Leap year
You are encouraged to solve this task according to the task description, using any language you may know.
Determine whether a given year is a leap year in the Gregorian calendar.
See Also
360 Assembly
This is a callable subroutine to determine whether or not a given zoned-decimal 4-digit year is a Leap Year. Leap years are "evenly divisible" by 4, except those which end in '00' and are not evenly divisible by 400. The subroutine receives two parameters:
(1) a 4-digit year (CCYY) (2) an 8-byte work area
The value returned in Register 15 (by convention the "return code") indicates whether the year is a Leap Year:
When R15 = zero, the year is a leap year. Otherwise it is not.
<lang 360 Assembly> LPCK CSECT
USING LPCK,15
STM 0,12,20(13) STORE CALLER REGS
LM 1,2,0(1) R1 -> CCYY, R2 -> DOUBLE-WORD WORK AREA
PACK 0(8,2),0(4,1) PACK CCYY INTO WORK AREA
CVB 0,0(2) CONVERT TO BINARY (R0 = CCYY)
SRDL 0,32 R0|R1 = CCYY
LA 2,100 R2 = 100
DR 0,2 DIVIDE BY 100: R0 = YY, R1 = CC
LTR 0,0 YY = 0?
BZ A YES: R0|R1 = CC
SRDL 0,32 NO: R0|R1 = YY
A LA 2,4 R2 = 4
DR 0,2 DIVIDE BY 4: R0 = REMAINDER, R1 = QUOTIENT
LR 15,0 LOAD REMAINDER: IF 0, THEN LEAP YEAR
LM 0,12,20(13) RESTORE REGS
BR 14
END
</lang>
Sample invocation from a COBOL program:
WORKING-STORAGE SECTION. 01 FILLER.
05 YEAR-VALUE PIC 9(4). 05 WKAREA PIC X(8).
PROCEDURE DIVISION.
MOVE 1936 TO YEAR-VALUE CALL 'LPCK' USING YEAR-VALUE, WKAREA PERFORM RESULT-DISPLAY MOVE 1900 TO YEAR-VALUE CALL 'LPCK' USING YEAR-VALUE, WKAREA PERFORM RESULT-DISPLAY GOBACK.
RESULT-DISPLAY.
IF RETURN-CODE = ZERO DISPLAY YEAR-VALUE ' IS A LEAP YEAR' ELSE DISPLAY YEAR-VALUE ' IS NOT A LEAP YEAR'.
ActionScript
<lang actionscript>public function isLeapYear(year:int):Boolean {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
Ada
<lang Ada>-- Incomplete code, just a sniplet to do the task. Can be used in any package or method. -- Adjust the type of Year if you use a different one. function Is_Leap_Year (Year : Integer) return Boolean is begin
if Year rem 100 = 0 then
return Year rem 400 = 0;
else
return Year rem 4 = 0;
end if;
end Is_Leap_Year;
-- An enhanced, more efficient version:
-- This version only does the 2 bit comparison (rem 4) if false.
-- It then checks rem 16 (a 4 bit comparison), and only if those are not
-- conclusive, calls rem 100, which is the most expensive operation.
-- I failed to be convinced of the accuracy of the algorithm at first,
-- so I rephrased it below.
-- FYI: 400 is evenly divisible by 16 whereas 100,200 and 300 are not. Ergo, the
-- set of integers evenly divisible by 16 and 100 are all evenly divisible by 400.
-- 1. If a year is not divisible by 4 => not a leap year. Skip other checks.
-- 2. If a year is evenly divisible by 16, it is either evenly divisible by 400 or
-- not evenly divisible by 100 => leap year. Skip further checks.
-- 3. If a year evenly divisible by 100 => not a leap year.
-- 4. Otherwise a leap year.
function Is_Leap_Year (Year : Integer) return Boolean is begin
return (Year rem 4 = 0) and then ((Year rem 16 = 0) or else (Year rem 100 /= 0));
end Is_Leap_Year;
-- To improve speed a bit more, use with pragma Inline (Is_Leap_Year);</lang>
ALGOL 68
<lang algol68>MODE YEAR = INT, MONTH = INT, DAY = INT;
PROC year days = (YEAR year)DAY: # Ignore 1752 CE for the moment #
( month days(year, 2) = 28 | 365 | 366 );
PROC month days = (YEAR year, MONTH month) DAY:
( month | 31,
28 + ABS (year MOD 4 = 0 AND year MOD 100 /= 0 OR year MOD 400 = 0),
31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
PROC is leap year = (YEAR year)BOOL: year days(year)=366;
test:(
[]INT test cases = (1900, 1994, 1996, 1997, 2000);
FOR i TO UPB test cases DO
YEAR year = test cases[i];
printf(($g(0)" is "b("","not ")"a leap year."l$, year, is leap year(year)))
OD
)</lang>
- Output:
1900 is not a leap year. 1994 is not a leap year. 1996 is a leap year. 1997 is not a leap year. 2000 is a leap year.
