ) ;
std::cout << d << std::endl ;
}
return 0 ;
}
Output:
2012-Jan-27
2012-Feb-24
2012-Mar-30
2012-Apr-27
2012-May-25
2012-Jun-29
2012-Jul-27
2012-Aug-31
2012-Sep-28
2012-Oct-26
2012-Nov-30
2012-Dec-28

## Clojure

Library: clj-time
(use '[clj-time.core :only [last-day-of-the-month day-of-week minus days]]
'[clj-time.format :only [unparse formatters]])

(defn last-fridays [year]
(let [last-days (map #(last-day-of-the-month year %) (range 1 13 1))
dow (map day-of-week last-days)
relation (zipmap last-days dow)]
(map #(minus (key %) (days (mod (+ (val %) 2) 7))) relation)))

(defn last-fridays-formatted [year]
(sort (map #(unparse (formatters :year-month-day) %) (last-fridays year))))
Output:
user=> (pprint (last-fridays-formatted 2012))
("2012-01-27"
"2012-02-24"
"2012-03-30"
"2012-04-27"
"2012-05-25"
"2012-06-29"
"2012-07-27"
"2012-08-31"
"2012-09-28"
"2012-10-26"
"2012-11-30"
"2012-12-28")

## COBOL

program-id. last-fri.
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 friday pic 9(4) value 5.
procedure division.
display "Enter a calendar year (1601 thru 9999): "
accept yr
if yr >= 1601 and <= 9999
continue
else
display "Invalid year"
stop run
end-if
perform 12 times
move 1 to da
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 - friday) 7)
subtract dow from da
display yr "-" mo "-" da
end-perform
stop run
.
end program last-fri.
Output:
2016-01-29
2016-02-26
2016-03-25
2016-04-29
2016-05-27
2016-06-24
2016-07-29
2016-08-26
2016-09-30
2016-10-28
2016-11-25
2016-12-30

## CoffeeScript

last_friday_of_month = (year, month) ->
# month is 1-based, JS API is 0-based, then we use
# non-positive indexes to work backward relative to the
# first day of the next month
i = 0
while true
last_day = new Date(year, month, i)
if last_day.getDay() == 5
return last_day.toDateString()
i -= 1

print_last_fridays_of_month = (year) ->
for month in [1..12]
console.log last_friday_of_month year, month

do ->
year = parseInt process.argv[2]
print_last_fridays_of_month year
Output:
> coffee last_friday.coffee 2012
Fri Jan 27 2012
Fri Feb 24 2012
Fri Mar 30 2012
Fri Apr 27 2012
Fri May 25 2012
Fri Jun 29 2012
Fri Jul 27 2012
Fri Aug 31 2012
Fri Sep 28 2012
Fri Oct 26 2012
Fri Nov 30 2012
Fri Dec 28 2012

## Common Lisp

Works with: CLISP

The command-line argument processing is the only CLISP-specific code.

(defun friday-before (year month day)
(let*
((timestamp (encode-universal-time 0 0 12 day month year))
(weekday (nth 6 (multiple-value-list (decode-universal-time timestamp))))
(fri (- timestamp (* (+ (mod (+ weekday 2) 7) 1) 86400))))
(multiple-value-bind (_ _ _ d m y) (decode-universal-time fri)
(list y m d))))

(defun last-fridays (year)
(append (loop for month from 2 to 12 collecting (friday-before year month 1))
(list (friday-before (1+ year) 1 1))))

(format t "~{~{~a-~2,'0d-~2,'0d~}~%~}" (last-fridays year)))

Sample run for the year 2015:

Output:
2015-01-30
2015-02-27
2015-03-27
2015-04-24
2015-05-29
2015-06-26
2015-07-31
2015-08-28
2015-09-25
2015-10-30
2015-11-27
2015-12-25

## D

import std.stdio, std.datetime, std.traits;

void lastFridays(in uint year) {
auto date = Date(year, 1, 1);
foreach (_; [EnumMembers!Month]) {
date.day(date.daysInMonth);
date.roll!"days"(-(date.dayOfWeek + 2) % 7);
writeln(date);
}
}

void main() {
lastFridays(2012);
}
2012-Jan-27
2012-Feb-24
2012-Mar-30
2012-Apr-27
2012-May-25
2012-Jun-29
2012-Jul-27
2012-Aug-31
2012-Sep-28
2012-Oct-26
2012-Nov-30
2012-Dec-28

## Delphi

Uses the standard Delphi library.

program LastFridayOfMonth;

{\$APPTYPE CONSOLE}

uses
System.SysUtils, System.DateUtils;

var
Year: Word;
Month: Word;
D1: TDateTime;
D2: Word;

begin
Write('Enter year: ');

for Month := MonthJanuary to MonthDecember do begin
D1 := EndOfAMonth(Year, Month);
D2 := DayOfTheWeek(D1);
while D2 <> DayFriday do begin
D1 := IncDay(D1, -1);
D2 := DayOfTheWeek(D1);
end;
WriteLn(DateToStr(D1));
end;
end.
Output:
Enter year: 2019
25.01.2019
22.02.2019
29.03.2019
26.04.2019
31.05.2019
28.06.2019
26.07.2019
30.08.2019
27.09.2019
25.10.2019
29.11.2019
27.12.2019

## Elixir

defmodule RC do
def lastFriday(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)
friday = lastday - rem(daynum + 2, 7)
{year, month, friday}
end)
end
end

y = String.to_integer(hd(System.argv))
Enum.each(RC.lastFriday(y), fn {year, month, day} ->
:io.format "~4b-~2..0w-~2..0w~n", [year, month, day]
end)
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Elm

import Html exposing (Html, Attribute, text, div, input)
import Html.App exposing (beginnerProgram)
import Html.Attributes exposing (placeholder, value, style)
import Html.Events exposing (onInput)
import String exposing (toInt)
import Maybe exposing (withDefault)
import List exposing (map, map2)
import List.Extra exposing (scanl1)

type Msg = SetYear String

lastFridays : Int -> List Int
lastFridays year =
let isLeap = (year % 400) == 0 || ( (year % 4) == 0 && (year % 100) /= 0 )
daysInMonth = [31, if isLeap then 29 else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
y = year-1
in scanl1 (+) daysInMonth
|> map2 (\len day -> len - (day + 2 + y + y//4 - y//100 + y//400) % 7) daysInMonth

lastFridayStrings : String -> List String
lastFridayStrings yearString =
let months= ["January ", "February ", "March ", "April ", "May ", "June ", "July ", "August ", "September ", "October ", "November ", "December "]
errString = "Only years after 1752 are valid."
in case toInt yearString of
Ok year ->
if (year < 1753)
then [errString]
else lastFridays year
|> map2 (\m d -> m ++ toString d ++ ", " ++ toString year) months
Err _ ->
[errString]

view :  String -> Html Msg
view yearString =
div []
([ input
[ placeholder "Enter a year."
, value yearString
, onInput SetYear
, myStyle
]
[]
] ++ (lastFridayStrings yearString
|> map (\date -> div [ myStyle ] [ text date ]) ))

myStyle : Attribute Msg
myStyle =
style
[ ("width", "100%")
, ("height", "20px")
, ("padding", "5px 0 0 5px")
, ("font-size", "1em")
, ("text-align", "left")
]

update : Msg -> String -> String
update msg _ =
case msg of
SetYear yearString -> yearString

main =
beginnerProgram
{ model = ""
, view = view
, update = update
}

Sample run for the year 2003; copied and pasted from web-page:

Output:
January 31, 2003
February 28, 2003
March 28, 2003
April 25, 2003
May 30, 2003
June 27, 2003
July 25, 2003
August 29, 2003
September 26, 2003
October 31, 2003
November 28, 2003
December 26, 2003

## Emacs Lisp

(require 'calendar)

(defun last-friday (year)
"Print the last Friday 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))
(lastfriday (1+ (cl-position 5 weekdays :from-end t))))
(insert (format "%i-%02i-%02i \n" year month lastfriday))))
(number-sequence 1 12)))

(last-friday 2012)
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Erlang

-module( last_date_each_month ).

-export( [monday/1, tuesday/1, wednesday/1, thursday/1, friday/1, saturday/1, sunday/1] ).

monday( Year ) -> last( Year, 1 ).
tuesday( Year ) -> last( Year, 2 ).
wednesday( Year ) -> last( Year, 3 ).
thursday( Year ) -> last( Year, 4 ).
friday( Year ) -> last( Year, 5 ).
saturday( Year ) -> last( Year, 6 ).
sunday( Year ) -> last( Year, 7 ).

last( Year, Week_day ) ->
Months = lists:seq( 1, 12 ),
Months_days = [{X, Y} || X <- Months, Y <- lists:seq(calendar:last_day_of_the_month(Year, X), calendar:last_day_of_the_month(Year, X) - 7, -1), calendar:valid_date(Year, X, Y), calendar:day_of_the_week(Year, X, Y) =:= Week_day],
[{Year, X, proplists:get_value(X, Months_days)} || X <- Months].
Output:
32> [io:fwrite("~B-~2.10.0B-~B~n", [Y,M,D]) || {Y,M,D} <- last_date_each_month:friday(2012)].
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Factor

Works with: Factor version 0.98

The last-friday-of-month word in the calendar vocabulary does most of the work. This program expects the year as a command line argument.

