Find the last Sunday of each month
You are encouraged to solve this task according to the task description, using any language you may know.
Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc).
Example of an expected output:
./last_sundays 2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
- Related tasks
11l
<lang 11l>F last_sundays(year)
[String] sundays
L(month) 1..12
V last_day_of_month = I month < 12 {Time(year, month + 1)} E Time(year + 1)
L
last_day_of_month -= TimeDelta(days' 1)
I last_day_of_month.strftime(‘%w’) == ‘0’
sundays [+]= year‘-’(‘#02’.format(month))‘-’last_day_of_month.strftime(‘%d’)
L.break
R sundays
print(last_sundays(2013).join("\n"))</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
360 Assembly
The program uses one ASSIST macro (XPRNT) to keep the code as short as possible. <lang 360asm>* Last Sunday of each month 31/01/2017 LASTSUND CSECT
USING LASTSUND,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) prolog
ST R13,4(R15) " <-
ST R15,8(R13) " ->
LR R13,R15 " addressability
L R4,YEAR year
SRDA R4,32 .
D R4,=F'400' year/400
LTR R4,R4 if year//400=0
BZ LEAP
L R4,YEAR year
SRDA R4,32 .
D R4,=F'4' year/4
LTR R4,R4 if year//4=0
BNZ NOTLEAP
L R4,YEAR year
SRDA R4,32 .
D R4,=F'100' year/400
LTR R4,R4 if year//100=0
BZ NOTLEAP
LEAP MVC DAYS+4(4),=F'29' days(2)=29 NOTLEAP L R8,YEAR year
BCTR R8,0 y=year-1
LA R7,42 42
AR R7,R8 +y
LR R3,R8 y
SRA R3,2 y/4
AR R7,R3 +y/4
LR R4,R8 y
SRDA R4,32 .
D R4,=F'100' y/100
LA R4,0 .
M R4,=F'6' *6
AR R7,R5 +6*(y/100)
LR R4,R8 y
SRDA R4,32 .
D R4,=F'400' y/100
AR R7,R5 k=42+y+y/4+6*(y/100)+y/400
LA R6,1 m=1
LOOPM C R6,=F'12' do m=1 to 12
BH ELOOPM
LR R1,R6 m
SLA R1,2 .
L R2,DAYS-4(R1) days(m)
AR R7,R2 k=k+days(m)
LR R4,R7 k
SRDA R4,32 .
D R4,=F'7' k/7
SR R2,R4 days(m)-k//7
LR R9,R2 d=days(m)-k//7
L R1,YEAR year
CVD R1,DW year: binary to packed
OI DW+7,X'0F' zap sign
UNPK PG(4),DW unpack (ZL4)
CVD R6,DW m : binary to packed
OI DW+7,X'0F' zap sign
UNPK PG+5(2),DW unpack (ZL2)
CVD R9,DW d: binary to packed
OI DW+7,X'0F' zap sign
UNPK PG+8(2),DW unpack (ZL2)
XPRNT PG,L'PG print buffer
LA R6,1(R6) m=m+1
B LOOPM
ELOOPM L R13,4(0,R13) epilog
LM R14,R12,12(R13) " restore
XR R15,R15 " rc=0
BR R14 exit
YEAR DC F'2013' <== input year DAYS DC F'31',F'28',F'31',F'30',F'31',F'30'
DC F'31',F'31',F'30',F'31',F'30',F'31'
PG DC CL80'YYYY-MM-DD' buffer XDEC DS CL12 temp DW DS D packed (PL8) 15num
YREGS
END LASTSUND</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Ada
The program from [[1]] solves this task, as well.
- Output:
>./last_weekday_in_month sunday 2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
ALGOL W
Uses the Day_of_week and isLeapYear procedures from the day-of-the-week and leap-year tasks - included here for convenience. <lang algolw>begin % find the last Sunday in each month of a year %
% returns true if year is a leap year, false otherwise %
% assumes year is in the Gregorian Calendar %
logical procedure isLeapYear ( integer value year ) ;
year rem 400 = 0 or ( year rem 4 = 0 and year rem 100 not = 0 );
% returns the day of the week of the specified date (d/m/y) %
% Sunday = 1 %
integer procedure Day_of_week ( integer value d, m, y );
begin
integer j, k, mm, yy;
mm := m;
yy := y;
if mm <= 2 then begin
mm := mm + 12;
yy := yy - 1;
end if_m_le_2;
j := yy div 100;
k := yy rem 100;
(d + ( ( mm + 1 ) * 26 ) div 10 + k + k div 4 + j div 4 + 5 * j ) rem 7
end Day_of_week;
% sets the elements of last to the day of the last Sunday %
% of each month in year %
procedure lastSundays ( integer value year
; integer array last ( * )
) ;
begin
integer array lastDays ( 1 :: 12 );
integer m;
% set ld to the day number od the last day of each %
% month in year %
m := 1;
for ld := 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 do begin
lastDays( m ) := ld;
m := m + 1
end for_ld ;
if isLeapYear( year ) then lastDays( 2 ) := 29;
% for each month, determine the day number of the %
% last Sunday %
for mPos := 1 until 12 do begin
integer dow;
dow := Day_of_week( lastDays( mPos ), mPos, year );
if dow = 0 % Saturday % then dow := 7;
% calculate the offset for the previous Sunday %
last( mPos ) := ( lastDays( mPos ) + 1 ) - dow
end for_mPos
end lastSundays ;
begin
% test the lastSundays procedure %
integer array last ( 1 :: 12 );
integer year;
year := 2020;
lastSundays( year, last );
i_w := 1; s_w := 0; % output formatting %
for mPos := 1 until 12 do write( year, if mPos < 10 then "-0" else "-1", mPos rem 10, "-", last( mPos ) )
end
end.</lang>
- Output:
2020-01-26 2020-02-23 2020-03-29 2020-04-26 2020-05-31 2020-06-28 2020-07-26 2020-08-30 2020-09-27 2020-10-25 2020-11-29 2020-12-27
AppleScript
<lang AppleScript>-- LAST SUNDAYS OF YEAR ------------------------------------------------------
-- lastSundaysOfYear :: Int -> [Date] on lastSundaysOfYear(y)
-- lastWeekDaysOfYear :: Int -> Int -> [Date]
script lastWeekDaysOfYear
on |λ|(intYear, iWeekday)
-- lastWeekDay :: Int -> Int -> Date
script lastWeekDay
on |λ|(iLastDay, iMonth)
set iYear to intYear
calendarDate(iYear, iMonth, iLastDay - ¬
(((weekday of calendarDate(iYear, iMonth, iLastDay)) as integer) + ¬
(7 - (iWeekday))) mod 7)
end |λ|
end script
map(lastWeekDay, lastDaysOfMonths(intYear))
end |λ|
-- isLeapYear :: Int -> Bool
on isLeapYear(y)
(0 = y mod 4) and (0 ≠ y mod 100) or (0 = y mod 400)
end isLeapYear
-- lastDaysOfMonths :: Int -> [Int]
on lastDaysOfMonths(y)
{31, cond(isLeapYear(y), 29, 28), 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
end lastDaysOfMonths
end script
lastWeekDaysOfYear's |λ|(y, Sunday as integer)
end lastSundaysOfYear
-- TEST ----------------------------------------------------------------------
on run argv
intercalate(linefeed, ¬
map(isoRow, ¬
transpose(map(lastSundaysOfYear, ¬
apply(cond(class of argv is list and argv ≠ {}, ¬
singleYearOrRange, fiveCurrentYears), argIntegers(argv))))))
end run
-- ARGUMENT HANDLING ---------------------------------------------------------
-- Up to two optional command line arguments: [yearFrom], [yearTo] -- (Default range in absence of arguments: from two years ago, to two years ahead)
-- ~ $ osascript ~/Desktop/lastSundays.scpt -- ~ $ osascript ~/Desktop/lastSundays.scpt 2013 -- ~ $ osascript ~/Desktop/lastSundays.scpt 2013 2016
-- singleYearOrRange :: [Int] -> [Int] on singleYearOrRange(argv)
apply(cond(length of argv > 0, my range, my fiveCurrentYears), argv)
end singleYearOrRange
-- fiveCurrentYears :: () -> [Int] on fiveCurrentYears(_)
set intThisYear to year of (current date) enumFromTo(intThisYear - 2, intThisYear + 2)
end fiveCurrentYears
-- argIntegers :: maybe [String] -> [Int] on argIntegers(argv)
if class of argv is list and argv ≠ {} then
{map(my parseInt, argv)}
else
{}
end if
end argIntegers
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- apply (a -> b) -> a -> b on apply(f, a)
mReturn(f)'s |λ|(a)
end apply
-- calendarDate :: Int -> Int -> Int -> Date on calendarDate(intYear, intMonth, intDay)
tell (current date)
set {its year, its month, its day, its time} to ¬
{intYear, intMonth, intDay, 0}
return it
end tell
end calendarDate
-- cond :: Bool -> (a -> b) -> (a -> b) -> (a -> b) on cond(bool, f, g)
if bool then
f
else
g
end if
end cond
-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo
-- intercalate :: Text -> [Text] -> Text on intercalate(strText, lstText)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- isoDateString :: Date -> String on isoDateString(dte)
(((year of dte) as string) & ¬
"-" & text items -2 thru -1 of ¬
("0" & ((month of dte) as integer) as string)) & ¬
"-" & text items -2 thru -1 of ¬
("0" & day of dte)
end isoDateString
-- isoRow :: [Date] -> String on isoRow(lstDate)
intercalate(tab, map(my isoDateString, lstDate))
end isoRow
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- parseInt :: String -> Int on parseInt(s)
s as integer
end parseInt
-- transpose :: a -> a on transpose(xss)
script column
on |λ|(_, iCol)
script row
on |λ|(xs)
item iCol of xs
end |λ|
end script
map(row, xss)
end |λ|
end script
map(column, item 1 of xss)
end transpose</lang>
- Output:
2015-01-25 2016-01-31 2017-01-29 2018-01-28 2019-01-27 2015-02-22 2016-02-28 2017-02-26 2018-02-25 2019-02-24 2015-03-29 2016-03-27 2017-03-26 2018-03-25 2019-03-31 2015-04-26 2016-04-24 2017-04-30 2018-04-29 2019-04-28 2015-05-31 2016-05-29 2017-05-28 2018-05-27 2019-05-26 2015-06-28 2016-06-26 2017-06-25 2018-06-24 2019-06-30 2015-07-26 2016-07-31 2017-07-30 2018-07-29 2019-07-28 2015-08-30 2016-08-28 2017-08-27 2018-08-26 2019-08-25 2015-09-27 2016-09-25 2017-09-24 2018-09-30 2019-09-29 2015-10-25 2016-10-30 2017-10-29 2018-10-28 2019-10-27 2015-11-29 2016-11-27 2017-11-26 2018-11-25 2019-11-24 2015-12-27 2016-12-25 2017-12-31 2018-12-30 2019-12-29
More straightforward solution:
AppleScript's weekday constants can be coerced either to English text or to the integers 1 (for Sunday) to 7 (Saturday).