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

(command-line) second string>number <year> 12 <iota>
[ months time+ last-friday-of-month ] with map
[ timestamp>ymd print ] each
Output:
>factor last-fridays.factor 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Fortran

Algorithm: compute day of week for last day of month, then subtract just enough to get to the preceding friday. Do this for each month. To simplify computations further, we only need to compute day of week of january 1st (the others are found by adding month lengths). Since day of week need only be known modulo 7, we do not compute modulo at all except once when subtracting.

program fridays
implicit none
integer :: days(1:12) = (/31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31/)
integer :: year, k, y, m
if (mod(year, 400) == 0 .or. (mod(year, 4) == 0 .and. mod(year, 100) /= 0)) days(2) = 29
y = year - 1
k = 44 + y + y/4 + 6*(y/100) + y/400
do m = 1, 12
k = k + days(m)
print "(I4,A1,I2.2,A1,I2)", year, '-', m, '-', days(m) - mod(k, 7)
end do
end program

## 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)
End With
Next

Do

Do
Print "For what year do you want to find the last Friday 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 Friday 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)
If x <> 5 Then d = d - IIf(x > 5, x - 5, x + 2)
Print d; " "; arr(i).m_name
Next

' empty key buffer
While InKey <> "" : keypress = InKey : Wend
Print : Print
Print "Find last Friday 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 Friday of the month
any number below 1800 stops program, year in YYYY format? 2017

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

## Frink

d = parseDate[ARGS@0]
for m = 1 to 12
{
d = beginningOfNextMonth[d]
n = d - (((parseInt[d -> ### u ###] + 1) mod 7) + 1) days
println[n -> ### yyyy-MM-dd ###]
}
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## 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.

## Gambas

Public Sub Form_Open()
Dim siYear As Short = InputBox("Please input a year", "Last Friday of each month")
Dim siMonth, siDay As Short
Dim dDay As Date

For siMonth = 1 To 12
For siDay = 31 DownTo 22
Try dDay = Date(siYear, siMonth, siDay)
If Error Then Continue
If WeekDay(dDay) = 5 Then
Print Format(dDay, "yyyy-mm-dd");;
Print Space(6) & Format(dDay, "dddd dd mmmm yyyy")
Break
End If
Next
Next

Me.Close

End

Output:

1925-01-30       Friday 30 January 1925
1925-02-27       Friday 27 February 1925
1925-03-27       Friday 27 March 1925
1925-04-24       Friday 24 April 1925
1925-05-29       Friday 29 May 1925
1925-06-26       Friday 26 June 1925
1925-07-31       Friday 31 July 1925
1925-08-28       Friday 28 August 1925
1925-09-25       Friday 25 September 1925
1925-10-30       Friday 30 October 1925
1925-11-27       Friday 27 November 1925
1925-12-25       Friday 25 December 1925

## Go

package main

import (
"fmt"
"os"
"strconv"
"time"
)

func main() {
y := time.Now().Year()
if len(os.Args) == 2 {
if i, err := strconv.Atoi(os.Args[1]); err == nil {
y = i
}
}
for m := time.January; m <= time.December; m++ {
d := time.Date(y, m+1, 1, 0, 0, 0, 0, time.UTC).Add(-24 * time.Hour)
d = d.Add(-time.Duration((d.Weekday()+7-time.Friday)%7) * 24 * time.Hour)
fmt.Println(d.Format("2006-01-02"))
}
}
Output:
> ./fridays 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Groovy

Solution: Same as Find last Sunday of each month

Test:

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

Execution (Cygwin on Windows 7):

[2273] groovy lastFridays.groovy 2012
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

import Data.Time.Calendar
import Data.Time.Calendar.WeekDate (toWeekDate)
import Data.List (transpose, intercalate)

-- [1 .. 7] for [Mon .. Sun]
findWeekDay :: Int -> Day -> Day
findWeekDay dayOfWeek date =
(filter
(\x ->
let (_, _, day) = toWeekDate x
in day == dayOfWeek)
((`addDays` date) <\$> [-6 .. 0]))

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

main :: IO ()
main =
mapM_
putStrLn
(intercalate "  " <\$> transpose (weekDayDates 5 <\$> [2012 .. 2017]))
Output:
2012-01-27  2013-01-25  2014-01-31  2015-01-30  2016-01-29  2017-01-27
2012-02-24  2013-02-22  2014-02-28  2015-02-27  2016-02-26  2017-02-24
2012-03-30  2013-03-29  2014-03-28  2015-03-27  2016-03-25  2017-03-31
2012-04-27  2013-04-26  2014-04-25  2015-04-24  2016-04-29  2017-04-28
2012-05-25  2013-05-31  2014-05-30  2015-05-29  2016-05-27  2017-05-26
2012-06-29  2013-06-28  2014-06-27  2015-06-26  2016-06-24  2017-06-30
2012-07-27  2013-07-26  2014-07-25  2015-07-31  2016-07-29  2017-07-28
2012-08-31  2013-08-30  2014-08-29  2015-08-28  2016-08-26  2017-08-25
2012-09-28  2013-09-27  2014-09-26  2015-09-25  2016-09-30  2017-09-29
2012-10-26  2013-10-25  2014-10-31  2015-10-30  2016-10-28  2017-10-27
2012-11-30  2013-11-29  2014-11-28  2015-11-27  2016-11-25  2017-11-24
2012-12-28  2013-12-27  2014-12-26  2015-12-25  2016-12-30  2017-12-29

## Icon and Unicon

This will write the last fridays for every year given as an argument. There is no error checking on the year.

procedure main(A)
every write(lastfridays(!A))
end

procedure lastfridays(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 = 4 do z -:=1 # backup to last friday=4
suspend sprintf("%d-%d-%d",year,m,d+z)
}
end

Output:
last_fridays.exe 2012
2012-1-27
2012-2-24
2012-3-30
2012-4-27
2012-5-25
2012-6-29
2012-7-27
2012-8-31
2012-9-28
2012-10-26
2012-11-30
2012-12-28

## J

require 'dates'
last_fridays=: 12 {. [: ({:/.~ }:"1)@(#~ 5 = weekday)@todate (i.366) + todayno@,&1 1

In other words, start from January 1 of the given year, and count forward for 366 days, keeping the Fridays. Then pick the last remaining day within each represented month (which will be a Friday because we only kept the Fridays). Then pick the first 12 (since on a non-leap year which ends on a Thursday we would get an extra Friday).

Example use:

last_fridays 2012
2012  1 27
2012  2 24
2012  3 30
2012  4 27
2012  5 25
2012  6 29
2012  7 27
2012  8 31
2012  9 28
2012 10 26
2012 11 30
2012 12 28

## Java

Works with: Java version 1.5+
import java.text.*;
import java.util.*;

public class LastFridays {

public static void main(String[] args) throws Exception {
int year = Integer.parseInt(args[0]);
GregorianCalendar c = new GregorianCalendar(year, 0, 1);

for (String mon : new DateFormatSymbols(Locale.US).getShortMonths()) {
if (!mon.isEmpty()) {
int totalDaysOfMonth = c.getActualMaximum(Calendar.DAY_OF_MONTH);
c.set(Calendar.DAY_OF_MONTH, totalDaysOfMonth);

int daysToRollBack = (c.get(Calendar.DAY_OF_WEEK) + 1) % 7;

int day = totalDaysOfMonth - daysToRollBack;
c.set(Calendar.DAY_OF_MONTH, day);

System.out.printf("%d %s %d\n", year, mon, day);

c.set(year, c.get(Calendar.MONTH) + 1, 1);
}
}
}
}
Output:
(for java LastFridays 2012)
2012 Jan 27
2012 Feb 24
2012 Mar 30
2012 Apr 27
2012 May 25
2012 Jun 29
2012 Jul 27
2012 Aug 31
2012 Sep 28
2012 Oct 26
2012 Nov 30
2012 Dec 28

## JavaScript

### ES5

#### Iteration

Works with: Nodejs
var last_friday_of_month, print_last_fridays_of_month;

last_friday_of_month = function(year, month) {
var i, last_day;
i = 0;
while (true) {
last_day = new Date(year, month, i);
if (last_day.getDay() === 5) {
return last_day.toDateString();
}
i -= 1;
}
};

print_last_fridays_of_month = function(year) {
var month, results;
results = [];
for (month = 1; month <= 12; ++month) {
results.push(console.log(last_friday_of_month(year, month)));
}
return results;
};

(function() {
var year;
year = parseInt(process.argv[2]);
return print_last_fridays_of_month(year);
})();
Output:
>node lastfriday.js  2015
Fri Jan 30 2015
Fri Feb 27 2015
Fri Mar 27 2015
Fri Apr 24 2015
Fri May 29 2015
Fri Jun 26 2015
Fri Jul 31 2015
Fri Aug 28 2015
Fri Sep 25 2015
Fri Oct 30 2015
Fri Nov 27 2015
Fri Dec 25 2015

#### Functional composition

(function () {
'use strict';

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

// 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(lastFridaysOfYear)
)
.map(function (row) {
return row
.map(isoDateString)
.join('\t');
})
.join('\n');
})();
Output:
2012-01-27	2013-01-25	2014-01-31	2015-01-30	2016-01-29
2012-02-24	2013-02-22	2014-02-28	2015-02-27	2016-02-26
2012-03-30	2013-03-29	2014-03-28	2015-03-27	2016-03-25
2012-04-27	2013-04-26	2014-04-25	2015-04-24	2016-04-29
2012-05-25	2013-05-31	2014-05-30	2015-05-29	2016-05-27
2012-06-29	2013-06-28	2014-06-27	2015-06-26	2016-06-24
2012-07-27	2013-07-26	2014-07-25	2015-07-31	2016-07-29
2012-08-31	2013-08-30	2014-08-29	2015-08-28	2016-08-26
2012-09-28	2013-09-27	2014-09-26	2015-09-25	2016-09-30
2012-10-26	2013-10-25	2014-10-31	2015-10-30	2016-10-28
2012-11-30	2013-11-29	2014-11-28	2015-11-27	2016-11-25
2012-12-28	2013-12-27	2014-12-26	2015-12-25	2016-12-30

### ES6

(() => {
"use strict";

// ------------ LAST FRIDAY OF EACH MONTH ------------

// lastWeekDaysOfYear :: Int -> Int -> [Date]
const lastWeekDaysOfYear = iWeekDay =>
y => {
const isLeap = n => (
(0 === n % 4) && (0 !== n % 100)) || (
0 === y % 400
);

return [
31, isLeap(y) ? 29 : 28,
31, 30, 31, 30, 31, 31, 30, 31, 30, 31
]
.map((d, m) =>
new Date(Date.UTC(
y, m, d - ((
new Date(Date.UTC(
y, m, d
))
.getDay() + (7 - iWeekDay)
) % 7)
))
);
};

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

// ---------------------- TEST -----------------------
const main = () =>
transpose(
enumFromTo(2015)(2019)
.map(lastWeekDaysOfYear(days.friday))
)
.map(
row => row.map(isoDateString).join("\t")
)
.join("\n");

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

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

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

// transpose :: [[a]] -> [[a]]
const transpose = rows =>
// The columns of the input transposed
// into new rows.
0 < rows.length ? rows[0].map(
(x, i) => rows.flatMap(
v => v[i]
)
) : [];

// MAIN ---
return main();
})();
Output:
2015-01-30    2016-01-29    2017-01-27    2018-01-26    2019-01-25
2015-02-27    2016-02-26    2017-02-24    2018-02-23    2019-02-22
2015-03-27    2016-03-25    2017-03-31    2018-03-30    2019-03-29
2015-04-24    2016-04-29    2017-04-28    2018-04-27    2019-04-26
2015-05-29    2016-05-27    2017-05-26    2018-05-25    2019-05-31
2015-06-26    2016-06-24    2017-06-30    2018-06-29    2019-06-28
2015-07-31    2016-07-29    2017-07-28    2018-07-27    2019-07-26
2015-08-28    2016-08-26    2017-08-25    2018-08-31    2019-08-30
2015-09-25    2016-09-30    2017-09-29    2018-09-28    2019-09-27
2015-10-30    2016-10-28    2017-10-27    2018-10-26    2019-10-25
2015-11-27    2016-11-25    2017-11-24    2018-11-30    2019-11-29
2015-12-25    2016-12-30    2017-12-29    2018-12-28    2019-12-27

## 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 is 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 ;

findLastFridays

# year and month are numbered conventionally
def findLastFriday(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) == 6 ; .-1);

# input: year
def findLastFridays:
def months:
["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"];
. as \$year
| "YEAR: \(.)",
(range(0;12) | "\(months[.]) \(findLastFriday(\$year; .+1))") ;

\$year|tonumber|findLastFridays
Output:
\$ jq --arg year 2012 -n -r -f findLastFridays.jq
YEAR: 2012
January 27
February 24
March 30
April 27
May 25
June 29
July 27
August 31
September 28
October 26
November 30
December 28

## Julia

using Dates

const wday = Dates.Fri
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> ")
0 < length(y) || break
y = try
parseint(y)
catch
println("Sorry, but \"", y, "\" 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
Output:
This script will print the last Fridays of each month of the year given.
(Leave input empty to quit.)

Year> 2012

2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

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

Year>

## K

/ List the dates of last Fridays of each month of
/ a given year
/ lastfridt.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)}
lfd: {[m;y]; wdme[m;y];:[mewd>3;jd::jd+(4-mewd);jd::jd-(3+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))}
lfd1: {[y];arr::(); m:1; do[12;lfd[m;y];m+:1]}
main: {[y]; lfd1[y];`0: ,"Dates of last Fridays of ",(\$y);12 10#arr}

The output of a session is given below:

Output:
K Console - Enter \ for help

\l lastfridt
main 2012
Dates of last Fridays of 2012
("2012- 1-27"
"2012- 2-24"
"2012- 3-30"
"2012- 4-27"
"2012- 5-25"
"2012- 6-29"
"2012- 7-27"
"2012- 8-31"
"2012- 9-28"
"2012-10-26"
"2012-11-30"
"2012-12-28")

## Kotlin

// version 1.0.6

import java.util.*

fun main(args: Array<String>) {
print("Enter a year : ")

println("The last Fridays 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)
var offset = calendar[Calendar.DAY_OF_WEEK] - Calendar.FRIDAY
if (offset < 0) offset += 7
val lastFriday = daysInMonth - offset
println("\$year-" + "%02d-".format(month) + "%02d".format(lastFriday))
if (month < 12) {
}
}
}
Output:
Enter a year : 2012
The last Fridays of each month in 2012 are as follows:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Lasso

define isLeapYear(y::integer) => {
#y % 400 == 0 ? return true
#y % 100 == 0 ? return false
#y % 4 == 0 ? return true
return false
}
define fridays(y::integer) => {
local(out = array)
loop(12) => {
local(last = 28)
loop_count == 2 && isLeapYear(#y) ? #last = 29
array(4,6,9,11) >> loop_count ? #last == 30
#last == 28 && loop_count != 2 ? #last = 31
local(start = date(-year=#y,-month=loop_count,-day=#last))
while(#start->dayofweek != 6) => {
#start->subtract(-day=1)
}
#out->insert(#start)
}
return #out
}
with f in fridays(2012) do => {^
#f->format('%Q') + '\r'
^}
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## LiveCode

function lastFriday yyyy
-- 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
-- 6 is friday
if item 7 of startDate is 6 and item 1 of startDate is yyyy and item 2 of startDate is m then
put item 3 of startDate into fridays[item 2 of startDate]
end if
end repeat
end repeat
combine fridays using cr and space
sort fridays ascending numeric
return fridays
end lastFriday
Example
put lastFriday("2012")
Output
1 27
2 24
3 30
4 27
5 25
6 29
7 27
8 31
9 28
10 26
11 30
12 28

## Logo

; Determine if a Gregorian calendar year is leap
to leap? :year
output (and
equal? 0 modulo :year 4
not member? modulo :year 400 [100 200 300]
)
end

; Convert Gregorian calendar date to a simple day count from
; RD 1 = January 1, 1 CE
to day_number :year :month :day
local "elapsed make "elapsed difference :year 1
output (sum  product 365 :elapsed
int quotient :elapsed 4
minus int quotient :elapsed 100
int quotient :elapsed 400
int quotient difference product 367 :month 362 12
ifelse lessequal? :month 2 0 ifelse leap? :year -1 -2
:day)
end

; Find the day of the week from a day number, 0 = Sunday through 6 = Saturday
to day_of_week :day_number
output modulo :day_number 7
end

; Find the date of the last Friday of a given month
to last_friday :year :month
local "zero make "zero day_number :year :month 0
local "last make "last day_number :year sum 1 :month 0
local "wday make "wday day_of_week :last
local "friday make "friday sum :last remainder difference -2 :wday 7
output difference :friday :zero
end

local "year
make "year ifelse empty? :command.line 2012 :command.line

repeat 12 [
local "month make "month #
local "day make "day last_friday :year :month
if (less? :month 10) [make "month word "0 :month]
print reduce [(word ?1 "- ?2)] (list :year :month :day)
]
bye
Output:
\$  logo last_fridays.lg - 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## 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("Friday", tonumber(arg[1]))