<lang AppleScript>on lastSundayOfEachMonthInYear(y) -- Initialise an AppleScript date to the first day of some month in the specified year. tell (current date) to set {firstDayOfNextMonth, its day, its year} to {it, 1, y}
-- Get a string representation of y, zero-padded if necessary, and initialise the output string. set y to text 2 thru 5 of ((10000 + y) as text) set outputText to "./last_sundays " & y repeat with nextMonth from 2 to 13 -- Yes! -- For each month in the year, get the first day of the following month. set firstDayOfNextMonth's month to nextMonth -- Calculate the date of the Sunday which occurs in the seven days prior to that -- by subtracting a day plus the difference between the previous day's weekday and the target weekday. set lastSundayOfThisMonth to firstDayOfNextMonth - (1 + ((firstDayOfNextMonth's weekday) + 5) mod 7) * days -- Append the required details to the output text. set {month:m, day:d} to lastSundayOfThisMonth tell (10000 + m * 100 + d) as text to ¬ set outputText to outputText & (linefeed & y & "-" & text 2 thru 3 & "-" & text 4 thru 5) end repeat
return outputText end lastSundayOfEachMonthInYear
lastSundayOfEachMonthInYear(2020)</lang>
- Output:
"./last_sundays 2020 2020-01-26 2020-02-23 2020-03-29 2020-04-26 2020-05-31 2020-06-28 2020-07-26 2020-08-30 2020-09-27 2020-10-25 2020-11-29 2020-12-27"
The above is hard-coded for Sundays. As with the "Last Friday of each month" task [Last Friday of each month], the handler can be made more flexible with a weekday constant parameter, so that it can be used for any last weekday of the months:
<lang AppleScript>on lastWeekdayWOfEachMonthInYear(w, y) -- Parameters: (AppleScript weekday constant, AD year number) -- Initialise an AppleScript date to the first day of some month in the specified year. tell (current date) to set {firstDayOfNextMonth, its day, its year} to {it, 1, y}
-- Get a string representation of y, zero-padded if necessary, and initialise the output string. set y to text 2 thru 5 of ((10000 + y) as text) set outputText to "./last_" & w & "s " & y repeat with nextMonth from 2 to 13 -- Yes! -- For each month in the year, get the first day of the following month. set firstDayOfNextMonth's month to nextMonth -- Calculate the date of the target weekday which occurs in the seven days prior to that. -- The calculation can be described in various ways, the simplest being the subtraction of a day plus the difference between the previous day's weekday and the target weekday: -- firstDayOfNextMonth - (1 + (((firstDayOfNextMonth's weekday) - 1) - w + 7) mod 7) * days -- But they all boil down to: set lastWOfThisMonth to firstDayOfNextMonth - (1 + ((firstDayOfNextMonth's weekday) - w + 6) mod 7) * days -- Get the day and month of the calculated date and append the required details to the output text. set {month:m, day:d} to lastWOfThisMonth tell (10000 + m * 100 + d) as text to ¬ set outputText to outputText & (linefeed & y & "-" & text 2 thru 3 & "-" & text 4 thru 5) end repeat
return outputText end lastWeekdayWOfEachMonthInYear
lastWeekdayWOfEachMonthInYear(Sunday, 2020)</lang>
- Output:
"./last_Sundays 2020 2020-01-26 2020-02-23 2020-03-29 2020-04-26 2020-05-31 2020-06-28 2020-07-26 2020-08-30 2020-09-27 2020-10-25 2020-11-29 2020-12-27"
Arturo
<lang arturo>lastSundayForMonth: @(m){ if [contains #(1 3 5 7 8 10 12) m] -> nofDays: 31 if [contains #(4 6 9 11) m] -> nofDays: 30 if m=2 -> nofDays: 27
loop nofDays..1 @(d){ dt: datetime "2013-"+m+"-"+d "yyyy-M-dd" if [day dt]="Sunday" -> return dt } }
loop 1..12 -> print [datetime [lastSundayForMonth &] "yyyy-MM-dd"]</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
AutoHotkey
<lang AutoHotkey>InputBox, Year, , Enter a year., , 300, 135 Date := Year . "0101"
while SubStr(Date, 1, 4) = Year {
FormatTime, WD, % Date, WDay
if (WD = 1)
MM := LTrim(SubStr(Date, 5, 2), "0"), Day%MM% := SubStr(Date, 7, 2)
Date += 1, Days
}
Gui, Font, S10, Courier New Gui, Add, Text, , % "Last Sundays of " Year ":`n---------------------"
Loop, 12 {
FormatTime, Month, % Year (A_Index > 9 ? "" : "0") A_Index, MMMM Gui, Add, Text, y+1, % Month (StrLen(Month) > 7 ? "" : "`t") "`t" Day%A_Index%
}
Gui, Show return</lang>
- Output:
Last Sundays of 2013: --------------------- January 27 February 24 March 31 April 28 May 26 June 30 July 28 August 25 September 29 October 27 November 24 December 29
AWK
<lang AWK>
- syntax: GAWK -f FIND_THE_LAST_SUNDAY_OF_EACH_MONTH.AWK [year]
BEGIN {
split("31,28,31,30,31,30,31,31,30,31,30,31",daynum_array,",") # days per month in non leap year
year = (ARGV[1] == "") ? strftime("%Y") : ARGV[1]
if (year % 400 == 0 || (year % 4 == 0 && year % 100 != 0)) {
daynum_array[2] = 29
}
for (m=1; m<=12; m++) {
for (d=daynum_array[m]; d>=1; d--) {
if (strftime("%a",mktime(sprintf("%d %d %d 0 0 0",year,m,d))) == "Sun") {
printf("%04d-%02d-%02d\n",year,m,d)
break
}
}
}
exit(0)
} </lang>
Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Batch File
Uses day of week, last day of month and leapyear routines <lang Batch File> @echo off setlocal enabledelayedexpansion set /p yr= Enter year: echo. call:monthdays %yr% list set mm=1 for %%i in (!list!) do (
call:calcdow !yr! !mm! %%i dow set/a lsu=%%i-dow set mf=0!mm! echo !yr!-!mf:~-2!-!lsu! set /a mm+=1
) pause exit /b
- monthdays yr &list
setlocal call:isleap %1 ly for /L %%i in (1,1,12) do (
set /a "nn = 30 + ^!(((%%i & 9) + 6) %% 7) + ^!(%%i ^^ 2) * (ly - 2) set list=!list! !nn!
) endlocal & set %2=%list% exit /b
- calcdow yr mt dy &dow :: 0=sunday
setlocal set/a a=(14-%2)/12,yr=%1-a,m=%2+12*a-2,"dow=(%3+yr+yr/4-yr/100+yr/400+31*m/12)%%7" endlocal & set %~4=%dow% exit /b
- isleap yr &leap :: remove ^ if not delayed expansion
set /a "%2=^!(%1%%4)+(^!^!(%1%%100)-^!^!(%1%%400))" exit /b </lang>
- Output:
Enter year: 2016 2016-01-31 2016-02-28 2016-03-27 2016-04-24 2016-05-29 2016-06-26 2016-07-31 2016-08-28 2016-09-25 2016-10-30 2016-11-27 2016-12-25
BBC BASIC
<lang bbcbasic> INSTALL @lib$+"DATELIB"
INPUT "What year to calculate (YYYY)? " Year%
PRINT '"Last Sundays in ";Year%;" are on:" FOR Month%=1 TO 12
PRINT Year% "-" RIGHT$("0"+STR$Month%,2) "-";FN_dim(Month%,Year%)-FN_dow(FN_mjd(FN_dim(Month%,Year%),Month%,Year%))
NEXT END </lang>
- Output:
What year to calculate (YYYY)? 2013
Last Sundays in 2013 are on:
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Befunge
This is essentially identical to Last Friday of each month except for the initial day offset. <lang befunge>":raeY",,,,,&>55+,:::45*:*%\"d"%!*\4%+!3v v2++6**"I"5\+/*:*54\-/"d"\/4::-1::p53+g5< >:00p5g4-+7%\:0\v>,"-",5g+:55+/68*+,55+%v ^<<_$$vv*86%+55:<^+*86%+55,+*86/+55:-1:<6 >$$^@$<>+\55+/:#^_$>:#,_$"-",\:04-\-00g^8 ^<# #"#"##"#"##!` +76:+1g00,+55,+*<</lang>
- Output:
Year:2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
C
Identical to Last Friday of each month except for removal of the offset day in the output.
<lang C>
- include <stdio.h>
- include <stdlib.h>
int main(int argc, char *argv[]) {
int days[] = {31,29,31,30,31,30,31,31,30,31,30,31};
int m, y, w;
if (argc < 2 || (y = atoi(argv[1])) <= 1752) return 1;
days[1] -= (y % 4) || (!(y % 100) && (y % 400));
w = y * 365 + 97 * (y - 1) / 400 + 4;
for(m = 0; m < 12; m++) {
w = (w + days[m]) % 7;
printf("%d-%02d-%d\n", y, m + 1,days[m] - w);
}
return 0;
} </lang>
C#
<lang csharp>using System;
namespace LastSundayOfEachMonth {
class Program
{
static void Main()
{
Console.Write("Year to calculate: ");
string strYear = Console.ReadLine();
int year = Convert.ToInt32(strYear);
DateTime date;
for (int i = 1; i <= 12; i++)
{
date = new DateTime(year, i, DateTime.DaysInMonth(year, i), System.Globalization.CultureInfo.CurrentCulture.Calendar);
/* Modification by Albert Zakhia on 2021-16-02
The below code is very slow due to the loop, we will go twice as fast
while (date.DayOfWeek != DayOfWeek.Sunday)
{
date = date.AddDays(-1);
}
*/
// The updated code
int daysOffset = date.DayOfWeek - dayOfWeek; // take the offset to subtract directly instead of looping
if (daysOffset < 0) daysOffset += 7; // if the code is negative, we need to normalize them
date = date.AddDays(-daysOffset ); // now just add the days offset
Console.WriteLine(date.ToString("yyyy-MM-dd"));
}
}
}
} </lang>
- Output:
Year to calculate: 2013 2013-Jan-27 2013-Feb-24 2013-Mar-31 2013-Apr-28 2013-May-26 2013-Jun-30 2013-Jul-28 2013-Aug-25 2013-Sep-29 2013-Oct-27 2013-Nov-24 2013-Dec-29
C++
<lang cpp>
- include <windows.h>
- include <iostream>
- include <string>
//-------------------------------------------------------------------------------------------------- using namespace std;
//-------------------------------------------------------------------------------------------------- class lastSunday { public:
lastSunday()
{
m[0] = "JANUARY: "; m[1] = "FEBRUARY: "; m[2] = "MARCH: "; m[3] = "APRIL: "; m[4] = "MAY: "; m[5] = "JUNE: "; m[6] = "JULY: "; m[7] = "AUGUST: "; m[8] = "SEPTEMBER: "; m[9] = "OCTOBER: "; m[10] = "NOVEMBER: "; m[11] = "DECEMBER: ";
}
void findLastSunday( int y )
{
year = y; isleapyear();
int days[] = { 31, isleap ? 29 : 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }, d; for( int i = 0; i < 12; i++ ) { d = days[i]; while( true ) { if( !getWeekDay( i, d ) ) break; d--; } lastDay[i] = d; }
display();
}
private:
void isleapyear()
{
isleap = false; if( !( year % 4 ) ) { if( year % 100 ) isleap = true; else if( !( year % 400 ) ) isleap = true; }
}
void display()
{
system( "cls" ); cout << " YEAR " << year << endl << "=============" << endl; for( int x = 0; x < 12; x++ ) cout << m[x] << lastDay[x] << endl;
cout << endl << endl;
}
int getWeekDay( int m, int d )
{
int y = year;
int f = y + d + 3 * m - 1; m++; if( m < 3 ) y--; else f -= int( .4 * m + 2.3 );
f += int( y / 4 ) - int( ( y / 100 + 1 ) * 0.75 ); f %= 7;
return f;
}
int lastDay[12], year; string m[12]; bool isleap;
}; //-------------------------------------------------------------------------------------------------- int main( int argc, char* argv[] ) {
int y; lastSunday ls;
while( true )
{
system( "cls" ); cout << "Enter the year( yyyy ) --- ( 0 to quit ): "; cin >> y; if( !y ) return 0;
ls.findLastSunday( y );
system( "pause" );
} return 0;
} //-------------------------------------------------------------------------------------------------- </lang>
- Output:
YEAR 2013 ============= JANUARY: 27 FEBRUARY: 24 MARCH: 31 APRIL: 28 MAY: 26 JUNE: 30 JULY: 28 AUGUST: 25 SEPTEMBER: 29 OCTOBER: 27 NOVEMBER: 24 DECEMBER: 29
Other solution, based on the Boost DateTime library: <lang C++>#include <iostream>
- include <boost/date_time/gregorian/gregorian.hpp>
- include <cstdlib>
int main( int argc , char* argv[ ] ) {
using namespace boost::gregorian ;
int year = std::atoi( argv[ 1 ] ) ;
for ( int i = 1 ; i < 13 ; i++ ) {
try {
date d( year , i , 1 ) ; d = d.end_of_month( ) ; day_iterator d_itr ( d ) ; while ( d_itr->day_of_week( ) != Sunday ) { --d_itr ; } std::cout << to_simple_string ( *d_itr ) << std::endl ;
} catch ( bad_year by ) {
std::cout << "Terminated because of " << by.what( ) << "\n" ;
} } return 0 ;
}</lang>
- Output:
2013-Jan-27 2013-Feb-24 2013-Mar-31 2013-Apr-28 2013-May-26 2013-Jun-30 2013-Jul-28 2013-Aug-25 2013-Sep-29 2013-Oct-27 2013-Nov-24 2013-Dec-29
Clojure
<lang clojure>(ns last-sundays.core
(:require [clj-time.core :as time]
[clj-time.periodic :refer [periodic-seq]]
[clj-time.format :as fmt])
(:import (org.joda.time DateTime DateTimeConstants))
(:gen-class))
(defn sunday? [t]
(= (.getDayOfWeek t) (DateTimeConstants/SUNDAY)))
(defn sundays [year]
(take-while #(= (time/year %) year)
(filter sunday? (periodic-seq (time/date-time year 1 1) (time/days 1)))))
(defn last-sundays-of-months [year]
(->> (sundays year)
(group-by time/month)
(vals)
(map (comp first #(sort-by time/day > %)))
(map #(fmt/unparse (fmt/formatters :year-month-day) %))
(interpose "\n")
(apply str)))
(defn -main [& args]
(println (last-sundays-of-months (Integer. (first args)))))
</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
COBOL
<lang COBOL>
program-id. last-sun.
data division.
working-storage section.
1 wk-date.
2 yr pic 9999.
2 mo pic 99 value 1.
2 da pic 99 value 1.
1 rd-date redefines wk-date pic 9(8).
1 binary.
2 int-date pic 9(8).
2 dow pic 9(4).
2 sunday pic 9(4) value 7.
procedure division.
display "Enter a calendar year (1601 thru 9999): "
with no advancing
accept yr
if yr >= 1601 and <= 9999
continue
else
display "Invalid year"
stop run
end-if
perform 12 times
move 1 to da
add 1 to mo
if mo > 12 *> to avoid y10k in 9999
move 12 to mo
move 31 to da
end-if
compute int-date = function
integer-of-date (rd-date)
if mo =12 and da = 31 *> to avoid y10k in 9999
continue
else
subtract 1 from int-date
end-if
compute rd-date = function
date-of-integer (int-date)
compute dow = function mod
((int-date - 1) 7) + 1
compute dow = function mod ((dow - sunday) 7)
subtract dow from da
display yr "-" mo "-" da
add 1 to mo
end-perform
stop run
.
end program last-sun.
</lang>
- Output:
2016-01-31 2016-02-28 2016-03-27 2016-04-24 2016-05-29 2016-06-26 2016-07-31 2016-08-28 2016-09-25 2016-10-30 2016-11-27 2016-12-25
Common Lisp
First, calculate the day of week of the 1st day of next month. Then figure out how many days you have to go back until the last Sunday of the month.
<lang lisp>(defun last-sundays (year)
(loop for month from 1 to 12
for last-month-p = (= month 12)
for next-month = (if last-month-p 1 (1+ month))
for year-of-next-month = (if last-month-p (1+ year) year)
for 1st-day-next-month = (encode-universal-time 0 0 0 1 next-month year-of-next-month 0)
for 1st-day-dow = (nth-value 6 (decode-universal-time 1st-day-next-month 0))
;; 0: monday, 1: tuesday, ... 6: sunday
for diff-to-last-sunday = (1+ 1st-day-dow)
for last-sunday = (- 1st-day-next-month (* diff-to-last-sunday 24 60 60))
do (multiple-value-bind (second minute hour date month year)
(decode-universal-time last-sunday 0)
(declare (ignore second minute hour))
(format t "~D-~2,'0D-~2,'0D~%" year month date))))
(last-sundays 2013)</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
D
<lang d>void lastSundays(in uint year) {
import std.stdio, std.datetime;
foreach (immutable month; 1 .. 13) {
auto date = Date(year, month, 1);
date.day(date.daysInMonth);
date.roll!"days"(-(date.dayOfWeek + 7) % 7);
date.writeln;
}
}
void main() {
lastSundays(2013);
}</lang>
- Output:
2013-Jan-27 2013-Feb-24 2013-Mar-31 2013-Apr-28 2013-May-26 2013-Jun-30 2013-Jul-28 2013-Aug-25 2013-Sep-29 2013-Oct-27 2013-Nov-24 2013-Dec-29
Elixir
<lang elixir>defmodule RC do
def lastSunday(year) do
Enum.map(1..12, fn month ->
lastday = :calendar.last_day_of_the_month(year, month)
daynum = :calendar.day_of_the_week(year, month, lastday)
sunday = lastday - rem(daynum, 7)
{year, month, sunday}
end)
end
end
y = String.to_integer(hd(System.argv)) Enum.each(RC.lastSunday(y), fn {year, month, day} ->
:io.format "~4b-~2..0w-~2..0w~n", [year, month, day]
end)</lang>
- Output:
C:\Elixir>elixir lastSunday.exs 2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Erlang
<lang Erlang> -module(last_sundays).
-export([in_year/1]).
% calculate all the last sundays in a particular year in_year(Year) ->
[lastday(Year, Month, 7) || Month <- lists:seq(1, 12)].
% calculate the date of the last occurrence of a particular weekday lastday(Year, Month, WeekDay) ->
Ldm = calendar:last_day_of_the_month(Year, Month),
Diff = calendar:day_of_the_week(Year, Month, Ldm) rem WeekDay,
{Year, Month, Ldm - Diff}.
</lang>
- Output:
30> [io:fwrite("~B-~2.10.0B-~B~n", [Y,M,D]) || {Y,M,D} <- last_sundays:in_year(2013)].
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
F#
We use a transformation from and to the Julian Day Number, see also PARI/GP or Fortran.
The formulars used here come from [2] (section "Calculation").
<lang fsharp>let jdn (year, month, day) =
let a = (14 - month) / 12 let y = year + 4800 - a let m = month + 12 * a - 3 day + (153*m+2)/5 + 365*y + y/4 - y/100 + y/400 - 32045
let date_from_jdn jdn =
let j = jdn + 32044 let g = j / 146097 let dg = j % 146097 let c = (dg / 36524 + 1) * 3 / 4 let dc = dg - c * 36524 let b = dc / 1461 let db = dc % 1461 let a = (db / 365 + 1) * 3 / 4 let da = db - a * 365 let y = g * 400 + c * 100 + b * 4 + a let m = (da * 5 + 308) / 153 - 2 let d = da - (m + 4) * 153 / 5 + 122 (y - 4800 + (m + 2) / 12, (m + 2) % 12 + 1, d + 1)
[<EntryPoint>] let main argv =
let year = System.Int32.Parse(argv.[0]) [for m in 1..12 do yield jdn (year,m+1,0)] |> List.map (fun x -> date_from_jdn (x - (x+1)%7)) |> List.iter (printfn "%A") 0</lang>
- Output:
RosettaCode 2016 (2016, 1, 31) (2016, 2, 28) (2016, 3, 27) (2016, 4, 24) (2016, 5, 29) (2016, 6, 26) (2016, 7, 31) (2016, 8, 28) (2016, 9, 25) (2016, 10, 30) (2016, 11, 27) (2016, 12, 25)
Factor
This program expects a year passed in via command line argument. In case you are wondering — yes — Factor has a last-sunday-of-month word in its calendar vocabulary. This is par for the course when it comes to Factor. <lang factor>USING: calendar calendar.format command-line io kernel math math.parser sequences ; IN: rosetta-code.last-sunday
- parse-year ( -- ts ) (command-line) second string>number <year> ;
- print-last-sun ( ts -- ) last-sunday-of-month (timestamp>ymd) nl ;
- inc-month ( ts -- ts' ) 1 months time+ ;
- process-month ( ts -- ts' ) dup print-last-sun inc-month ;
- main ( -- ) parse-year 12 [ process-month ] times drop ;
MAIN: main</lang>
- Output:
>factor last-sunday.factor 2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
FBSL
<lang qbasic>#APPTYPE CONSOLE
DIM date AS INTEGER, dayname AS STRING FOR DIM i = 1 TO 12
FOR DIM j = 31 DOWNTO 1
date = 20130000 + (i * 100) + j
IF CHECKDATE(i, j, 2013) THEN
dayname = DATECONV(date, "dddd")
IF dayname = "Sunday" THEN
PRINT 2013, " ", i, " ", j
EXIT FOR
END IF
END IF
NEXT
NEXT
PAUSE </lang>
- Output:
2013-1-27 2013-2-24 2013-3-31 2013-4-28 2013-5-26 2013-6-30 2013-7-28 2013-8-25 2013-9-29 2013-10-27 2013-11-24 2013-12-29 Press any key to continue...
Fortran
Having already functions that calculate a daynumber from a date and the reverse, the notorious routines of H. F. Fliegel and T. C. van Flandern, the problem becomes simple. Their routine calculates the Julian day number but my version adjusts so that day zero is the thirty-first of December 1899, which is a Sunday. Thus, the day of the week is given by MOD(Daynumber,7) with zero for Sunday up to six for Saturday, as per Genesis.
Their routine is not only fast, but flexible, in particular allowing Month + 1 so that DAYNUM(Y,M + 1,0) gives the last day of month M for M = 1 to 12, with no agony over which month has what last day and leap years or not. If W is the day of week desired, all that remains is to use MOD(D - W,7) to determine the offset back to the desired day of the week and make that shift. There is a problem with the MOD function when negative numbers are involved: it may or may not (depending on computer, compiler, language) return a negative result; by adding 7 and taking a MOD again, this variation is suppressed.
The source uses the MODULE protocol for convenience in its usage in larger projects (which contain a calculation for Easter, daylight savings changeover dates, solar azimuth and altitude, etc. which have been removed here), but requires only F77 features, except for the annoyance of the MUNYAD function returning a TYPE(DATEBAG) result to get the three parts. This is an example where a palindromic programming protocol would be so much better. One would write <lang Fortran> D = DAYNUM(Y,M,D) !Daynumber from date.
DAYNUM(Y,M,D) = D !Date parts from a day number.</lang>
But alas, only pl/i offers palindromic functions, and only for SUBSTR.
<lang Fortran>
MODULE DATEGNASH
C Calculate conforming to complex calendarical contortions. C Astronomer Simon Newcomb determined that the tropical year of 1900 c contained 31556925.9747 seconds, or 365.24219879 days. c Subsequent definitions involve "no measurable differences", c whereas in 45BC (when the Julian calendar was adopted), the year was c 365.24232 days long, going by modern calculations. c The "tropical" year is the time between the same equinoxes, and thus c contains the effect of the precession of the Earth's axis, which would c otherwise cause the seasons to likewise precess around the "fixed star" c year, that being the time for midnight to point in the same direction c amongst the "fixed" stars. Specifically, the vernal equinox c (the northern hemisphere's spring equinox: cultural colonialism) c is meant to hover around the 21'st of March, although it may fall c within the 19'th or 20'th, which last has been the most popular in c the 20'th century, until the leap year of 2000 resynchronised c the civil calendar. C By contrast, the computations of astrologers are still based on the c constellations as oriented in Babylonian times...
c So, to add .24219879 to 365 days... c Adjustment per year Nett Discrepancy remaining. c +1/4 +.25 +.25 -.00780121 5h 48m 45.98s c -1/100 -.01 .24 +.00219879 -11m 14.02s c +1/400 +.0025 .2425 -.00030121 -26.02s c -1/4000 -.00025 .24225 -.00005121 -4.42s c c The remnant of -.00005121 (meaning that the calendar year is too long) c amounts to needing to drop one day on 19,527 years, and while this could be c accommodated nicely enough by -1/20000 to give a calendar year of c 365.24420 days with a remaining discrepancy of -.00000121 or -.01sec/year, c there is a problem. The Earth's spin is slowing by a similar amount. c Similar to leap years and leap days are the leap seconds that since the c development of clocks based on atomic oscillations, have been added or c sometimes removed to keep clock time aligned with astronomical observations. c This additional confusion is not further considered.
TYPE DateBag !Pack three parts into one.
INTEGER DAY,MONTH,YEAR !The usual suspects.
END TYPE DateBag !Simple enough.
CHARACTER*9 MONTHNAME(12),DAYNAME(0:6) !Re-interpretations.
PARAMETER (MONTHNAME = (/"January","February","March","April",
1 "May","June","July","August","September","October","November",
2 "December"/))
PARAMETER (DAYNAME = (/"Sunday","Monday","Tuesday","Wednesday",
1 "Thursday","Friday","Saturday"/)) !Index this array with DayNum mod 7.
CHARACTER*3 MTHNAME(12) !The standard abbreviations.
PARAMETER (MTHNAME = (/"JAN","FEB","MAR","APR","MAY","JUN",
1 "JUL","AUG","SEP","OCT","NOV","DEC"/))
INTEGER*4 JDAYSHIFT !INTEGER*2 just isn't enough.
PARAMETER (JDAYSHIFT = 2415020) !Thus shall 31/12/1899 give 0, a Sunday, via DAYNUM.
DOUBLE PRECISION DAYSINYEAR !A real ache.
PARAMETER (DAYSINYEAR = 365.24219879D0) !The "D0" demands DOUBLE PRECISION precision.
INTEGER*4 SECONDSINDAY !This has its uses.
PARAMETER (SECONDSINDAY = 24*60*60) !86400. Disregarding "leap" seconds.
INTEGER NOTADAYNUMBER !Might as well settle on one value.
PARAMETER (NOTADAYNUMBER = -2147483648) !And everyone use it.
PARAMETER NZBASEPLACE = "Mt. Cook Trig, Wellington." !Name the place.
DOUBLE PRECISION NZBASELAT,NZBASELONG !Alas, DATA statements do not allow arithmetic expressions.
PARAMETER (NZBASELAT = -((59.3 D0/60 + 17)/60 + 41)) !Degrees South, thus negative.
PARAMETER (NZBASELONG = ((34.65D0/60 + 46)/60 + 174)) !Degrees East are oriented as positive: left to right with N up.
C Seconds Minutes Degrees. C This is the location of the Mt. Cook trigonometrical base point for New Zealand. C (It's in the foyer of what used to be the Dominion Museum, Wellington) C Determined by Henry Jackson, chief surveyor, in 1870.
TYPE Terroir !Collate the attributes of location, as so far needed.
CHARACTER*28 PLACENAME !Name the location.
DOUBLE PRECISION LATITUDE,LONGITUDE !Locate the location.
DOUBLE PRECISION ZONETIME !The time zone of its civil clock, not necessarily a whole hour.
END TYPE Terroir !The nature of the climate, soil, etc. is not yet involved.
TYPE(Terroir) BASE !Righto, let's have one of them.
DATA BASE/Terroir("Mt. Cook Trig, Wellington.", !The compiler bungles if NZBASEPLACE is used here.
1 NZBASELAT,NZBASELONG,+12.0)/ !Where it's at. +12 hours ahead = +180 degrees Eastward of Greenwhich.
Careful! New Zealand is not centred on longitude 180, but its civil clock's time zone is.
DOUBLE PRECISION SINBASELAT,COSBASELAT !Calculated in SOLARDIRECTION and ZAPME.
CONTAINS !Let the madness begin.
INTEGER*4 FUNCTION DAYNUM(YY,M,D) !Computes (JDayN - JDayShift), not JDayN.
C Conversion from a Gregorian calendar date to a Julian day number, JDayN. C Valid for any Gregorian calendar date producing a Julian day number C greater than zero, though remember that the Gregorian calendar C was not used before y1582m10d15 and often, not after that either. C thus in England (et al) when Wednesday 2'nd September 1752 (Julian style) C was followed by Thursday the 14'th, occasioning the Eleven Day riots C because creditors demanded a full month's payment instead of 19/30'ths. C The zero of the Julian day number corresponds to the first of January C 4713BC on the *Julian* calendar's naming scheme, as extended backwards C with current usage into epochs when it did not exist: the proleptic Julian calendar. c This function employs the naming scheme of the *Gregorian* calendar, c and if extended backwards into epochs when it did not exist (thus the c proleptic Gregorian calendar) it would compute a zero for y-4713m11d24 *if* c it is supposed there was a year zero between 1BC and 1AD (as is convenient c for modern mathematics and astronomers and their simple calculations), *but* c 1BC immediately precedes 1AD without any year zero in between (and is a leap year) c thus the adjustment below so that the date is y-4714m11d24 or 4714BCm11d24, c not that this name was in use at the time... c Although the Julian calendar (introduced by himself in what we would call 45BC, c which was what the Romans occasionally called 709AUC) was provoked by the c "years of confusion" resulting from arbitrary application of the rules c for the existing Roman calendar, other confusions remain unresolved, c so precise dating remains uncertain despite apparently precise specifications c (and much later, Dennis the Short chose wrongly for the birth of Christ) c and the Roman practice of inclusive reckoning meant that every four years c was interpreted as every third (by our exclusive reckoning) so that the c leap years were not as we now interpret them. This was resolved by Augustus c but exactly when (and what date name is assigned) and whose writings used c which system at the time of writing is a matter of more confusion, c and this has continued for centuries. C Accordingly, although an algorithm may give a regular sequence of date names, c that does not mean that those date names were used at the time even if the c calendar existed then, because the interpretation of the algorithm varied. c This in turn means that a date given as being on the Julian calendar c prior to about 10AD is not as definite as it may appear and its alignment c with the astronomical day number is uncertain even though the calculation c is quite definite. c C Computationally, year 1 is preceded by year 0, in a smooth progression. C But there was never a year zero despite what astronomers like to say, C so the formula's year 0 corresponds to 1BC, year -1 to 2BC, and so on back. C Thus y-4713 in this counting would be 4714BC on the Gregorian calendar, C were it to have existed then which it didn't. C To conform to the civil usage, the incoming YY, presumed a proper BC (negative) C and AD (positive) year is converted into the computational counting sequence, Y, C and used in the formula. If a YY = 0 is (improperly) offered, it will manifest C as 1AD. Thus YY = -4714 will lead to calculations with Y = -4713. C Thus, 1BC is a leap year on the proleptic Gregorian calendar. C For their convenience, astronomers decreed that a day starts at noon, so that C in Europe, observations through the night all have the same day number. C The current Western civil calendar however has the day starting just after midnight C and that day's number lasts until the following midnight. C C There is no constraint on the values of D, which is just added as it stands. C This means that if D = 0, the daynumber will be that of the last day of the C previous month. Likewise, M = 0 or M = 13 will wrap around so that Y,M + 1,0 C will give the last day of month M (whatever its length) as one day before C the first day of the next month. C C Example: Y = 1970, M = 1, D = 1; JDAYN = 2440588, a Thursday but MOD(2440588,7) = 3. C and with the adjustment JDAYSHIFT, DAYNUM = 25568; mod 7 = 4 and DAYNAME(4) = "Thursday". C The Julian Day number 2440588.0 is for NOON that Thursday, 2440588.5 is twelve hours later. C And Julian Day number 2440587.625 is for three a.m. Thursday. C C DAYNUM and MUNYAD are the infamous routines of H. F. Fliegel and T.C. van Flandern, C presented in Communications of the ACM, Vol. 11, No. 10 (October, 1968). Carefully typed in again by R.N.McLean (whom God preserve) December XXMMIIX. C Though I remain puzzled as to why they used I,J,K for Y,M,D, C given that the variables were named in the INTEGER statement anyway.