Command line session:

>lua lastFridays.lua 2012
2012-1-27
2012-2-24
2012-3-30
2012-4-27
2012-5-25
2012-6-29
2012-7-27
2012-8-31
2012-9-28
2012-10-26
2012-11-30
2012-12-28

>

## Maple

fridays := 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) = 6) then
change := -(Calendar:-DayOfWeek(dt) mod 7)-1;
end if;
printf("%d-%d-%d\n", year, Month(dt), DayOfMonth(dt));
end do;
end proc;

fridays(2012);
Output:
2012-1-27
2012-2-24
2012-3-30
2012-4-27
2012-5-25
2012-6-29
2012-7-27
2012-8-31
2012-9-28
2012-10-26
2012-11-30
2012-12-28

## Mathematica/Wolfram Language

FridaysOfYear[Y_] :=
NestWhile[(DaysPlus[#, - 1]) &, #, (DateString[#, "DayName"] != "Friday") &] & /@
Most@Reverse@NestList [DaysPlus[# /. {x_, y_, X_} -> {x, y, 1}, - 1] &, {Y + 1, 1, 1}, 12]
Column@FridaysOfYear[2012]
Output:
{2012,1,27}
{2012,2,24}
{2012,3,30}
{2012,4,27}
{2012,5,25}
{2012,6,29}
{2012,7,27}
{2012,8,31}
{2012,9,28}
{2012,10,26}
{2012,11,30}
{2012,12,28}

## MATLAB / Octave

function t = last_fridays_of_year(y)
t1 = datenum([y,1,1,0,0,0]);
t2 = datenum([y,12,31,0,0,0]);
t  = datevec(t1:t2);
t  = t(strmatch('Friday', datestr(t,'dddd')), :);     % find all Fridays
t  = t([find(diff(t(:,2)) > 0); end], :);     % find Fridays before change of month
end;

datestr(last_fridays_of_year(2012),'yyyy-mm-dd')
Output:
ans =
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Maxima

weekday(year,  month,  day) := block([m: month,  y: year,  k],
if m < 3 then (m: m + 12,  y: y - 1),
k: 1 + remainder(day + quotient((m + 1)*26,  10) + y + quotient(y,  4)
+ 6*quotient(y,  100) + quotient(y,  400) + 5,  7),
['monday,  'tuesday,  'wednesday,  'thurdsday,  'friday,  'saturday,  'sunday][k]
)\$

leapyearp(year) := is(mod(year,  4) = 0 and (mod(year,  100) # 0 or mod(year,  400) = 0))\$

lastfridays(year) := block(
[m: [31,  if leapyearp(year) then 29 else 28,  31,  30,  31,  30,  31,  31,  30,  31,  30,  31],  v: [ ]],
for month thru 12 do v: endcons(sconcat(year,  "-",  month,  "-",
lmax(sublist(makelist(i,  i,  1,  m[month]),  lambda([day],  weekday(year,  month,  day) = 'friday)))),  v),
v
)\$

lastfridays(2012);
["2012-1-27", "2012-2-24", "2012-3-30", "2012-4-27", "2012-5-25", "2012-6-29",
"2012-7-27","2012-8-31", "2012-9-28", "2012-10-26", "2012-11-30", "2012-12-28"]

## Nanoquery

import Nanoquery.Util

// a function to check if a year is a leap year
def isLeapYear(year)
if (year % 100 = 0)
return (year % 400 = 0)
else
return (year % 4 = 0)
end
end

// a function to format 1-digit numbers as "0x"
def form(num)
if (num > 9)
return str(num)
else
return "0" + str(num)
end
end

// get a year from the console
print "enter year: "
year = int(input())

// build a list of the expected amount of days for each month
days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
if isLeapYear(\$year)
days[1] = 29
end

// loop through each month
for month in range(1, len(\$days))
// loop through each day of the month
friday = null
for day in range(1, days[month - 1])
// create a date object for this date
date = new(Date)
date.setYear(year).setMonth(month).setDay(day)

// check if it's a friday
if (date.getDayOfWeek() = "Friday")
// if it is, keep it
friday = new(Date, date)
end
end for

// display the last friday found
print   friday.getYear() + "-"
print   form(friday.getMonth()) + "-"
println form(friday.getDay())
end

## NetRexx

Translation of: Java
Translation of: C

Implements the algorithms from both the Java and C implementations.

/* NetRexx */
options replace format comments java crossref symbols nobinary

import java.text.

runSample(arg)
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method lastFridayByLib(year) public static

cal = GregorianCalendar(year, 0, 1)

loop mon over DateFormatSymbols().getShortMonths()
if \mon.isEmpty() then do
totalDaysOfMonth = cal.getActualMaximum(Calendar.DAY_OF_MONTH)
cal.set(Calendar.DAY_OF_MONTH, totalDaysOfMonth)

daysToRollBack = (cal.get(Calendar.DAY_OF_WEEK) + 1) // 7

day = totalDaysOfMonth - daysToRollBack
cal.set(Calendar.DAY_OF_MONTH, day)

say year.right(4, 0) mon day.right(2, 0)

cal.set(year, cal.get(Calendar.MONTH) + 1, 1)
end
end mon
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method lastFridayCalc(year) public static binary signals BadArgumentException

if year <= 1700 then do
signal BadArgumentException(year 'is out of range')
end

wk  = int
mth = int
yr  = int year
days = [int 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]  -- days in month
days[1] = days[1] - ((yr // 4) | \(yr // 100) & (yr // 400)) -- adjust for leap year

wk = yr * 365 + (yr - 1) % 4 - (yr - 1) % 100 + (yr - 1) % 400 + 6 -- week number

loop mth = 0 to 11
wk = (wk + days[mth]) // 7
wx = int
if wk < 5 then wx = -2
else wx = 5
yy = Rexx(yr)
mm = Rexx(mth + 1)
dd = Rexx(days[mth] + wx - wk)
say yy.right(4, 0)'-'mm.right(2, 0)'-'dd.right(2, 0)
end mth
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
do
parse arg year .
if year = '' | year = '.' then year = 2012
dlm = '-'
dlm = dlm.left(60, dlm)
say
say 'Using Java calendar libraries'
say dlm
lastFridayByLib(year)
say
say 'Calculated'
say dlm
lastFridayCalc(year)
catch ex = Exception
ex.printStackTrace
end
return
Output:
Using Java calendar libraries
------------------------------------------------------------
2012 Jan 27
2012 Feb 24
2012 Mar 30
2012 Apr 27
2012 May 25
2012 Jun 29
2012 Jul 27
2012 Aug 31
2012 Sep 28
2012 Oct 26
2012 Nov 30
2012 Dec 28

Calculated
------------------------------------------------------------
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## 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] = [3, 4, 5, 6, 0, 1, 2]

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")
Output:

Sample usage: ./lastfridays 2012

2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## OCaml

Using the module Unix from the standard OCaml library:

open Unix

let usage() =
Printf.eprintf "%s <year>\n" Sys.argv.(0);
exit 1

let print_date t =
Printf.printf "%d-%02d-%02d\n" (t.tm_year + 1900) (t.tm_mon + 1) t.tm_mday

let is_date_ok tm t =
(tm.tm_year = t.tm_year &&
tm.tm_mon  = t.tm_mon  &&
tm.tm_mday = t.tm_mday)

let () =
let _year =
try int_of_string Sys.argv.(1)
with _ -> usage()
in
let year = _year - 1900 in
let fridays = Array.make 12 (Unix.gmtime 0.0) in
for month = 0 to 11 do
for day_of_month = 1 to 31 do
let tm = { (Unix.gmtime 0.0) with
tm_year = year;
tm_mon = month;
tm_mday = day_of_month;
} in
let _, t = Unix.mktime tm in
if is_date_ok tm t  (* check for months that have less than 31 days *)
&& t.tm_wday = 5  (* is a friday *)
then fridays.(month) <- t
done;
done;
Array.iter print_date fridays
Output:
\$ ocaml last_fridays.ml 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

### With a dedicated library

open CalendarLib

let usage() =
Printf.eprintf "%s <year>\n" Sys.argv.(0);
exit 1

let print_date (year, month, day) =
Printf.printf "%d-%02d-%02d\n" year month day

let () =
let year =
try int_of_string Sys.argv.(1)
with _ -> usage()
in
let fridays = ref [] in
for month = 1 to 12 do
let num_days = Date.days_in_month (Date.make_year_month year month) in
let rec aux day =
if Date.day_of_week (Date.make year month day) = Date.Fri
then fridays := (year, month, day) :: !fridays
else aux (pred day)
in
aux num_days
done;
List.iter print_date (List.rev !fridays)