INTEGER*4 JDAYN !Without rebasing, this won't fit in INTEGER*2.
INTEGER YY,Y,M,MM,D !NB! Full year number, so 1970, not 70.
Caution: integer division in Fortran does not produce fractional results. C The fractional part is discarded so that 4/3 gives 1 and -4/3 gives -1. C Thus 4/3 might be Trunc(4/3) or 4 div 3 in other languages. Beware of negative numbers!
Y = YY !I can fiddle this copy without damaging the original's value.
IF (Y.LT.1) Y = Y + 1 !Thus YY = -2=2BC, -1=1BC, +1=1AD, ... becomes Y = -1, 0, 1, ...
MM = (M - 14)/12 !Calculate once. Note that this is integer division, truncating.
JDAYN = D - 32075 !This is the proper astronomer's Julian Day Number.
a + 1461*(Y + 4800 + MM)/4
b + 367*(M - 2 - MM*12)/12
c - 3*((Y + 4900 + MM)/100)/4
DAYNUM = JDAYN - JDAYSHIFT !Thus, *NOT* the actual *Julian* Day Number.
END FUNCTION DAYNUM !But one such that Mod(n,7) gives day names.
Could compute the day of the year somewhat as follows... c DN:=D + (61*Month + (Month div 8)) div 2 - 30 c + if Month > 2 then FebLength - 30 else 0;
TYPE(DATEBAG) FUNCTION MUNYAD(DAYNUM) !Oh for palindromic programming!
Conversion from a Julian day number to a Gregorian calendar date. See JDAYN/DAYNUM.
INTEGER*4 DAYNUM,JDAYN !Without rebasing, this won't fit in INTEGER*2.
INTEGER Y,M,D,L,N !Y will be a full year number: 1950 not 50.
JDAYN = DAYNUM + JDAYSHIFT !Revert to a proper Julian day number.
L = JDAYN + 68569 !Further machinations of H. F. Fliegel and T.C. van Flandern.
N = 4*L/146097
L = L - (146097*N + 3)/4
Y = 4000*(L + 1)/1461001
L = L - 1461*Y/4 + 31
M = 80*L/2447
D = L - 2447*M/80
L = M/11
M = M + 2 - 12*L
Y = 100*(N - 49) + Y + L
IF (Y.LT.1) Y = Y - 1 !The other side of conformity to BC/AD, as in DAYNUM.
MUNYAD%YEAR = Y !Now place for the world to see.
MUNYAD%MONTH = M
MUNYAD%DAY = D
END FUNCTION MUNYAD !A year has 365.2421988 days...
CHARACTER*10 FUNCTION SLASHDATE(DAYNUM) !This is relatively innocent.
Caution! The Gregorian calendar did not exist prior to 15/10/1582! Confine expected operation to four-digit years, since fixed-field sizes are in mind. Can use this function in WRITE statements with FORMAT, since this function does not use them. Compilers of lesser merit can concoct code that bungles such double usage otherwise.
INTEGER*4 DAYNUM !-32768 to 32767 is just not adequate.
TYPE(DATEBAG) D !Though these numbers are more restrained.
INTEGER N,L !Workers.
IF (DAYNUM.EQ.NOTADAYNUMBER) THEN !Perhaps some work can be dodged.
SLASHDATE = " Undated!!" !No proper day number has been placed.
RETURN !So give up, rather than show odd results.
END IF !So much for confusion.
D = MUNYAD(DAYNUM) !Get the pieces.
IF (D%DAY.GT.9) THEN !Here we go.
SLASHDATE(1:1) = CHAR(D%DAY/10 + ICHAR("0")) !Faster than a table look-up?
ELSE !Even if not,
SLASHDATE(1:1) = " " !This should be quick.
END IF !So much for the tens digit.
SLASHDATE(2:2) = CHAR(MOD(D%DAY,10) + ICHAR("0")) !The units digit.
SLASHDATE(3:3) = "/" !Enough of the day number. The separator.
IF (D%MONTH.GT.9) THEN !Now for the month.
SLASHDATE(4:4) = CHAR(D%MONTH/10 + ICHAR("0")) !The tens digit.
ELSE !Not so often used. A table beckons...
SLASHDATE(4:4) = " " !Some might desire leading zeroes here.
END IF !Enough of October, November and December.
SLASHDATE(5:5) = CHAR(MOD(D%MONTH,10) + ICHAR("0")) !The units digit.
SLASHDATE(6:6) = "/" !Enough of the month number. The separator.
L = 10 !The year value deserves a loop, it having four digits.
N = ABS(D%YEAR) !Should never be zero. 1BC is year -1 and 1AD is year = +1.
1 SLASHDATE(L:L) = CHAR(MOD(N,10) + ICHAR("0")) !But if it is, this will place a zero.
N = N/10 !Drop a power of ten.
L = L - 1 !Step back for the next digit.
IF (L.GT.6) GO TO 1 !Thus always four digits, even if they lead with zero.
IF (N.GT.0) SLASHDATE(7:7) = "?" !Y > 9999? Might as well do something.
IF (D%YEAR.LT.0) SLASHDATE(7:7) = "-" !Years BC? Rather than give no indication.
c WRITE (SLASHDATE,1) D%DAY,D%MONTH,D%YEAR !Some compilers will bungle this. c 1 FORMAT (I2,"/",I2,"/",I4) !If so, a local variable must be used.
RETURN !Enough. !As when SLASHDATE is invoked in a WRITE statement.
END FUNCTION SLASHDATE !Simple enough.
END MODULE DATEGNASH
PROGRAM LASTSUNDAY
USE DATEGNASH
INTEGER D,W,M,Y
WRITE (6,1)
1 FORMAT ("Employs the Gregorian calendar pattern.",/,
1 "You specify a day of the week, then nominate a year.",/,
2 "For each month of that year, this calculates the date ",
3 "of the last such day.",//
4 "So, what day (0 = Sunday, 6 = Saturday):",$)
READ (5,*) W
IF (W.LT.0 .OR. W.GT.6) STOP "Not a good week day number!"
10 WRITE (6,11)
11 FORMAT ("What year (non-positive to quit):",$)
READ (5,*) Y
IF (Y.LE.0) STOP
DO M = 1,12
D = DAYNUM(Y,M + 1,0) !Zeroth day = last day of previous month.
D = D - MOD(MOD(D - W,7) + 7,7) !Protect against MOD(D,7) giving negative values for D < 0.
WRITE (6,*) SLASHDATE(D)," : ",DAYNAME(MOD(D,7))
END DO
GO TO 10
END
</lang> And after all that, results:
Employs the Gregorian calendar pattern. You specify a day of the week, then nominate a year. For each month of that year, this calculates the date of the last such day. So, what day (0 = Sunday, 6 = Saturday):0 What year (non-positive to quit):2013 27/ 1/2013 : Sunday 24/ 2/2013 : Sunday 31/ 3/2013 : Sunday 28/ 4/2013 : Sunday 26/ 5/2013 : Sunday 30/ 6/2013 : Sunday 28/ 7/2013 : Sunday 25/ 8/2013 : Sunday 29/ 9/2013 : Sunday 27/10/2013 : Sunday 24/11/2013 : Sunday 29/12/2013 : Sunday What year (non-positive to quit):0
FreeBASIC
<lang FreeBASIC>' version 23-06-2015 ' compile with: fbc -s console
- Ifndef TRUE ' define true and false for older freebasic versions
#Define FALSE 0 #Define TRUE Not FALSE
- EndIf
Function leapyear(Year_ As Integer) As Integer
' from the leapyear entry If (Year_ Mod 4) <> 0 Then Return FALSE If (Year_ Mod 100) = 0 AndAlso (Year_ Mod 400) <> 0 Then Return FALSE Return TRUE
End Function
Function wd(m As Integer, d As Integer, y As Integer) As Integer
' Zellerish ' 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday ' 4 = Thursday, 5 = Friday, 6 = Saturday
If m < 3 Then ' If m = 1 Or m = 2 Then
m += 12
y -= 1
End If
Return (y + (y \ 4) - (y \ 100) + (y \ 400) + d + ((153 * m + 8) \ 5)) Mod 7
End Function
' ------=< MAIN >=------
Type month_days
m_name As String days As UByte
End Type
Dim As month_days arr(1 To 12) Data "January", 31, "February", 28, "March", 31, "April", 30 Data "May", 31, "June", 30, "July", 31, "August", 31 Data "September", 30, "October", 31, "November", 30, "December", 31
Dim As Integer yr, d, i, x Dim As String keypress
For i = 1 To 12
With arr(i)
Read .m_name
Read .days
End With
Next
Do
Do
Print "For what year do you want to find the last Sunday of the month"
Input "any number below 1800 stops program, year in YYYY format";yr
' empty input also stops
If yr < 1800 Then
End
Else
Exit Do
End If
Loop
Print : Print Print "Last Sunday of the month for"; yr
For i = 1 To 12
d = arr(i).days
If i = 2 AndAlso leapyear(yr) = TRUE Then d = d + 1
x = wd(i, d, yr)
d = d - x ' don't test it just do it
Print d; " "; arr(i).m_name
Next
' empty key buffer While Inkey <> "" : keypress = Inkey : Wend Print : Print Print "Find last Sunday for a other year [Y|y], anything else stops" keypress ="" While keypress = "" : keypress = Inkey : Wend If LCase(keypress) <> "y" Then Exit Do Print : Print
Loop End</lang>
- Output:
For what year do you want to find the last Sunday of the month any number below 1800 stops program, year in YYYY format? 2017 Last Sunday of the month for 2017 29 January 26 February 26 March 30 April 28 May 25 June 30 July 27 August 24 September 29 October 26 November 31 December
Gambas
<lang gambas>Public Sub Form_Open() Dim sYear As String 'To store the year chosen Dim siDay, siMonth, siWeekday As Short 'Day, Month and Weekday
sYear = InputBox("Input year", "Last Sunday of each month") 'Get the user to enter a year Print "Last Sundays in " & sYear 'Print a heading
For siMonth = 1 To 12 'Loop for each month
For siDay = 31 DownTo 23 'Count backwards from 31 to 23 (Sunday 23rd February 1930)
siWeekday = 7 'Set the Weekday to Saturday in case of error in the next line
Try siWeekday = WeekDay(Date(Val(sYear), siMonth, siDay)) 'TRY and get the Weekday. If there is an error it will be ignored e.g. 31 February
If siWeekday = 0 Then 'If Weekday is Sunday then..
Print Format(Date(Val(sYear), siMonth, siDay), "dddd dd mmmm yyyy") 'Print the date
Break 'Jump out of this loop
End If
Next
Next
End</lang> Output:
Last Sundays in 2013 Sunday 27 January 2013 Sunday 24 February 2013 Sunday 31 March 2013 Sunday 28 April 2013 Sunday 26 May 2013 Sunday 30 June 2013 Sunday 28 July 2013 Sunday 25 August 2013 Sunday 29 September 2013 Sunday 27 October 2013 Sunday 24 November 2013 Sunday 29 December 2013
Go
This is different from the Go code for Last Friday of each month. It uses the fact that time functions in Go correct for dates: if you enter 32nd of october, Go corrects to 1st of November. So that if 29th of February is corrected to 1st of March, chosen year is not a leap year.
<lang go>package main
import ( "fmt" "time" )
func main() {
var year int var t time.Time var lastDay = [12]int { 31,29,31,30,31,30,31,31,30,31,30,31 }
for { fmt.Print("Please select a year: ") _, err := fmt.Scanf("%d", &year) if err != nil { fmt.Println(err) continue } else { break } }
fmt.Println("Last Sundays of each month of", year) fmt.Println("==================================")
for i := 1;i < 13; i++ { j := lastDay[i-1] if i == 2 { if time.Date(int(year), time.Month(i), j, 0, 0, 0, 0, time.UTC).Month() == time.Date(int(year), time.Month(i), j-1, 0, 0, 0, 0, time.UTC).Month() { j = 29 } else { j = 28 } } for { t = time.Date(int(year), time.Month(i), j, 0, 0, 0, 0, time.UTC) if t.Weekday() == 0 { fmt.Printf("%s: %d\n", time.Month(i), j) break } j = j - 1 } } } </lang>
Please select a year: 2013 Last Sundays of each month of 2013 ================================== January: 27 February: 24 March: 31 April: 28 May: 26 June: 30 July: 28 August: 25 September: 29 October: 27 November: 24 December: 29
Groovy
Solution: <lang groovy>enum Day {
Sun, Mon, Tue, Wed, Thu, Fri, Sat
static Day valueOf(Date d) { Day.valueOf(d.format('EEE')) }
}
def date = Date.&parse.curry('yyyy-MM-dd') def month = { it.format('MM') } def days = { year -> (date("${year}-01-01")..<date("${year+1}-01-01")) } def weekDays = { dayOfWeek, year -> days(year).findAll { Day.valueOf(it) == dayOfWeek } }
def lastWeekDays = { dayOfWeek, year ->
weekDays(dayOfWeek, year).reverse().inject([:]) { months, sunday ->
def monthStr = month(sunday)
!months[monthStr] ? months + [(monthStr):sunday] : months
}.values().sort()
}</lang>
Test: <lang groovy>def ymd = { it.format('yyyy-MM-dd') } def lastSundays = lastWeekDays.curry(Day.Sun) lastSundays(args[0] as int).each { println (ymd(it)) }</lang>
Execution (Cygwin on Windows 7):
[2201] groovy lastSundays.groovy 2013
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Haskell
<lang Haskell>import Data.Time.Calendar
(Day, addDays, fromGregorian, gregorianMonthLength, showGregorian)
import Data.Time.Calendar.WeekDate (toWeekDate) import Data.List (find, intercalate, transpose) import Data.Maybe (fromJust)
-- [1 .. 7] for [Mon .. Sun] findWeekDay :: Int -> Day -> Day findWeekDay dayOfWeek date =
fromJust $
find
(\x ->
let (_, _, day) = toWeekDate x
in dayOfWeek == day)
((`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 " " <$> transpose (weekDayDates 7 <$> [2013 .. 2017]))</lang>
- Output:
2018-01-28 2019-01-27 2020-01-26 2021-01-31 2022-01-30 2023-01-29 2018-02-25 2019-02-24 2020-02-23 2021-02-28 2022-02-27 2023-02-26 2018-03-25 2019-03-31 2020-03-29 2021-03-28 2022-03-27 2023-03-26 2018-04-29 2019-04-28 2020-04-26 2021-04-25 2022-04-24 2023-04-30 2018-05-27 2019-05-26 2020-05-31 2021-05-30 2022-05-29 2023-05-28 2018-06-24 2019-06-30 2020-06-28 2021-06-27 2022-06-26 2023-06-25 2018-07-29 2019-07-28 2020-07-26 2021-07-25 2022-07-31 2023-07-30 2018-08-26 2019-08-25 2020-08-30 2021-08-29 2022-08-28 2023-08-27 2018-09-30 2019-09-29 2020-09-27 2021-09-26 2022-09-25 2023-09-24 2018-10-28 2019-10-27 2020-10-25 2021-10-31 2022-10-30 2023-10-29 2018-11-25 2019-11-24 2020-11-29 2021-11-28 2022-11-27 2023-11-26 2018-12-30 2019-12-29 2020-12-27 2021-12-26 2022-12-25 2023-12-31
Icon and Unicon
This is a trivial adaptation of the solution to the "Last Friday of each month" task and works in both languages: <lang unicon>procedure main(A) every write(lastsundays(!A)) end
procedure lastsundays(year) every m := 1 to 12 do {
d := case m of {
2 : if IsLeapYear(year) then 29 else 28
4|6|9|11 : 30
default : 31
} # last day of month
z := 0
j := julian(m,d,year) + 1 # first day of next month
until (j-:=1)%7 = 6 do z -:=1 # backup to last sunday (6)
suspend sprintf("%d-%d-%d",year,m,d+z)
}
end
link datetime, printf</lang>
Sample run:
->lsem 2013 2014 2013-1-27 2013-2-24 2013-3-31 2013-4-28 2013-5-26 2013-6-30 2013-7-28 2013-8-25 2013-9-29 2013-10-27 2013-11-24 2013-12-29 2014-1-26 2014-2-23 2014-3-30 2014-4-27 2014-5-25 2014-6-29 2014-7-27 2014-8-31 2014-9-28 2014-10-26 2014-11-30 2014-12-28 ->
J
Same solution as for Last_Friday_of_each_month#J <lang j>require'dates' last_sundays=: 12 {. [: ({:/.~ }:"1)@(#~ 0 = weekday)@todate (i.366) + todayno@,&1 1</lang>
Example use: <lang j> last_sundays 2013 2013 1 27 2013 2 24 2013 3 31 2013 4 28 2013 5 26 2013 6 30 2013 7 28 2013 8 25 2013 9 29 2013 10 27 2013 11 24 2013 12 29</lang>
Java
<lang java>import java.util.Scanner;
public class LastSunday { static final String[] months={"January","February","March","April","May","June","July","August","September","October","November","December"};
public static int[] findLastSunday(int year) { boolean isLeap = isLeapYear(year);
int[] days={31,isLeap?29:28,31,30,31,30,31,31,30,31,30,31}; int[] lastDay=new int[12];
for(int m=0;i<12;i++) { int d; for(d=days[m]; getWeekDay(year,m,d)!=0; d--) ; lastDay[m]=d; }
return lastDay; }
private static boolean isLeapYear(int year) { if(year%4==0) { if(year%100!=0) return true; else if (year%400==0) return true; } return false; }
private static int getWeekDay(int y, int m, int d) { int f=y+d+3*m-1; m++;
if(m<3) y--; else f-=(int)(0.4*m+2.3);
f+=(int)(y/4)-(int)((y/100+1)*0.75); f%=7;
return f; }
private static void display(int year, int[] lastDay) { System.out.println("\nYEAR: "+year); for(int m=0;i<12;i++) System.out.println(months[m]+": "+lastDay[m]); }
public static void main(String[] args) throws Exception { System.out.print("Enter year: "); Scanner s=new Scanner(System.in);
int y=Integer.parseInt(s.next());
int[] lastDay = findLastSunday(y); display(y, lastDay);
s.close(); } }</lang>
- Output:
Enter year: 2013 YEAR: 2013 January: 27 February: 24 March: 31 April: 28 May: 26 June: 30 July: 28 August: 25 September: 29 October: 27 November: 24 December: 29
Java 8
<lang Java>import java.time.*; import java.util.stream.*; import static java.time.temporal.TemporalAdjusters.*;
public class FindTheLastSundayOfEachMonth {
public static Stream<LocalDate> lastSundaysOf(int year) {
return IntStream.rangeClosed(1, 12).mapToObj(month ->
LocalDate.of(year, month, 1).with(lastDayOfMonth())
.with(previousOrSame(DayOfWeek.SUNDAY))
);
}
public static java.util.List<LocalDate> listLastSundaysOf(int year) {
return lastSundaysOf(year).collect(Collectors.toList());
}
public static void main(String[] args) throws Exception {
int year = args.length > 0 ? Integer.parseInt(args[0]) : LocalDate.now().getYear();
for (LocalDate d : listLastSundaysOf(year)) {
System.out.println(d);
};
String result = lastSundaysOf(2013).map(LocalDate::toString).collect(Collectors.joining("\n"));
String test = "2013-01-27\n2013-02-24\n2013-03-31\n2013-04-28\n2013-05-26\n2013-06-30\n2013-07-28\n2013-08-25\n2013-09-29\n2013-10-27\n2013-11-24\n2013-12-29";
if (!test.equals(result)) throw new AssertionError("test failure");
}
}</lang>
JavaScript
ES5
Iteration
<lang javascript>function lastSundayOfEachMonths(year) { var lastDay = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]; var sundays = []; var date, month; if (year % 4 === 0 && (year % 100 !== 0 || year % 400 === 0)) { lastDay[2] = 29; } for (date = new Date(), month = 0; month < 12; month += 1) { date.setFullYear(year, month, lastDay[month]); date.setDate(date.getDate() - date.getDay()); sundays.push(date.toISOString().substring(0, 10)); } return sundays; }
console.log(lastSundayOfEachMonths(2013).join('\n'));</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Functional composition
<lang JavaScript>(function () {
'use strict';
// lastSundaysOfYear :: Int -> [Date]
function lastSundaysOfYear(y) {
return lastWeekDaysOfYear(y, days.sunday);
}
// lastWeekDaysOfYear :: Int -> Int -> [Date]
function lastWeekDaysOfYear(y, iWeekDay) {
return [
31,
0 === y % 4 && 0 !== y % 100 || 0 === y % 400 ? 29 : 28,
31, 30, 31, 30, 31, 31, 30, 31, 30, 31
]
.map(function (d, m) {
var dte = new Date(Date.UTC(y, m, d));
return new Date(Date.UTC(
y, m, d - (
(dte.getDay() + (7 - iWeekDay)) % 7
)
));
});
}
// isoDateString :: Date -> String
function isoDateString(dte) {
return dte.toISOString()
.substr(0, 10);
}
// range :: Int -> Int -> [Int]
function range(m, n) {
return Array.apply(null, Array(n - m + 1))
.map(function (x, i) {
return m + i;
});
}
// transpose :: a -> a function transpose(lst) { return lst[0].map(function (_, iCol) { return lst.map(function (row) { return row[iCol]; }); }); }
var days = {
sunday: 0,
monday: 1,
tuesday: 2,
wednesday: 3,
thursday: 4,
friday: 5,
saturday: 6
}
// TEST
return transpose(
range(2012, 2016)
.map(lastSundaysOfYear)
)
.map(function (row) {
return row
.map(isoDateString)
.join('\t');
})
.join('\n');
})();</lang>
- Output:
2013-01-27 2014-01-26 2015-01-25 2016-01-31 2017-01-29 2013-02-24 2014-02-23 2015-02-22 2016-02-28 2017-02-26 2013-03-31 2014-03-30 2015-03-29 2016-03-27 2017-03-26 2013-04-28 2014-04-27 2015-04-26 2016-04-24 2017-04-30 2013-05-26 2014-05-25 2015-05-31 2016-05-29 2017-05-28 2013-06-30 2014-06-29 2015-06-28 2016-06-26 2017-06-25 2013-07-28 2014-07-27 2015-07-26 2016-07-31 2017-07-30 2013-08-25 2014-08-31 2015-08-30 2016-08-28 2017-08-27 2013-09-29 2014-09-28 2015-09-27 2016-09-25 2017-09-24 2013-10-27 2014-10-26 2015-10-25 2016-10-30 2017-10-29 2013-11-24 2014-11-30 2015-11-29 2016-11-27 2017-11-26 2013-12-29 2014-12-28 2015-12-27 2016-12-25 2017-12-31
ES6
<lang JavaScript>(() => {
'use strict'
// MAIN -----------------------------------------------
// main :: IO ()
const main = () =>
console.log(unlines(
map(
compose(
intercalate('\t'),
map(isoDateString)
)
)(
transpose(
map(lastWeekDaysOfYear(days.sunday))(
enumFromTo(2019)(2022)
)
)
)
));
// WEEKDAYS -------------------------------------------
// lastWeekDaysOfYear :: Int -> Int -> [Date]
const lastWeekDaysOfYear = iWeekDay =>
y => map((d, m) =>
new Date(Date.UTC(
y, m, d - ((new Date(Date.UTC(y, m, d))
.getDay() + (7 - iWeekDay)) % 7))))([
31,
0 === y % 4 && 0 !== y % 100 || 0 === y % 400 ? 29 : 28,
31, 30, 31, 30, 31, 31, 30, 31, 30, 31
]);
const days = {
sunday: 0,
monday: 1,
tuesday: 2,
wednesday: 3,
thursday: 4,
friday: 5,
saturday: 6
};
// GENERIC FUNCTIONS-----------------------------------
// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
x => fs.reduceRight((a, f) => f(a), x);
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m => n =>
Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// intercalate :: String -> [String] -> String
const intercalate = s => xs =>
xs.join(s);
// isoDateString :: Date -> String
const isoDateString = dte =>
dte.toISOString()
.substr(0, 10);
// map :: (a -> b) -> [a] -> [b]
const map = f => xs =>
(Array.isArray(xs) ? (
xs
) : xs.split()).map(f);
// If some of the rows are shorter than the following rows, // their elements are skipped: // > transpose [[10,11],[20],[],[30,31,32]] == [[10,20,30],[11,31],[32]]
// transpose :: a -> a const transpose = xss => { const go = xss => 0 < xss.length ? (() => { const h = xss[0], t = xss.slice(1); return 0 < h.length ? ( [ [h[0]].concat(t.reduce( (a, xs) => a.concat( 0 < xs.length ? ( [xs[0]] ) : [] ), [] )) ].concat(go([h.slice(1)].concat( t.map(xs => xs.slice(1)) ))) ) : go(t); })() : []; return go(xss); };
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// MAIN --- return main();
})();</lang>
- Output:
2019-01-27 2020-01-26 2021-01-31 2022-01-30 2019-02-24 2020-02-23 2021-02-28 2022-02-27 2019-03-31 2020-03-29 2021-03-28 2022-03-27 2019-04-28 2020-04-26 2021-04-25 2022-04-24 2019-05-26 2020-05-31 2021-05-30 2022-05-29 2019-06-30 2020-06-28 2021-06-27 2022-06-26 2019-07-28 2020-07-26 2021-07-25 2022-07-31 2019-08-25 2020-08-30 2021-08-29 2022-08-28 2019-09-29 2020-09-27 2021-09-26 2022-09-25 2019-10-27 2020-10-25 2021-10-31 2022-10-30 2019-11-24 2020-11-29 2021-11-28 2022-11-27 2019-12-29 2020-12-27 2021-12-26 2022-12-25
jq
Foundations <lang jq># In case your jq does not have "until" defined: def until(cond; next):
def _until: if cond then . else (next|_until) end; _until;
- Zeller's Congruence from Day_of_the_week#jq
- Use Zeller's Congruence to determine the day of the week, given
- year, month and day as integers in the conventional way.