Run this script with the command:

ocaml unix.cma str.cma -I +calendar calendarLib.cma last_fridays.ml 2012

## Oforth

import: date

: lastFridays(y)
| m |
Date.JANUARY Date.DECEMBER for: m [
Date newDate(y, m, Date.DaysInMonth(y, m))
while(dup dayOfWeek Date.FRIDAY <>) [ addDays(-1) ]
println
] ;
Output:
2012-01-27 00:00:00,000
2012-02-24 00:00:00,000
2012-03-30 00:00:00,000
2012-04-27 00:00:00,000
2012-05-25 00:00:00,000
2012-06-29 00:00:00,000
2012-07-27 00:00:00,000
2012-08-31 00:00:00,000
2012-09-28 00:00:00,000
2012-10-26 00:00:00,000
2012-11-30 00:00:00,000
2012-12-28 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([2012,m+1,0]); print(njdate(a-(a+1)%7)))
Output:
[2012, 1, 27]
[2012, 2, 24]
[2012, 3, 30]
[2012, 4, 27]
[2012, 5, 25]
[2012, 6, 29]
[2012, 7, 27]
[2012, 8, 31]
[2012, 9, 28]
[2012, 10, 26]
[2012, 11, 30]
[2012, 12, 28]

## Pascal

Works with: Free Pascal

Using Free Pascal's DateUtils library would dramatically simplify the coding (see the Delphi example) but for older Pascal implementations the needed routines are the programmer's responsibility.

program LastFriday;

{\$mode objfpc}{\$H+}

uses
SysUtils;

type
weekdays = (Sun,Mon,Tue,Wed,Thu,Fri,Sat);

var
m, d, y : integer;

function IsLeapYear(Year : integer) : boolean;
begin
if Year mod 4 <> 0  { quick exit in most likely case }
then IsLeapYear := false
else if Year mod 400 = 0
then IsLeapYear := true
else if Year mod 100 = 0
then IsLeapYear := false
else { non-century year and divisible by 4 }
IsLeapYear := true;
end;

function DaysInMonth(Month, Year : integer) : integer;
const
LastDay : array[1..12] of integer =
(31,28,31,30,31,30,31,31,30,31,30,31);
begin
if (Month = 2) and (IsLeapYear(Year)) then
DaysInMonth := 29
else
DaysInMonth := LastDay[Month];
end;

{ return day of week (Sun = 0, Mon = 1, etc.) for a }
{ given mo, da, and yr using Zeller's congruence    }
function DayOfWeek(mo, da, yr : integer) : weekdays;
var
y, c, z : integer;
begin
if mo < 3 then
begin
mo := mo + 10;
yr := yr - 1
end
else mo := mo - 2;
y := yr mod 100;
c := yr div 100;
z := (26 * mo - 2) div 10;
z := z + da + y + (y div 4) + (c div 4) - 2 * c + 777;
DayOfWeek := weekdays(z mod 7);
end;

{ return the calendar day of the last occurrence of the }
{ specified weekday in the given month and year         }
function LastWeekday(k : weekdays; m, y : integer) : integer;
var
d : integer;
w : weekdays;
begin
{ determine weekday for the last day of the month }
d := DaysInMonth(m, y);
w := DayOfWeek(m, d, y);
{ back up as needed to desired weekday }
if w >= k then
LastWeekday := d - (ord(w) - ord(k))
else
LastWeekday := d - (7 - ord(k)) - ord(w);
end;

begin { main program }
write('Find last Fridays in what year? ');
writeln;
writeln('Month  Last Fri');
for m := 1 to 12 do
begin
d  := LastWeekday(Fri, m, y);
writeln(m:5,'   ',d:5);
end;
end.
Output:
Find last Fridays in what year? 2020
Month  Last Fri
1      31
2      28
3      27
4      24
5      29
6      26
7      31
8      28
9      25
10      30
11      27
12      25

## Perl

#!/usr/bin/perl -w
use strict ;
use DateTime ;
use feature qw( say ) ;

foreach my \$month ( 1..12 ) {
my \$dt = DateTime->last_day_of_month( year => \$ARGV[ 0 ] , month => \$month ) ;
while ( \$dt->day_of_week != 5 ) {
\$dt->subtract( days => 1 ) ;
}
say \$dt->ymd ;
}
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Phix

with javascript_semantics
procedure last_day_of_month(integer y, dow)
for m=1 to 12 do
integer d = days_in_month(y,m),
a = dow-day_of_week(y,m,d)
printf(1,"%4d-%02d-%02d\n",{y,m,d+a-7*(a>0)})
end for
end procedure
constant FRIDAY=5
--prompt_number() is not compatible with pwa/p2js
--last_day_of_month(prompt_number("Year:",{1752,9999}),FRIDAY)
last_day_of_month(2012,FRIDAY)
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## PHP

PHP is generally used for web apps, so I am not implementing the command-line component of this task.

<?php
function last_friday_of_month(\$year, \$month) {
\$day = 0;
while(True) {
\$last_day = mktime(0, 0, 0, \$month+1, \$day, \$year);
if (date("w", \$last_day) == 5) {
return date("Y-m-d", \$last_day);
}
\$day -= 1;
}
}

function print_last_fridays_of_month(\$year) {
foreach(range(1, 12) as \$month) {
echo last_friday_of_month(\$year, \$month), "<br>";
}
}

date_default_timezone_set("GMT");
\$year = 2012;
print_last_fridays_of_month(\$year);
?>
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Picat

% for command line argument
main(ARGV) =>
if ARGV.length > 0 then
Year = ARGV[1].to_integer(),
show_year(Year),
nl
end.

% Without command line argument
main => go.

go =>
show_year(2022),
nl.

% Show the months
show_year(Year) =>
foreach(Date in get_months(Year))
println(format_date(Date))
end,
nl.

% Format date to YYYY-DD-MM
format_date(Date) = to_fstring("%4d-%02d-%02d",Date[1],Date[2],Date[3]).

% Return the last Fridays of each month for year Year
get_months(Year) =
[ [ [Year,Month,Day] : Day in 1..max_days_in_month(Year,Month),
dow(Year, Month, Day) == 5].last() : Month in 1..12].

% Day of week, Sakamoto's method
dow(Y, M, D) = R =>
T = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4],
if M < 3 then
Y := Y - 1
end,
R = (Y + Y // 4 - Y // 100 + Y // 400 + T[M] + D) mod 7.

% Maximum days in month
max_days_in_month(Year,Month) = Days =>
if member(Month, [1,3,5,7,8,10,12]) then
Days = 31
elseif member(Month,[4,6,9,11]) then
Days = 30
else
if leap_year(Year) then
Days = 29
else
Days = 28
end
end.

% Is Year a leap year?
leap_year(Year) =>
(Year mod 4 == 0, Year mod 100 != 0)
;
Year mod 400 == 0.

### Running the program

There are several ways to run this program; let's call it "last_friday_of_each_month.pi":

### From Picat's shell

\$ picat
Picat> cl(last_friday_of_each_month)
Picat> show_year(2022)
2022-01-28
2022-02-25
2022-03-25
2022-04-29
2022-05-27
2022-06-24
2022-07-29
2022-08-26
2022-09-30
2022-10-28
2022-11-25
2022-12-30

### From the command line, year as parameter

Via main(ARGV).

\$ picat last_friday_of_each_month.pi 2022

### From the command line, as a goal

\$ picat -g "show_year(2022)" last_friday_of_each_month.pi

### Run the default goal (go/0))

\$ picat last_friday_of_each_month.pi

## PicoLisp

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

Test:

: (lastFridays 2012)
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Pike

int(0..1) last_friday(object day)
{
return day->week_day() == 5 &&
day->month_day() > day->month()->number_of_days()-7;
}

int main(int argc, array argv)
{
array days = filter(Calendar.Year((int)argv[1])->months()->days()[*], last_friday);
write("%{%s\n%}", days->format_ymd());
return 0;
}

## PL/I

Fridays: procedure (year) options (main); /* 8 January 2013 */
declare year character (4) varying;
declare start fixed binary (31);
declare months fixed decimal (2) initial (0);
declare (current_month, month_one_week_hence) character (2);

put list ('Last Fridays in each month for the year ' || year || ':' );
start = days('0101' || year, 'DDMMYYYY');
/* Find first Friday */
do while (weekday(start) ^= 6); start = start + 1; end;

do until (months=12);
current_month = substr (daystodate(start, 'MMDDYYYY'), 1, 2 );
month_one_week_hence = substr (daystodate(start+7, 'MMDDYYYY'), 1, 2 );
if current_month ^= month_one_week_hence then
do;
months = months + 1;
put skip list (daystodate(start, 'DDMmmYYYY'));
end;
start = start + 7;
end;
end Fridays;

The command: FRIDAYS /2008 produces:

Last Fridays in each month for the year 2008:
25Jan2008
29Feb2008
28Mar2008
25Apr2008
30May2008
27Jun2008
25Jul2008
29Aug2008
26Sep2008
31Oct2008
28Nov2008
26Dec2008
Output:
for 2013
Last Fridays in each month for the year 2013:
25Jan2013
22Feb2013
29Mar2013
26Apr2013
31May2013
28Jun2013
26Jul2013
30Aug2013
27Sep2013
25Oct2013
29Nov2013
27Dec2013

## 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) {
}
}
last-dayofweek 2012 "Friday"

Output:

2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

### 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 2012 -Last -Friday
Output:
Friday, January 27, 2012 12:00:00 AM
Friday, February 24, 2012 12:00:00 AM
Friday, March 30, 2012 12:00:00 AM
Friday, April 27, 2012 12:00:00 AM
Friday, May 25, 2012 12:00:00 AM
Friday, June 29, 2012 12:00:00 AM
Friday, July 27, 2012 12:00:00 AM
Friday, August 31, 2012 12:00:00 AM
Friday, September 28, 2012 12:00:00 AM
Friday, October 26, 2012 12:00:00 AM
Friday, November 30, 2012 12:00:00 AM
Friday, December 28, 2012 12:00:00 AM

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

1..12 | Get-Date0fDayOfWeek -Year 2012 -Last -Friday -AsString
Output:
27-Jan-2012
24-Feb-2012
30-Mar-2012
27-Apr-2012
25-May-2012
29-Jun-2012
27-Jul-2012
31-Aug-2012
28-Sep-2012
26-Oct-2012
30-Nov-2012
28-Dec-2012

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

1..12 | Get-Date0fDayOfWeek -Year 2012 -Last -Friday -AsString -Format yyyy-MM-dd
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## PureBasic

Procedure LastFridayOfEachMonth(yyyy.i,List lfem.i())
Define dv.i=ParseDate("%yyyy",Str(yyyy)), mv.i=1
NewList d.i()
For d=1 To 365
If DayOfWeek(dv)=5
EndIf
Next
dv=0
For mv=1 To 12
ForEach d()
If dv<d() And Month(d())=mv
dv=d()
EndIf
Next
Next
EndProcedure

NewList lf.i()
Define y.i
OpenConsole("Last Friday of each month")
Print("Input Year [ 1971 < y < 2038 ]: ")
y=Val(Input())
If y>1971 And y<2038
PrintN("Last Friday of each month...")
LastFridayOfEachMonth(y,lf())
ForEach lf()
PrintN(FormatDate("%dd.%mm.%yyyy",lf()))
Next
EndIf
Print("...End")
Input()
Output:
Input Year [ 1971 < y < 2038 ]: 2017
Last Friday of each month...
27.01.2017
24.02.2017
31.03.2017
28.04.2017
26.05.2017
30.06.2017
28.07.2017
25.08.2017
29.09.2017
27.10.2017
24.11.2017
29.12.2017
...End

## Python

import calendar

def last_fridays(year):
for month in range(1, 13):
last_friday = max(week[calendar.FRIDAY]
for week in calendar.monthcalendar(year, month))
print('{:4d}-{:02d}-{:02d}'.format(year, month, last_friday))
Output:
>>> last_fridays(2012)
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

Another solution

import calendar
c=calendar.Calendar()
fridays={}
year=raw_input("year")
for item in c.yeardatescalendar(int(year)):
for i1 in item:
for i2 in i1:
for i3 in i2:
if "Fri" in i3.ctime() and year in i3.ctime():
month,day=str(i3).rsplit("-",1)
fridays[month]=day

for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item

Using reduce

import calendar
c=calendar.Calendar()
fridays={}
year=raw_input("year")

if "Fri" in day.ctime() and year in day.ctime():
month,day=str(day).rsplit("-",1)
fridays[month]=day

for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item

using itertools

import calendar
from itertools import chain
f=chain.from_iterable
c=calendar.Calendar()
fridays={}
year=raw_input("year")

for day in f(f(f(c.yeardatescalendar(int(year))))):

if "Fri" in day.ctime() and year in day.ctime():
month,day=str(day).rsplit("-",1)
fridays[month]=day

for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item

## 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 -->        )

[ 5 lastwkdays ]                 is lastfridays (           year -->        )

2012 lastfridays
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## R

year = commandArgs(T)
d = as.Date(paste0(year, "-01-01"))
fridays = d + seq(by = 7,
(5 - as.POSIXlt(d)\$wday) %% 7,
364 + (months(d + 30 + 29) == "February"))
message(paste(collapse = "\n", fridays[tapply(
seq_along(fridays), as.POSIXlt(fridays)\$mon, max)]))

## 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-fridays y)
(for/list ([m (in-range 1 13)])
(prev-friday (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-friday d)
(if (eq? (week-day d) 'fri)
d
(prev-friday (prev-day d))))

(for ([d (last-fridays 2012)])
(displayln (~a (date->string d "~a ~d ~b ~Y"))))
Output:
Fri 27 Jan 2012
Fri 24 Feb 2012
Fri 30 Mar 2012
Fri 27 Apr 2012
Fri 25 May 2012
Fri 29 Jun 2012
Fri 27 Jul 2012
Fri 31 Aug 2012
Fri 28 Sep 2012
Fri 26 Oct 2012
Fri 30 Nov 2012
Fri 28 Dec 2012

## Raku

(formerly Perl 6)

sub MAIN (Int \$year = Date.today.year) {
my @fri;
for Date.new("\$year-01-01") .. Date.new("\$year-12-31") {
@fri[.month] = .Str if .day-of-week == 5;
}
.say for @fri[1..12];
}

Example:

\$ ./lastfri 2038
2038-01-29
2038-02-26
2038-03-26
2038-04-30
2038-05-28
2038-06-25
2038-07-30
2038-08-27
2038-09-24
2038-10-29
2038-11-26
2038-12-31

A solution without a result array to store things in:

sub MAIN (Int \$year = Date.today.year) {
say ~.value.reverse.first: *.day-of-week == 5
for classify *.month, Date.new("\$year-01-01") .. Date.new("\$year-12-31");
}

Here, classify sorts the dates into one bin per month (but preserves the order in each bin). We then take the list inside each bin (.value) and find the last (.reverse.first) date which is a Friday.

Another variation where the data flow can be read left to right using feed operators:

sub MAIN (Int \$year = Date.today.year) {
.say for Date.new("\$year-01-01") .. Date.new("\$year-12-31") ==> classify *.month ==>
map *.value.reverse.first: *.day-of-week == 5
}

## REBOL

The longer version:

leap-year?:  function [year] [to-logic attempt [to-date reduce [29 2 year]]]

days-in-feb: function [year] [either leap-year? year [29] [28]]

days-in-month: function [month year] [
do pick [31 (days-in-feb year) 31 30 31 30 31 31 30 31 30 31] month
]

last-day-of-month: function [month year] [
to-date reduce [year month  days-in-month month year]
]

last-weekday-of-month: function [weekday month year] [
d: last-day-of-month month year
while [d/weekday != weekday] [d/day: d/day - 1]
d
]

last-friday-of-month: function [month year] [last-weekday-of-month 5 month year]

year: to-integer input
repeat month 12 [print last-friday-of-month month year]
Output:
rebol last-fridays.reb <<< 2012
27-Jan-2012
24-Feb-2012
30-Mar-2012
27-Apr-2012
25-May-2012
29-Jun-2012
27-Jul-2012
31-Aug-2012
28-Sep-2012
26-Oct-2012
30-Nov-2012
28-Dec-2012

A shorter version:

last-fridays-of-year: function [year] [
collect [
repeat month 12 [
d: to-date reduce [1 month year]
d/month: d/month + 1                      ; start of next month
until [d/day: d/day - 1  d/weekday = 5]   ; go backwards until find a Friday
keep d
]
]
]

foreach friday last-fridays-of-year to-integer input [print friday]

NB. See "Find the last Sunday of each month" Rosetta for alternative (even more succinct) solution

## REXX

This REXX program will find the last day-of-week (for any day) of all the months for any year.
It wasn't optimized just to find a particular day-of-week.

The documentation for the   lastDOW   function   (used in the REXX program below):

╔════════════════════════════════════════════════════════════════════════════════════════════════╗
║ 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.:    TU  for Tuesday.    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     4=April     5=May     6=June    ...   12=December  ║
║ or the English  name  of the month,  with a minimum length that causes no ambiguity.           ║
║ I.E.:    JA  for January,   AP  for April,   JUN  for June,   JUL  for July,   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.                                                                      ║
║                                                                                                ║
║ The method used is:  find the "day number" of the 1st of the next month and 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.                    ║
╚════════════════════════════════════════════════════════════════════════════════════════════════╝
/*REXX program displays the dates of the  last Fridays of each month for any given year.*/
parse arg yyyy                                   /*obtain optional argument from the CL.*/
do j=1  for 12                  /*traipse through all the year's months*/
say lastDOW('Friday', j, yyyy)  /*find last Friday for the  Jth  month.*/
end  /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
lastDOW: procedure;  arg dow .,mm .,yy .;      parse arg a.1,a.2,a.3 /*DOW = day of week*/
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   /*no month given ? */
if arg()>3                 then call .er 'arguments specified',  4   /*too many args  ? */

do j=1  for 3                                                      /*any plural args ?*/
if words( arg(j) ) > 1   then call .er 'is illegal:',   j          /*check if plural. */
end
/*find DOW in list.*/
dw= pos(' 'dow, \$)                                                   /*find  day-of-week*/
if dw==0                   then call .er 'is invalid:'  , 1          /*no DOW was found?*/
if dw\==lastpos(' 'dow,\$)  then call .er 'is ambiguous:', 1          /*check min length.*/

if datatype(mm, 'M')  then do                                        /*is MM alphabetic?*/
m= pos(' 'mm, !)                          /*maybe its good...*/
if m==0                   then call .er 'is invalid:'  ,   1
if m\==lastpos(' 'mm,!)   then call .er 'is ambiguous:',   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   /*MM (mon) ¬integer*/
if \datatype(yy, 'W')   then call .er "isn't an integer:",       3   /*YY (yr)  ¬integer*/
if mm<1 | mm>12         then call .er "isn't in range 1──►12:",  2   /*MM out─of─range. */
if yy=0                 then call .er "can't be 0 (zero):",      3   /*YY can't be zero.*/
if yy<0                 then call .er "can't be negative:",      3   /* "   "    " neg. */
if yy>9999              then call .er "can't be > 9999:",        3   /* "   "    " huge.*/

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._  /*plug in a choice.*/
say;      exit 13                                               /*··· then exit.   */
output   when using the following input of:     2012     or     12
Friday 27 Jan 2012
Friday 24 Feb 2012
Friday 30 Mar 2012
Friday 27 Apr 2012
Friday 25 May 2012
Friday 29 Jun 2012
Friday 27 Jul 2012
Friday 31 Aug 2012
Friday 28 Sep 2012
Friday 26 Oct 2012
Friday 30 Nov 2012
Friday 28 Dec 2012

## Ring

see "What year to calculate (yyyy) : "
give year
see "Last Friday 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]
if year%4 = 0 and n<3
x = (month[n] + i) % 7 - 1
else x = (month[n] + i) % 7 ok
if  n < 10 strn = "0" + string(n) else strn = string(n) ok
if x = 2 see year + "-" + strn + "-" + string(i) + nl ok
next
next

Output:

What year to calculate (yyyy) : 2012
Last Fridays in 2012 are on :
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Ruby

require 'date'

def last_friday(year, month)
# Last day of month: Date.new interprets a negative number as a relative month/day from the end of year/month.
d = Date.new(year, month, -1)
d -= (d.wday - 5) % 7  # Subtract days after Friday.
end

year = Integer(ARGV.shift)
(1..12).each {|month| puts last_friday(year, month)}

Friday is d.wday == 5; the expression (d.wday - 5) % 7 counts days after Friday.

Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

Or get the last day of the month and go to the previous day until it's a Friday.

require 'date'

def last_friday(year, month)
d = Date.new(year, month, -1)
d = d.prev_day until d.friday?
d
end

## Run BASIC

input "Year:";yr
dayOne\$ = "01-01-";yr
n1	= date\$(dayOne\$)
for i = 1 to 12
n1  = n1 + 26
m1\$ = left\$(date\$(n1),2)
while  m1\$ = left\$(date\$(n1),2) ' find end of month
n1 = n1 + 1
wend
n1 = n1 -1
while (n1 Mod 7) <> 3 	  ' find Friday
n1 = n1 - 1
wend
print date\$(n1)		  ' print last Friday's date
next i
Year:?2013
01/25/2013
02/22/2013
03/29/2013
04/26/2013
05/31/2013
06/28/2013
07/26/2013
08/30/2013
09/27/2013
10/25/2013
11/29/2013
12/27/2013

## 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) % 7) + 1);
println!("{}", date - days_back);
});
}
Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Scala

import java.util.Calendar
import java.text.SimpleDateFormat

object Fridays {

def lastFridayOfMonth(year:Int, month:Int)={
val cal=Calendar.getInstance
cal.set(Calendar.YEAR, year)
cal.set(Calendar.MONTH, month)
cal.set(Calendar.DAY_OF_WEEK, Calendar.FRIDAY)
cal.set(Calendar.DAY_OF_WEEK_IN_MONTH, -1)
cal.getTime
}

def fridaysOfYear(year:Int)=for(month <- 0 to 11) yield lastFridayOfMonth(year, month)

def main(args:Array[String]){
val year=args(0).toInt
val formatter=new SimpleDateFormat("yyyy-MMM-dd")
fridaysOfYear(year).foreach{date=>
println(formatter.format(date))
}
}
}
Output:
2012-Jan-27
2012-Feb-24
2012-Mrz-30
2012-Apr-27
2012-Mai-25
2012-Jun-29
2012-Jul-27
2012-Aug-31
2012-Sep-28
2012-Okt-26
2012-Nov-30
2012-Dez-28

## Seed7

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

\$ include "seed7_05.s7i";
include "time.s7i";
include "duration.s7i";

const proc: main is func
local
var integer: weekday is 1; # 1 for monday, 2 for tuesday, and so on up to 7 for sunday.
var integer: year is 0;
var integer: month is 1;
var time: 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
end if;
end while;
writeln(strDate(selected));
end for;
end if;
end func;
Output:
when called with s7 rosetta/lastWeekdayInMonth 5 2013
2013-01-25
2013-02-22
2013-03-29
2013-04-26
2013-05-31
2013-06-28
2013-07-26
2013-08-30
2013-09-27
2013-10-25
2013-11-29
2013-12-27

## SenseTalk

put it into year

put !"The last Fridays of each month in [[year]] are:"

set lastDayOfMonth to year & "-01-31" -- start with January 31 of the year

repeat 12 times
set lastFriday to lastDayOfMonth
repeat until weekdayName of lastFriday is "Friday"
subtract a day from lastFriday -- work back to Friday
end repeat

put the monthName of lastFriday && ordinalWords of the day of lastFriday

add a month to lastDayOfMonth -- advance to last day of next month
end repeat
Output:
The last Fridays of each month in 2008 are:
January twenty-fifth
February twenty-ninth
March twenty-eighth
April twenty-fifth
May thirtieth
June twenty-seventh
July twenty-fifth
August twenty-ninth
September twenty-sixth
October thirty-first
November twenty-eighth
December twenty-sixth

## Sidef

Translation of: Perl
require('DateTime')
var (year=2016) = ARGV.map{.to_i}...

for month (1..12) {
var dt = %O<DateTime>.last_day_of_month(year => year, month => month)
while (dt.day_of_week != 5) {
dt.subtract(days => 1)
}
say dt.ymd
}
Output:
\$ sidef lastfriday.sf 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Smalltalk

Pharo Smalltalk

[ :yr | | firstDay firstFriday |
firstDay := Date year: yr month: 1 day: 1.
firstFriday := firstDay addDays: (6 - firstDay dayOfWeek).
(0 to: 53)
collect: [ :each | firstFriday addDays: (each * 7) ]
thenSelect: [ :each |
(((Date daysInMonth: each monthIndex forYear: yr) - each dayOfMonth) <= 6) and: [	each year = yr ] ] ]
Output:
Send value: 2012 to the above block to return an array:
(27 January 2012 24 February 2012 30 March 2012 27 April 2012 25 May 2012 29 June 2012 27 July 2012 31 August 2012 28 September 2012 26 October 2012 30 November 2012 28 December 2012)

## SQL

select to_char( next_day( last_day( add_months( to_date(
:yr||'01','yyyymm' ),level-1))-7,'Fri') ,'yyyy-mm-dd Dy') lastfriday
from dual
connect by level <= 12;
LASTFRIDAY
-----------------------
2012-01-27 Fri
2012-02-24 Fri
2012-03-30 Fri
2012-04-27 Fri
2012-05-25 Fri
2012-06-29 Fri
2012-07-27 Fri
2012-08-31 Fri
2012-09-28 Fri
2012-10-26 Fri
2012-11-30 Fri
2012-12-28 Fri

12 rows selected.