- If iso == "iso" or "ISO", then emit an integer in 1 -- 7 where
- 1 represents Monday, 2 Tuesday, etc;
- otherwise emit 0 for Saturday, 1 for Sunday, etc.
def day_of_week(year; month; day; iso):
if month == 1 or month == 2 then [year - 1, month + 12, day] else [year, month, day] end | .[2] + (13*(.[1] + 1)/5|floor) + (.[0]%100) + ((.[0]%100)/4|floor) + (.[0]/400|floor) - 2*(.[0]/100|floor) | if iso == "iso" or iso == "ISO" then 1 + ((. + 5) % 7) else . % 7 end ;</lang>
findLastSundays <lang jq># year and month are numbered conventionally def findLastSunday(year; month):
def isLeapYear: year%4 == 0 and ( year%100!=0 or year%400==0 ) ; def days: if month == 2 then (if isLeapYear then 29 else 28 end) else [31, 28, 31,30,31,30,31,31,30,31,30,31][month-1] end; year as $year | month as $month | days | until( day_of_week($year; $month; .; null) == 1 ; .-1);
- input: year
def findLastSundays:
def months: ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]; . as $year | "YEAR: \(.)", (range(0;12) | "\(months[.]) \(findLastSunday($year; .+1))") ;
$year|tonumber|findLastSundays </lang> Example: <lang sh>$ jq --arg year 2013 -n -r -f findLastSundays.jq YEAR: 2013 January 27 February 24 March 31 April 28 May 26 June 30 July 28 August 25 September 29 October 27 November 24 December 29</lang>
Julia
<lang Julia> isdefined(:Date) || using Dates
const wday = Dates.Sun const lo = 1 const hi = 12
print("\nThis script will print the last ", Dates.dayname(wday)) println("s of each month of the year given.") println("(Leave input empty to quit.)")
while true
print("\nYear> ")
y = chomp(readline())
0 < length(y) || break
y = try
parseint(y)
catch
println("Sorry, but that does not compute as a year.")
continue
end
println()
for m in Date(y, lo):Month(1):Date(y, hi)
println(" ", tolast(m, wday))
end
end </lang>
This code uses the Dates module, which is being incorporated into Julian's standard library with the current development version (0.4). I've used isdefined to make this code good for the current stable version (0.3) as well as for future releases. If Dates is not installed on your instance of Julian try Pkg.add("Dates") from the REPL.
- Output:
This script will print the last Sundays of each month of the year given.
(Leave input empty to quit.)
Year> 2013
2013-01-27
2013-02-24
2013-03-31
2013-04-28
2013-05-26
2013-06-30
2013-07-28
2013-08-25
2013-09-29
2013-10-27
2013-11-24
2013-12-29
Year> this year
Sorry, but that does not compute as a year.
Year>
K
<lang K> / List the dates of last Sundays of each month of / a given year / lastsundt.k
isleap: {(+/~x!' 4 100 400)!2} wd: {(_jd x)!7} dom: (31;28;31;30;31;30;31;31;30;31;30;31) init: {:[isleap x;dom[1]::29;dom[1]::28]} wdme: {[m;y]; init y; dt:(10000*y)+(100*m)+dom[m-1];jd::(_jd dt);mewd::(wd dt)} lsd: {[m;y]; wdme[m;y];:[mewd>5;jd::jd+(6-mewd);jd::jd-(1+mewd)];dt:_dj(jd);yy:$(yr:dt%10000);dd:$(d:dt!100);mm:$(mo:((dt-yr*10000)%100));arr::arr,$(yy,"-",(2$mm),"-",(2$dd))} lsd1: {[y];arr::(); m:1; do[12;lsd[m;y];m+:1]} main: {[y]; lsd1[y];`0: ,"Dates of last Sundays of ",($y); 12 10#arr}
</lang> The output of a session with the above script is given below:
- Output:
K Console - Enter \ for help
\l lastsundt
main 2013
Dates of last Sundays of 2013
("2013- 1-27"
"2013- 2-24"
"2013- 3-31"
"2013- 4-28"
"2013- 5-26"
"2013- 6-30"
"2013- 7-28"
"2013- 8-25"
"2013- 9-29"
"2013-10-27"
"2013-11-24"
"2013-12-29")
Kotlin
<lang scala>// version 1.0.6
import java.util.*
fun main(args: Array<String>) {
print("Enter a year : ")
val year = readLine()!!.toInt()
println("The last Sundays of each month in $year are as follows:")
val calendar = GregorianCalendar(year, 0, 31)
for (month in 1..12) {
val daysInMonth = calendar.getActualMaximum(Calendar.DAY_OF_MONTH)
val lastSunday = daysInMonth - (calendar[Calendar.DAY_OF_WEEK] - Calendar.SUNDAY)
println("$year-" + "%02d-".format(month) + "%02d".format(lastSunday))
if (month < 12) {
calendar.add(Calendar.DAY_OF_MONTH, 1)
calendar.add(Calendar.MONTH, 1)
calendar.add(Calendar.DAY_OF_MONTH, -1)
}
}
}</lang> Sample input/output:
- Output:
Enter a year : 2013 The last Sundays of each month in 2013 are as follows: 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Lasso
<lang Lasso>local( year = integer(web_request -> param('year') || 2013), date = date(#year + '-1-1'), lastsu = array, lastday )
with month in generateseries(1,12) do { #date -> day = 1 #date -> month = #month #lastday = #date -> month(-days) #date -> day = #lastday loop(7) => { if(#date -> dayofweek == 1) => { #lastsu -> insert(#date -> format(`dd MMMM`)) loop_abort } #date -> day-- } }
- lastsu -> join('
')</lang>
27 January 24 February 31 March 28 April 26 May 30 June 28 July 25 August 29 September 27 October 24 November 29 December
Liberty BASIC
This program goes a step further and provides a function definition, NthDayOfMonth(), that returns the date of any occurrence of a day in any given month and year. For example: The third Monday in February, the last Monday in May, the fourth Thursday in November, etc. <lang lb> yyyy=2013: if yyyy<1901 or yyyy>2099 then end nda$="Lsu" print "The last Sundays of "; yyyy for mm=1 to 12
x=NthDayOfMonth(yyyy, mm, nda$)
select case mm
case 1: print " January "; x
case 2: print " February "; x
case 3: print " March "; x
case 4: print " April "; x
case 5: print " May "; x
case 6: print " June "; x
case 7: print " July "; x
case 8: print " August "; x
case 9: print " September "; x
case 10: print " October "; x
case 11: print " November "; x
case 12: print " December "; x
end select
next mm end
function NthDayOfMonth(yyyy, mm, nda$) ' nda$ is a two-part code. The first character, n, denotes ' first, second, third, fourth, and last by 1, 2, 3, 4, or L. ' The last two characters, da, denote the day of the week by ' mo, tu, we, th, fr, sa, or su. For example: ' the nda$ for the second Monday of a month is "2mo"; ' the nda$ for the last Thursday of a month is "Lth".
if yyyy<1900 or yyyy>2099 or mm<1 or mm>12 then
NthDayOfMonth=0: exit function
end if
nda$=lower$(trim$(nda$))
if len(nda$)<>3 then NthDayOfMonth=0: exit function
n$=left$(nda$,1): nC$="1234l"
da$=right$(nda$,2): daC$="tuwethfrsasumotuwethfrsasumo"
if not(instr(nC$,n$)) or not(instr(daC$,da$)) then
NthDayOfMonth=0: exit function
end if
NthDayOfMonth=1
mm$=str$(mm): if mm<10 then mm$="0"+mm$
db$=DayOfDate$(str$(yyyy)+mm$+"01")
if da$<>db$ then
x=instr(daC$,db$): y=instr(daC$,da$,x): NthDayOfMonth=1+(y-x)/2
end if
dim MD(12)
MD(1)=31: MD(2)=28: MD(3)=31: MD(4)=30: MD(5)=31: MD(6)=30
MD(7)=31: MD(8)=31: MD(9)=30: MD(10)=31: MD(11)=30: MD(12)=31
if yyyy mod 4 = 0 then MD(2)=29
if n$<>"1" then
if n$<>"l" then
NthDayOfMonth=NthDayOfMonth+((val(n$)-1)*7)
else
if NthDayOfMonth+27<MD(mm) then
NthDayOfMonth=NthDayOfMonth+28
else
NthDayOfMonth=NthDayOfMonth+21
end if
end if
end if
end function
function DayOfDate$(ObjectDate$) 'yyyymmdd format
if ObjectDate$="" then 'today
DaysSince1900 = date$("days")
else
DaysSince1900 = date$(mid$(ObjectDate$,5,2)+"/"+right$(ObjectDate$,2)_
+"/"+left$(ObjectDate$,4))
end if
DayOfWeek = DaysSince1900 mod 7
select case DayOfWeek
case 0: DayOfDate$="tu"
case 1: DayOfDate$="we"
case 2: DayOfDate$="th"
case 3: DayOfDate$="fr"
case 4: DayOfDate$="sa"
case 5: DayOfDate$="su"
case 6: DayOfDate$="mo"
end select
end function </lang>
- Output:
The last Sundays of 2013 January 27 February 24 March 31 April 28 May 26 June 30 July 28 August 25 September 29 October 27 November 24 December 29
LiveCode
Abstracted version of "last friday of each month". LiveCode dateItem item 7 is day of week. It is numbered 1 (Sunday) to 7 (Saturday). <lang LiveCode>function lastDay yyyy, dayofweek
-- year,month num,day of month,hour in 24-hour time,minute,second,numeric day of week.
convert the long date to dateitems
put 1 into item 2 of it
put 1 into item 3 of it
put yyyy into item 1 of it
put it into startDate
convert startDate to dateItems
repeat with m = 1 to 12
put m into item 2 of startDate
repeat with d = 20 to 31
put d into item 3 of startDate
convert startDate to dateItems
-- 1 is Sunday through to 7 Saturday
if item 7 of startDate is dayofweek and item 1 of startDate is yyyy and item 2 of startDate is m then
put item 3 of startDate into mydays[item 2 of startDate]
end if
end repeat
end repeat
combine mydays using cr and space
sort mydays ascending numeric
return mydays
end lastDay</lang>Example<lang LiveCode>put lastDay(2013, 1)</lang>Output
1 27 2 24 3 31 4 28 5 26 6 30 7 28 8 25 9 29 10 27 11 24 12 29
Lua
<lang Lua>function isLeapYear (y)
return (y % 4 == 0 and y % 100 ~=0) or y % 400 == 0
end
function dayOfWeek (y, m, d)
local t = os.time({year = y, month = m, day = d})
return os.date("%A", t)
end
function lastWeekdays (wday, year)
local monthLength, day = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}
if isLeapYear(year) then monthLength[2] = 29 end
for month = 1, 12 do
day = monthLength[month]
while dayOfWeek(year, month, day) ~= wday do day = day - 1 end
print(year .. "-" .. month .. "-" .. day)
end
end
lastWeekdays("Sunday", tonumber(arg[1]))</lang> Command line session:
>lua lastSundays.lua 2013 2013-1-27 2013-2-24 2013-3-31 2013-4-28 2013-5-26 2013-6-30 2013-7-28 2013-8-25 2013-9-29 2013-10-27 2013-11-24 2013-12-29 >
Maple
<lang Maple>sundays := proc(year) local i, dt, change, last_days; last_days := [31,28,31,30,31,30,31,31,30,31,30,31]; if (Calendar:-IsLeapYear(year)) then last_days[2] := 28; end if; for i to 12 do dt := Date(year, i, last_days[i]); change := 0; if not(Calendar:-DayOfWeek(dt) = 1) then change := -Calendar:-DayOfWeek(dt) + 1; end if; dt := Calendar:-AdjustDateField(dt, "date", change); printf("%d-%d-%d\n", year, Month(dt), DayOfMonth(dt)); end do; end proc;
sundays(2013);</lang>
- Output:
2013-1-27 2013-2-24 2013-3-31 2013-4-28 2013-5-26 2013-6-30 2013-7-28 2013-8-25 2013-9-29 2013-10-27 2013-11-24 2013-12-29
Mathematica / Wolfram Language
<lang mathematica>LastSundays[year_] :=
Table[Last@
DayRange[{year, i},
DatePlus[{year, i}, {{1, "Month"}, {-1, "Day"}}], Sunday], {i,
12}]
LastSundays[2013]</lang>
- Output:
{{2013, 1, 27}, {2013, 2, 24}, {2013, 3, 31}, {2013, 4, 28}, {2013, 5,
26}, {2013, 6, 30}, {2013, 7, 28}, {2013, 8, 25}, {2013, 9,
29}, {2013, 10, 27}, {2013, 11, 24}, {2013, 12, 29}}
Nim
<lang nim>import os, strutils, times
const
DaysInMonth: array[Month, int] = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] DayDiffs: array[WeekDay, int] = [1, 2, 3, 4, 5, 6, 0]
let year = paramStr(1).parseInt
for month in mJan..mDec:
var lastDay = DaysInMonth[month]
if month == mFeb and year.isLeapYear: lastDay = 29
var date = initDateTime(lastDay, month, year, 0, 0, 0)
date = date - days(DayDiffs[date.weekday])
echo date.format("yyyy-MM-dd")</lang>
Sample usage:
./lastsunday 2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
OCaml
<lang OCaml> let is_leap_year y =
(* See OCaml solution on Rosetta Code for
determing if it's a leap year *)
if (y mod 100) = 0
then (y mod 400) = 0
else (y mod 4) = 0;;
let get_days y =
if is_leap_year y
then
[31;29;31;30;31;30;31;31;30;31;30;31]