## Stata

program last_fridays
args year
clear
qui set obs 12
gen day=dofm(mofd(mdy(_n,1,`year'))+1)-1
qui replace day=day-mod(dow(day)-5,7)
format %td day
end

last_fridays 2012

+-----------+
| 27jan2012 |
| 24feb2012 |
| 30mar2012 |
| 27apr2012 |
| 25may2012 |
| 29jun2012 |
|-----------|
| 27jul2012 |
| 31aug2012 |
| 28sep2012 |
| 26oct2012 |
| 30nov2012 |
| 28dec2012 |
+-----------+

## Swift

import Foundation

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

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

for month in 2...13 {

let lastDayOfMonth = DateComponents(calendar: calendar,
year: year,
month: month,
day: 0,
hour: 12)

let date = calendar.date(from: lastDayOfMonth)!

let isFriday = calendar.component(.weekday, from: date) == 6

if isFriday {

dates.append(calendar.date(from: lastDayOfMonth)!)

} else {

let lastWeekofMonth = calendar.ordinality(of: .weekOfMonth,
in: .month,
for: date)!

let lastWithFriday = lastWeekofMonth - (calendar.component(.weekday, from: date) > 6 ? 0 : 1)

let lastFridayOfMonth = DateComponents(calendar: calendar,
year: year,
month: month - 1,
hour: 12,
weekday: 6,
weekOfMonth: lastWithFriday)

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

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

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

## Tcl

package require Tcl 8.5
set year [lindex \$argv 0]
foreach dm {02/1 03/1 04/1 05/1 06/1 07/1 08/1 09/1 10/1 11/1 12/1 12/32} {
# The [clock scan] code is unhealthily clever; use it for our own evil purposes
set t [clock scan "last friday" -base [clock scan \$dm/\$year -gmt 1] -gmt 1]
# Print the interesting part
puts [clock format \$t -format "%Y-%m-%d" -gmt 1]
}

Sample execution:

\$ tclsh8.5 lastfri.tcl 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## TUSCRIPT

\$\$ MODE TUSCRIPT
year=2012
LOOP month=1,12
LOOP day=31,22,-1
dayofweek=DATE (number,day,month,year,nummer)
IF (dayofweek==5) THEN
PRINT year,"-",month,"-",day
EXIT
ENDIF
ENDLOOP
ENDLOOP
Output:
2012-1-27
2012-2-24
2012-3-30
2012-4-27
2012-5-25
2012-6-29
2012-7-27
2012-8-31
2012-9-28
2012-10-26
2012-11-30
2012-12-28

## UNIX Shell

Using ncal. Will switch to Julian calender as ncal sees fit, and will not calculate past year 9999 (chances are you'll be too dead by then to worry about weekends anyway).

#!/bin/sh

if [ -z \$1 ]; then exit 1; fi

# weed out multiple erros due to bad year
ncal 1 \$1 > /dev/null && \
for m in 01 02 03 04 05 06 07 08 09 10 11 12; do
echo \$1-\$m-`ncal \$m \$1 | grep Fr | sed 's/.* \([0-9]\)/\1/'`
done

For systems without ncal:

#!/bin/sh

# usage: last_fridays [ year]

year=\${1:-`date +%Y`}    # default to current year
month=1
while [ 12 -ge \$month ]; do
# Ensure 2 digits: if we try to strip off 2 characters but it still
# looks the same, that means there was only 1 char, so we'll pad it.
[ "\$month" = "\${month%??}" ] && month=0\$month

cal \$month \$year | awk '{print \$6}' | grep . | tail -1 \
| sed "s@^@\$year-\$month-@"

# Strip leading zeros to avoid octal interpretation
month=\$(( 1 + \${month#0} ))
done

Using date --date from GNU date??? This code is not portable.

#!/bin/sh

# Free code, no limit work
# \$Id: lastfridays,v 1.1 2011/11/10 00:48:16 gilles Exp gilles \$

# usage :
# lastfridays 2012 # prints last fridays of months of year 2012

debug=\${debug:-false}
#debug=true

epoch_year_day() {
#set -x
x_epoch=`expr \${2:-0} '*' 86400 + 43200`
date --date="\${1:-1970}-01-01 UTC \$x_epoch seconds" +%s
}

year_of_epoch() {
date --date="1970-01-01 UTC \${1:-0} seconds" +%Y
}
day_of_epoch() {
LC_ALL=C date --date="1970-01-01 UTC \${1:-0} seconds" +%A
}
date_of_epoch() {
date --date="1970-01-01 UTC \${1:-0} seconds" "+%Y-%m-%d"
}
month_of_epoch() {
date --date="1970-01-01 UTC \${1:-0} seconds" "+%m"
}

last_fridays() {
year=\${1:-2012}

next_year=`expr \$year + 1`
\$debug && echo "next_year \$next_year"

current_year=\$year
day=0
previous_month=01

while test \$current_year != \$next_year; do

\$debug && echo "day \$day"

current_epoch=`epoch_year_day \$year \$day`
\$debug && echo "current_epoch \$current_epoch"

current_year=`year_of_epoch \$current_epoch`

current_day=`day_of_epoch \$current_epoch`
\$debug && echo "current_day \$current_day"

test \$current_day = 'Friday' && current_friday=`date_of_epoch \$current_epoch`
\$debug && echo "current_friday \$current_friday"

current_month=`month_of_epoch \$current_epoch`
\$debug && echo "current_month \$current_month"

# Change of month => previous friday is the last of month
test "\$previous_month" != "\$current_month" \
&& echo \$previous_friday

previous_month=\$current_month
previous_friday=\$current_friday
day=`expr \$day + 1`
done
}

# main
last_fridays \${1:-2012}

Sample execution:

lastfridays 2012
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## Visual FoxPro

*!* OOP implementaion
LOCAL lnYear As Integer, oCalc As fricalc
CLEAR
lnYear = VAL(INPUTBOX("Year", "Year"))
oCalc = NEWOBJECT("fricalc")
oCalc.LastFriday(lnYear)

DEFINE CLASS fricalc As Session
DataSession = 2	&& Private

PROCEDURE Init
*!* These date settings are private to this class
SET DATE YMD
SET CENTURY ON
SET MARK TO "-"
ENDPROC

FUNCTION LastFriday(tnYear As Integer) As VOID
LOCAL i As Integer, ldDate As Date
CLEAR
? "Last Fridays in the year " + TRANSFORM(tnYear)
FOR i = 1 TO 12
ldDate = DATE(tnYear, i, 1)	&& 1st of month
ldDate = GOMONTH(ldDate, 1) - 1	&& last day of month
*!* Use the built in function to return the day of the week
*!* 6 is Friday
DO WHILE DOW(ldDate) # 6
ldDate = ldDate - 1
ENDDO
? ldDate
ENDFOR
ENDFUNC

ENDDEFINE

## V (Vlang)

import time
import os

fn main() {
mut year := 0
mut t := time.now()
year = os.input("Please select a year: ").int()
println("Last Fridays of each month of \$year")
println("==================================")
for i in 1..13 {
mut j := time.month_days[i-1]
if i == 2 {
if time.is_leap_year(year) {j = 29}
}
for {
t = time.parse('\$year-\${i:02}-\$j 12:30:00')!
if t.weekday_str() == 'Fri' {
println("\${time.long_months[i-1]}: \$j")
break
}
j--
}
}
}
Output:
Last Fridays of each month of 2012
==================================
January: 27
February: 24
March: 30
April: 27
May: 25
June: 29
July: 27
August: 31
September: 28
October: 26
November: 30
December: 28

## 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 Fridays 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 == 5) {
System.print(dt)
} else if (wd > 5) {
} else {
}
}
Output:
\$ wren last_friday.wren 2012
The dates of the last Fridays in the month for 2012 are:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

\$ wren last_friday.wren 2020
The dates of the last Fridays in the month for 2020 are:
2020-01-31
2020-02-28
2020-03-27
2020-04-24
2020-05-29
2020-06-26
2020-07-31
2020-08-28
2020-09-25
2020-10-30
2020-11-27
2020-12-25

## XPL0

include c:\cxpl\codes; \intrinsic 'code' declarations

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);
WD:= WD - 5;
if WD < 0 then WD:= WD + 7;
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:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28

## zkl

Gregorian calendar

var [const] D=Time.Date;
fcn lastDay(y,d){
[1..12].pump(List,'wrap(m){  // 12 months, closure for y & d
[D.daysInMonth(y,m)..1,-1].pump(Void,'wrap(_d){  // work backwards
D.weekDay(y,m,_d) :
if (_==d) return(Void.Stop,D.toYMDString(y,m,_d))
})
})
}
lastDay(2012,D.Friday).concat("\n").println();

For each month in year y, count back from the last day in the month until a Friday is found and print that date. A pump is a loop over a sequence and Void.Stop stops the pump with a value. The first parameter to a pump is the sink. All the imperative loop constructs are available but I didn't feel like using them. A wrap is a function closure over unknown values in the function, necessary because functions are not lexically scoped.

Output:
2012-01-27
2012-02-24
2012-03-30
2012-04-27
2012-05-25
2012-06-29
2012-07-27
2012-08-31
2012-09-28
2012-10-26
2012-11-30
2012-12-28