else
[31;28;31;30;31;30;31;31;30;31;30;31];;
let print_date = Printf.printf "%d/%d/%d\n";;
let get_day_of_week y m d =
let y = if m > 2 then y else y - 1 in let c = y / 100 in let y = y mod 100 in let m_shifted = float_of_int ( ((m + 9) mod 12) + 1) in let m_factor = int_of_float (2.6 *. m_shifted -. 0.2) in let leap_factor = 5 * (y mod 4) + 3 * (y mod 7) + 5 * (c mod 4) in (d + m_factor + leap_factor) mod 7;;
let get_shift y m last_day =
get_day_of_week y m last_day;;
let print_last_sunday y m =
let days = get_days y in let last_day = List.nth days (m - 1) in let last_sunday = last_day - (get_shift y m last_day) in print_date y m last_sunday;;
let print_last_sundays y =
let months = [1;2;3;4;5;6;7;8;9;10;11;12] in List.iter (print_last_sunday y) months;;
match (Array.length Sys.argv ) with
2 -> print_last_sundays( int_of_string (Sys.argv.(1))); |_ -> invalid_arg "Please enter a year";
</lang> Sample usage:
ocaml sundays.ml 2013 2013/01/27 2013/02/24 2013/03/31 2013/04/28 2013/05/26 2013/06/30 2013/07/28 2013/08/25 2013/09/29 2013/10/27 2013/11/24 2013/12/29
Oforth
<lang Oforth>import: date
- lastSunday(y)
| m |
Date.JANUARY Date.DECEMBER for: m [
Date newDate(y, m, Date.DaysInMonth(y, m))
while(dup dayOfWeek Date.SUNDAY <>) [ addDays(-1) ] println
] ;</lang>
- Output:
2013-01-27 00:00:00,000 2013-02-24 00:00:00,000 2013-03-31 00:00:00,000 2013-04-28 00:00:00,000 2013-05-26 00:00:00,000 2013-06-30 00:00:00,000 2013-07-28 00:00:00,000 2013-08-25 00:00:00,000 2013-09-29 00:00:00,000 2013-10-27 00:00:00,000 2013-11-24 00:00:00,000 2013-12-29 00:00:00,000
PARI/GP
<lang parigp>\\ Normalized Julian Day Number from date njd(D) = {
my (m = D[2], y = D[1]); if (D[2] > 2, m++, y--; m += 13); (1461 * y) \ 4 + (306001 * m) \ 10000 + D[3] - 694024 + 2 - y \ 100 + y \ 400
}
\\ Date from Normalized Julian Day Number njdate(J) = {
my (a = J + 2415019, b = (4 * a - 7468865) \ 146097, c, d, m, y);
a += 1 + b - b \ 4 + 1524; b = (20 * a - 2442) \ 7305; c = (1461 * b) \ 4; d = ((a - c) * 10000) \ 306001; m = d - 1 - 12 * (d > 13); y = b - 4715 - (m > 2); d = a - c - (306001 * d) \ 10000;
[y, m, d]
}
for (m=1, 12, a=njd([2013,m+1,0]); print(njdate(a-(a+6)%7)))</lang>
Output:
[2013, 1, 27] [2013, 2, 24] [2013, 3, 31] [2013, 4, 28] [2013, 5, 26] [2013, 6, 30] [2013, 7, 28] [2013, 8, 25] [2013, 9, 29] [2013, 10, 27] [2013, 11, 24] [2013, 12, 29]
Perl
<lang Perl>#!/usr/bin/perl use strict ; use warnings ; use DateTime ;
for my $i( 1..12 ) {
my $date = DateTime->last_day_of_month( year => $ARGV[ 0 ] ,
month => $i ) ;
while ( $date->dow != 7 ) {
$date = $date->subtract( days => 1 ) ;
}
my $ymd = $date->ymd ;
print "$ymd\n" ;
}</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Phix
Requires 0.8.1 (day_of_week() is now ISO 8601 compliant) <lang Phix>include timedate.e
constant SUNDAY=7
procedure showlast(integer dow, integer doy, timedate td)
td = adjust_timedate(td,timedelta(days:=doy-1))
integer {year,month,day} = td
while day_of_week(year,month,day)!=dow do day-=1 end while
printf(1,"%4d-%02d-%02d\n",{year,month,day})
end procedure
procedure last_day_of_month(integer year, integer dow) integer doy timedate first = {year,1,1,0,0,0,0,0}
-- start by finding the 1st of the next month, less 1
for i=1 to 11 do
doy = day_of_year(year,i+1,1)-1
showlast(dow,doy,first)
end for
-- do December separately, as 1st would be next year
doy = day_of_year(year,12,31)
showlast(dow,doy,first)
end procedure last_day_of_month(2013,SUNDAY)</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
PHP
<lang PHP><?php // Created with PHP 7.0
function printLastSundayOfAllMonth($year) {
$months = array(
'January', 'February', 'March', 'April', 'June', 'July',
'August', 'September', 'October', 'November', 'December');
foreach ($months as $month) {
echo $month . ': ' . date('Y-m-d', strtotime('last sunday of ' . $month . ' ' . $year)) . "\n";
}
}
printLastSundayOfAllMonth($argv[1]); </lang>
- Output:
January: 2013-01-27 February: 2013-02-24 March: 2013-03-31 April: 2013-04-28 June: 2013-06-30 July: 2013-07-28 August: 2013-08-25 September: 2013-09-29 October: 2013-10-27 November: 2013-11-24 December: 2013-12-29
PicoLisp
<lang PicoLisp>(de lastSundays (Y)
(for M 12
(prinl
(dat$
(find '((D) (= "Sunday" (day D)))
(mapcar '((D) (date Y M D)) `(range 31 22)) )
"-" ) ) ) )</lang>
Test: <lang PicoLisp>: (lastSundays 2013) 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29</lang>
PowerShell
<lang PowerShell> function last-dayofweek {
param(
[Int][ValidatePattern("[1-9][0-9][0-9][0-9]")]$year,
[String][validateset('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday')]$dayofweek
)
$date = (Get-Date -Year $year -Month 1 -Day 1)
while($date.DayOfWeek -ne $dayofweek) {$date = $date.AddDays(1)}
while($date.year -eq $year) {
if($date.Month -ne $date.AddDays(7).Month) {$date.ToString("yyyy-dd-MM")}
$date = $date.AddDays(7)
}
} last-dayofweek 2013 "Sunday" </lang> Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Alternate Version
This script finds the first and/or last or all dates of any of the days of week; accepts [Int32] and [DateTime] values for Month and Year parameters; outputs [DateTime] objects by default but has an option to output time strings in various formats. This script also allows for pipeline input based mainly upon the Month parameter.
This script has a syntax as complex as any PowerShell Cmdlet because it attempts to do everything.
<lang PowerShell>
function Get-Date0fDayOfWeek
{
[CmdletBinding(DefaultParameterSetName="None")]
[OutputType([datetime])]
Param
(
[Parameter(Mandatory=$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
}
}
}
}
</lang>
The default is to return [DateTime] objects:
<lang PowerShell>
1..12 | Get-Date0fDayOfWeek -Year 2013 -Last -Sunday
</lang>
- Output:
Sunday, January 27, 2013 12:00:00 AM Sunday, February 24, 2013 12:00:00 AM Sunday, March 31, 2013 12:00:00 AM Sunday, April 28, 2013 12:00:00 AM Sunday, May 26, 2013 12:00:00 AM Sunday, June 30, 2013 12:00:00 AM Sunday, July 28, 2013 12:00:00 AM Sunday, August 25, 2013 12:00:00 AM Sunday, September 29, 2013 12:00:00 AM Sunday, October 27, 2013 12:00:00 AM Sunday, November 24, 2013 12:00:00 AM Sunday, December 29, 2013 12:00:00 AM
Return the [DateTime] objects as strings (using the default string format):
<lang PowerShell>
1..12 | Get-Date0fDayOfWeek -Year 2013 -Last -Sunday -AsString
</lang>
- Output:
27-Jan-2013 24-Feb-2013 31-Mar-2013 28-Apr-2013 26-May-2013 30-Jun-2013 28-Jul-2013 25-Aug-2013 29-Sep-2013 27-Oct-2013 24-Nov-2013 29-Dec-2013
Return the [DateTime] objects as strings (specifying the string format):
<lang PowerShell>
1..12 | Get-Date0fDayOfWeek -Year 2013 -Last -Sunday -AsString -Format yyyy-MM-dd
</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
PureBasic
<lang PureBasic>Procedure LastSundayOfEachMonth(yyyy.i,List lfem.i())
Define dv.i=ParseDate("%yyyy",Str(yyyy)), mv.i=1
NewList d.i()
For d=1 To 365
dv=AddDate(dv,#PB_Date_Day,1)
If DayOfWeek(dv)=0
AddElement(d()) : d()=dv
EndIf
Next
dv=0
For mv=1 To 12
ForEach d()
If dv<d() And Month(d())=mv
dv=d()
EndIf
Next
AddElement(lfem()) : lfem()=dv
Next
EndProcedure
NewList lf.i() Define y.i OpenConsole("Last Sunday of each month") Print("Input Year [ 1971 < y < 2038 ]: ") y=Val(Input()) If y>1971 And y<2038
PrintN("Last Sunday of each month...")
LastSundayOfEachMonth(y,lf())
ForEach lf()
PrintN(FormatDate("%dd.%mm.%yyyy",lf()))
Next
EndIf Print("...End") Input()</lang>
- Output:
Input Year [ 1971 < y < 2038 ]: 2013 Last Sunday of each month... 27.01.2013 24.02.2013 31.03.2013 28.04.2013 26.05.2013 30.06.2013 28.07.2013 25.08.2013 29.09.2013 27.10.2013 24.11.2013 29.12.2013 ...End
Python
<lang python> import sys import calendar
year = 2013 if len(sys.argv) > 1:
try:
year = int(sys.argv[-1])
except ValueError:
pass
for month in range(1, 13):
last_sunday = max(week[-1] for week in calendar.monthcalendar(year, month))
print('{}-{}-{:2}'.format(year, calendar.month_abbr[month], last_sunday))
</lang>
Output: <lang> 2013-Jan-27 2013-Feb-24 2013-Mar-31 2013-Apr-28 2013-May-26 2013-Jun-30 2013-Jul-28 2013-Aug-25 2013-Sep-29 2013-Oct-27 2013-Nov-24 2013-Dec-29 </lang>
R
<lang rsplus> last_sundays <- function(year) { for (month in 1:12) { if (month == 12) { date <- as.Date(paste0(year,"-",12,"-",31)) } else { date <- as.Date(paste0(year,"-",month+1,"-",1))-1 } while (weekdays(date) != "Sunday") { date <- date - 1 } print(date) } } last_sundays(2004) </lang>
- Output:
[1] "2004-01-25" [1] "2004-02-29" [1] "2004-03-28" [1] "2004-04-25" [1] "2004-05-30" [1] "2004-06-27" [1] "2004-07-25" [1] "2004-08-29" [1] "2004-09-26" [1] "2004-10-31" [1] "2004-11-28" [1] "2004-12-26"
Racket
<lang racket>
- lang racket
(require srfi/19 math)
(define (days-in-month m y)
(define lengths #(0 31 #f 31 30 31 30 31 31 30 31 30 31)) (define d (vector-ref lengths m)) (or d (days-in-feb y)))
(define (leap-year? y)
(and (divides? 4 y)
(or (not (divides? 100 y))
(divides? 400 y))))
(define (days-in-feb y)
(if (leap-year? y) 29 28))
(define (last-day-in-month m y)
(make-date 0 0 0 0 (days-in-month m y) m y 0))
(define (week-day date)
(define days #(sun mon tue wed thu fri sat)) (vector-ref days (date-week-day date)))
(define (last-sundays y)
(for/list ([m (in-range 1 13)]) (prev-sunday (last-day-in-month m y))))
(define 24hours (make-time time-duration 0 (* 24 60 60)))
(define (prev-day d)
(time-utc->date (subtract-duration (date->time-utc d) 24hours)))
(define (prev-sunday d)
(if (eq? (week-day d) 'sun)
d
(prev-sunday (prev-day d))))
(for ([d (last-sundays 2013)])
(displayln (~a (date->string d "~a ~d ~b ~Y"))))
</lang>
- Output:
Sun 27 Jan 2013 Sun 24 Feb 2013 Sun 31 Mar 2013 Sun 28 Apr 2013 Sun 26 May 2013 Sun 30 Jun 2013 Sun 28 Jul 2013 Sun 25 Aug 2013 Sun 29 Sep 2013 Sun 27 Oct 2013 Sun 24 Nov 2013 Sun 29 Dec 2013
Raku
(formerly Perl 6)
<lang perl6>sub MAIN ($year = Date.today.year) {
for 1..12 -> $month {
my $month-end = Date.new($year, $month, Date.new($year,$month,1).days-in-month);
say $month-end - $month-end.day-of-week % 7;
}
}</lang>
- Output — for input
2018:
2018-01-28 2018-02-25 2018-03-25 2018-04-29 2018-05-27 2018-06-24 2018-07-29 2018-08-26 2018-09-30 2018-10-28 2018-11-25 2018-12-30
REBOL
<lang REBOL>#!/usr/bin/env rebol
last-sundays-of-year: function [
"Return series of last sunday (date!) for each month of the year"
year [integer!] "which year?"
][
d: to-date reduce [1 1 year] ; start with first day of year
collect [
repeat month 12 [
d/month: month + 1 ; move to start of next month
keep d - d/weekday ; calculate last sunday & keep
]
]
]
foreach sunday last-sundays-of-year to-integer system/script/args [print sunday] </lang>
- Output:
./last-sundays.reb 2013 27-Jan-2013 24-Feb-2013 31-Mar-2013 28-Apr-2013 26-May-2013 30-Jun-2013 28-Jul-2013 25-Aug-2013 29-Sep-2013 27-Oct-2013 24-Nov-2013 29-Dec-2013
REXX
This REXX example is an exact replication of the Rosetta Code
- find last Fridays of each month for any year
except for the innards of the first DO loop.
The lastDOW subroutine can be used for any day-of-the-week for any month for any year.
<lang rexx>/*REXX program displays dates of last Sundays of each month for any year*/
parse arg yyyy
do j=1 for 12
_ = lastDOW('Sunday', j, yyyy)
say right(_,4)'-'right(j,2,0)"-"left(word(_,2),2)
end /*j*/
exit /*stick a fork in it, we're done.*/ /*┌────────────────────────────────────────────────────────────────────┐
│ lastDOW: procedure to return the date of the last day-of-week of │ │ any particular month of any particular year. │ │ │ │ The day-of-week must be specified (it can be in any case, │ │ (lower-/mixed-/upper-case) as an English name of the spelled day │ │ of the week, with a minimum length that causes no ambiguity. │ │ I.E.: W for Wednesday, Sa for Saturday, Su for Sunday ... │ │ │ │ The month can be specified as an integer 1 ──► 12 │ │ 1=January 2=February 3=March ... 12=December │ │ or the English name of the month, with a minimum length that │ │ causes no ambiguity. I.E.: Jun for June, D for December. │ │ If omitted [or an asterisk(*)], the current month is used. │ │ │ │ The year is specified as an integer or just the last two digits │ │ (two digit years are assumed to be in the current century, and │ │ there is no windowing for a two-digit year). │ │ If omitted [or an asterisk(*)], the current year is used. │ │ Years < 100 must be specified with (at least 2) leading zeroes.│ │ │ │ Method used: find the "day number" of the 1st of the next month, │ │ then subtract one (this gives the "day number" of the last day of │ │ the month, bypassing the leapday mess). The last day-of-week is │ │ then obtained straightforwardly, or via subtraction. │ └────────────────────────────────────────────────────────────────────┘*/
lastdow: procedure; arg dow .,mm .,yy . /*DOW = day of week*/ parse arg a.1,a.2,a.3 /*orig args, errmsg*/ if mm== | mm=='*' then mm=left(date('U'),2) /*use default month*/ if yy== | yy=='*' then yy=left(date('S'),4) /*use default year */ if length(yy)==2 then yy=left(date('S'),2)yy /*append century. */
/*Note mandatory leading blank in strings below.*/
$=" Monday TUesday Wednesday THursday Friday SAturday SUnday" !=" JAnuary February MARch APril MAY JUNe JULy AUgust September",
" October November December"
upper $ ! /*uppercase strings*/ if dow== then call .er "wasn't specified",1 if arg()>3 then call .er 'arguments specified',4
do j=1 for 3 /*any plural args ?*/ if words(arg(j))>1 then call .er 'is illegal:',j end
dw=pos(' 'dow,$) /*find day-of-week*/ if dw==0 then call .er 'is invalid:',1 if dw\==lastpos(' 'dow,$) then call .er 'is ambigious:',1
if datatype(mm,'month') then /*if MM is alpha...*/
do
m=pos(' 'mm,!) /*maybe its good...*/
if m==0 then call .er 'is invalid:',1
if m\==lastpos(' 'mm,!) then call .er 'is ambigious:',2
mm=wordpos(word(substr(!,m),1),!)-1 /*now, use true Mon*/
end
if \datatype(mm,'W') then call .er "isn't an integer:",2 if \datatype(yy,'W') then call .er "isn't an integer:",3 if mm<1 | mm>12 then call .er "isn't in range 1──►12:",2 if yy=0 then call .er "can't be 0 (zero):",3 if yy<0 then call .er "can't be negative:",3 if yy>9999 then call .er "can't be > 9999:",3
tdow=wordpos(word(substr($,dw),1),$)-1 /*target DOW, 0──►6*/
/*day# of last dom.*/
_=date('B',right(yy+(mm=12),4)right(mm//12+1,2,0)"01",'S')-1 ?=_//7 /*calc. DOW, 0──►6*/ if ?\==tdow then _=_-?-7+tdow+7*(?>tdow) /*not DOW? Adjust.*/ return date('weekday',_,"B") date(,_,'B') /*return the answer*/
.er: arg ,_;say; say '***error!*** (in LASTDOW)';say /*tell error, and */
say word('day-of-week month year excess',arg(2)) arg(1) a._
say; exit 13 /*... then exit. */</lang>
- output when using the default input (the current year, 2013):
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Ring
<lang ring> see "What year to calculate (yyyy) : " give year see "Last Sundays in " + year + " are on :" + nl month = list(12) mo = [4,0,0,3,5,1,3,6,2,4,0,2] mon = [31,28,31,30,31,30,31,31,30,31,30,31] if year < 2100 leap = year - 1900 else leap = year - 1904 ok m = ((year-1900)%7) + floor(leap/4) % 7 for n = 1 to 12
month[n] = (mo[n] + m) % 7
next for n = 1 to 12
for i = (mon[n] - 6) to mon[n]
x = (month[n] + i) % 7
if n < 10 strn = "0" + string(n) else strn = string(n) ok
if x = 4 see year + "-" + strn + "-" + string(i) + nl ok
next
next </lang> Output:
What year to calculate (yyyy) : 2013 Last Sundays in 2013 are on : 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Ruby
<lang ruby>require 'date'
def last_sundays_of_year(year = Date.today.year)
(1..12).map do |month| d = Date.new(year, month, -1) # -1 means "last". d - d.wday end
end
puts last_sundays_of_year(2013)</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Results before the year 1581 may differ from other languages - the Date library takes the Julian reform into account.
Run BASIC
<lang runbasic>input "What year to calculate (yyyy) : ";year print "Last Sundays in ";year;" are:" dim month(12) mo$ = "4 0 0 3 5 1 3 6 2 4 0 2" mon$ = "31 28 31 30 31 30 31 31 30 31 30 31"
if year < 2100 then leap = year - 1900 else leap = year - 1904 m = ((year-1900) mod 7) + int(leap/4) mod 7 for n = 1 to 12
month(n) = (val(word$(mo$,n)) + m) mod 7 month(n) = (val(word$(mo$,n)) + m) mod 7
next for n = 1 to 12
for i = (val(word$(mon$,n)) - 6) to val(word$(mon$,n))
x = (month(n) + i) mod 7
if x = 4 then print year ; "-";right$("0"+str$(n),2);"-" ; i
next
next</lang>
What year to calculate (yyyy) : ?2013 Last Sundays in 2013 are: 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Rust
<lang rust>use std::env::args; use time::{Date, Duration};
fn main() {
let year = args().nth(1).unwrap().parse::<i32>().unwrap();
(1..=12)
.map(|month| Date::try_from_ymd(year + month / 12, ((month % 12) + 1) as u8, 1))
.filter_map(|date| date.ok())
.for_each(|date| {
let days_back =
Duration::days(((date.weekday().number_from_sunday() as i64 + 5) % 7) + 1);
println!("{}", date - days_back);
});
}</lang>
- Output:
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Scala
<lang Scala>object FindTheLastSundayOfEachMonth extends App {
import java.util.Calendar._ val cal = getInstance
def lastSundaysOf(year: Int) =
(JANUARY to DECEMBER).map{month =>
cal.set(year, month + 1, 1) // first day of next month
(1 to 7).find{_ => cal.add(DAY_OF_MONTH, -1); cal.get(DAY_OF_WEEK) == SUNDAY}
cal.getTime
}
val year = args.headOption.map(_.toInt).getOrElse(cal.get(YEAR))
val fmt = new java.text.SimpleDateFormat("yyyy-MM-dd")
println(lastSundaysOf(year).map(fmt.format) mkString "\n")
}</lang>
Java 8
<lang Scala>object FindTheLastSundayOfEachMonth extends App {
def lastSundaysOf(year: Int) = (1 to 12).map{month =>
import java.time._; import java.time.temporal.TemporalAdjusters._
LocalDate.of(year, month, 1).`with`(lastDayOfMonth).`with`(previousOrSame(DayOfWeek.SUNDAY))}
val year = args.headOption.map(_.toInt).getOrElse(java.time.LocalDate.now.getYear) println(lastSundaysOf(year) mkString "\n")
}</lang>
Seed7
Uses the libraries time.s7i and duration.s7i. Applicable to any day of the week, cf. [[3]].
<lang seed7>$ include "seed7_05.s7i";
include "time.s7i"; include "duration.s7i";
const proc: main is func
local
var integer: weekday is 1; # 1 for monday, 2 for tuesday, and so on up to 7 for sunday.
var integer: year is 0;
var integer: month is 1;
var time: aDate is time.value;
var time: selected is time.value;
begin
if length(argv(PROGRAM)) <> 2 then
writeln("usage: lastWeekdayInMonth weekday year");
writeln(" weekday: 1 for monday, 2 for tuesday, and so on up to 7 for sunday.");
else
weekday := integer parse (argv(PROGRAM)[1]);
year := integer parse (argv(PROGRAM)[2]);
for month range 1 to 12 do
aDate := date(year, month, 1);
while aDate.month = month do
if dayOfWeek(aDate) = weekday then
selected := aDate;
end if;
aDate +:= 1 . DAYS;
end while;
writeln(strDate(selected));
end for;
end if;
end func;</lang>
- Output:
when called with s7 rosetta/lastWeekdayInMonth 7 2013
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Sidef
<lang ruby>var dt = require('DateTime'); var (year=2016) = ARGV.map{.to_i}...
for i in (1 .. 12) {
var date = dt.last_day_of_month(
year => year,
month => i
);
while (date.dow != 7) {
date = date.subtract(days => 1);
}
say date.ymd;
}</lang>
- Output:
$ sidef last_sunday.sf 2013 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
Smalltalk
<lang smalltalk> Pharo Smalltalk
[ :yr | | firstDay firstSunday |
firstDay := Date year: yr month: 1 day: 1.
firstSunday := firstDay addDays: (1 - firstDay dayOfWeek).
(0 to: 53)
collect: [ :each | firstSunday addDays: (each * 7) ]
thenSelect: [ :each |
(((Date daysInMonth: each monthIndex forYear: yr) - each dayOfMonth) <= 6) and: [ each year = yr ] ] ]
</lang>
- Output:
Send value: 2013 to the above block to return an array: (27 January 2013 24 February 2013 31 March 2013 28 April 2013 26 May 2013 30 June 2013 28 July 2013 25 August 2013 29 September 2013 27 October 2013 24 November 2013 29 December 2013)
Stata
<lang stata>program last_sundays args year clear qui set obs 12 gen day=dofm(mofd(mdy(_n,1,`year'))+1)-1 qui replace day=day-mod(dow(day),7) format %td day list, noobs noheader sep(6) end
last_sundays 2013
+-----------+ | 27jan2013 | | 24feb2013 | | 31mar2013 | | 28apr2013 | | 26may2013 | | 30jun2013 | |-----------| | 28jul2013 | | 25aug2013 | | 29sep2013 | | 27oct2013 | | 24nov2013 | | 29dec2013 | +-----------+</lang>
Swift
<lang Swift>import Foundation
func lastSundays(of year: Int) -> [Date] {
let calendar = Calendar.current var dates = [Date]()
for month in 1...12 {
var dateComponents = DateComponents(calendar: calendar, year: year, month: month + 1, day: 0, hour: 12)
let date = calendar.date(from: dateComponents)! let weekday = calendar.component(.weekday, from: date)
if weekday != 1 { dateComponents.day! -= weekday - 1 }
dates.append(calendar.date(from: dateComponents)!) } return dates }
var dateFormatter = DateFormatter() dateFormatter.dateStyle = .short
print(lastSundays(of: 2013).map(dateFormatter.string).joined(separator: "\n"))</lang>
- Output:
1/27/13 2/24/13 3/31/13 4/28/13 5/26/13 6/30/13 7/28/13 8/25/13 9/29/13 10/27/13 11/24/13 12/29/13
Tcl
<lang tcl>proc lastSundays Template:Year "" {
if {$year eq ""} {
set year [clock format [clock seconds] -gmt 1 -format "%Y"]
}
foreach month {2 3 4 5 6 7 8 9 10 11 12 13} {
set d [clock add [clock scan "$month/1/$year" -gmt 1] -1 day] while {[clock format $d -gmt 1 -format "%u"] != 7} { set d [clock add $d -1 day] } lappend result [clock format $d -gmt 1 -format "%Y-%m-%d"]
} return $result
} puts [join [lastSundays {*}$argv] "\n"]</lang>
- Output:
When called as: “tclsh lastSundays.tcl 2013” (or with the year argument omitted during 2013)
2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-20 2013-11-24 2013-12-29
UNIX Shell
This uses the programs awk and cal, which are usually installed on any POSIXlike system.
<lang sh>last_sundays() {
local y=$1
for (( m=1; m<=12; ++m )); do
cal $m $y | awk -vy=$y -vm=$m '/^.[0-9]/ {d=$1} END {print y"-"m"-"d}'
done
}</lang>
- Output:
$ last_sundays 2013 2013-1-27 2013-2-24 2013-3-31 2013-4-28 2013-5-26 2013-6-30 2013-7-28 2013-8-25 2013-9-29 2013-10-27 2013-11-24 2013-12-29
VBScript
<lang VBScript> strYear = WScript.StdIn.ReadLine
For i = 1 To 12 d = DateSerial(strYear, i + 1, 1) - 1 WScript.Echo d - Weekday(d) + 1 Next </lang>
Wren
<lang ecmascript>import "os" for Process import "/date" for Date
var args = Process.arguments if (args.count != 1) {
Fiber.abort("Please pass just the year to be processed.")
}
var year = Num.fromString(args[0]) System.print("The dates of the last Sundays in the month for %(year) are:") Date.default = Date.isoDate for (m in 1..12) {
var d = Date.monthLength(year, m)
var dt = Date.new(year, m, d)
var wd = dt.dayOfWeek
if (wd == 7) {
System.print(dt)
} else {
System.print(dt.addDays(-wd))
}
}</lang>
- Output:
$ wren last_sunday.wren 2013 The dates of the last Sundays in the month for 2013 are: 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29 $ wren last_sunday.wren 2020 The dates of the last Sundays in the month for 2020 are: 2020-01-26 2020-02-23 2020-03-29 2020-04-26 2020-05-31 2020-06-28 2020-07-26 2020-08-30 2020-09-27 2020-10-25 2020-11-29 2020-12-27
zkl
The program from [[4]] solves this task, as well.
- Output:
var [const] D=Time.Date; lastDay(2013,D.Sunday) 2013-01-27 2013-02-24 2013-03-31 2013-04-28 2013-05-26 2013-06-30 2013-07-28 2013-08-25 2013-09-29 2013-10-27 2013-11-24 2013-12-29
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