ALGOL W
<lang algolw>begin
% 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 );
% some test cases %
for year := 1899, 1900, 1901, 1902, 1903, 1904, 1905, 1999, 2000, 2001, 2002, 2003, 2004 do begin
write( i_w := 1, s_w := 0
, year
, " is "
, if isLeapYear( year ) then "" else "not "
, " a leap year"
)
end for_year
end.</lang>
APL
<lang apl> ⍝returns 1 if leapyear, 0 otherwise: ∇Z←LEAPYEAR YEAR Z←(0=4|YEAR)∧(0=400|YEAR)∨~0=100|YEAR ∇ </lang>
AppleScript
<lang applescript>on leap_year(y)
return year mod 4 is equal to 0 and (year mod 100 is not equal to 0 or year mod 400 is equal to 0)
end leap_year
leap_year(1900)</lang>
AutoHotkey
<lang autohotkey>leapyear(year) {
if (Mod(year, 100) = 0)
return (Mod(year, 400) = 0)
return (Mod(year, 4) = 0)
}
MsgBox, % leapyear(1604)</lang>
- Output:
Returns 1 if year is a leap year
or <lang autohotkey>IsLeapYear(Year) {
return !Mod(Year, 4) && Mod(Year, 100) || !Mod(Year, 400)
}
MsgBox % "The year 1604 was " (IsLeapYear(1604) ? "" : "not ") "a leap year"</lang>
- Output:
The year 1600 was a leap year The year 1601 was not a leap year The year 1604 was a leap year
AutoIt
<lang AutoIt>; AutoIt Version: 3.3.8.1 $Year = 2012 $sNot = " not"
If IsLeapYear($Year) Then $sNot = "" ConsoleWrite ($Year & " is" & $sNot & " a leap year." & @LF)
Func IsLeapYear($_year) Return Not Mod($_year, 4) And (Mod($_year, 100) Or Not Mod($_year, 400)) EndFunc
- == But it exists the standard UDF "Date.au3" with this function
- "_IsLeapYear($Year)"
</lang>
- Output:
2012 is a leap year.
--BugFix (talk) 16:18, 16 November 2013 (UTC)
AWK
<lang AWK>function leapyear( year ) {
if ( year % 100 == 0 )
return ( year % 400 == 0 )
else
return ( year % 4 == 0 )
}</lang>
Bash
<lang Bash>#!/bin/bash year=$(date +"%Y") if (( $year % 4 == 0 )) then
if (( $year % 400 == 0 ))
then
echo "This is a leap year"
else
if (( $year % 100 == 0 ))
then
echo "This is not a leap year"
else
echo "This is a leap year"
fi
fi
else
echo "This is not a leap year"
fi </lang>
BASIC
Note that the year% function is not needed for most modern BASICs.
<lang qbasic>DECLARE FUNCTION diy% (y AS INTEGER)
DECLARE FUNCTION isLeapYear% (yr AS INTEGER)
DECLARE FUNCTION year% (date AS STRING)
PRINT isLeapYear(year(DATE$))
FUNCTION diy% (y AS INTEGER)
IF y MOD 4 THEN
diy = 365
ELSEIF y MOD 100 THEN
diy = 366
ELSEIF y MOD 400 THEN
diy = 365
ELSE
diy = 366
END IF
END FUNCTION
FUNCTION isLeapYear% (yr AS INTEGER)
isLeapYear = (366 = diy(yr))
END FUNCTION
FUNCTION year% (date AS STRING)
year% = VAL(RIGHT$(date, 4))
END FUNCTION</lang>
An old-timey solution:
<lang BASIC>10 DEF FNLY(Y)=(Y/4=INT(Y/4))*((Y/100<>INT(Y/100))+(Y/400=INT(Y/400)))</lang>
BaCon
From the Ada shortcut calculation <lang qbasic>' Leap year FUNCTION leapyear(NUMBER y) TYPE NUMBER
RETURN IIF(MOD(y, 4) = 0, IIF(MOD(y, 16) = 0, IIF(MOD(y, 100) != 0, TRUE, FALSE), TRUE), FALSE)
END FUNCTION
READ y WHILE y != 0
PRINT y, ": ", IIF$(leapyear(y), "", "not a "), "leapyear" READ y
WEND
DATA 1600, 1700, 1800, 1900, 1901, 1996, 2000, 2001, 2004, 0</lang>
- Output:
1600: not a leapyear 1700: leapyear 1800: leapyear 1900: leapyear 1901: not a leapyear 1996: leapyear 2000: not a leapyear 2001: not a leapyear 2004: leapyear
Batch File
<lang dos>@echo off
- The Main Thing...
for %%x in (1900 2046 2012 1600 1800 2031 1952) do ( call :leap %%x ) echo. pause exit/b
- /The Main Thing...
- The Function...
- leap
set year=%1 set /a op1=%year%%%4 set /a op2=%year%%%100 set /a op3=%year%%%400 if not "%op1%"=="0" (goto :no) if not "%op2%"=="0" (goto :yes) if not "%op3%"=="0" (goto :no)
- yes
echo. echo %year% is a leap year. goto :EOF
- no
echo. echo %year% is NOT a leap year. goto :EOF
- /The Function...</lang>
- Output:
1900 is NOT a leap year. 2046 is NOT a leap year. 2012 is a leap year. 1600 is a leap year. 1800 is NOT a leap year. 2031 is NOT a leap year. 1952 is a leap year. Press any key to continue . . .
BBC BASIC
<lang bbcbasic> REPEAT
INPUT "Enter a year: " year%
IF FNleap(year%) THEN
PRINT ;year% " is a leap year"
ELSE
PRINT ;year% " is not a leap year"
ENDIF
UNTIL FALSE
END
DEF FNleap(yr%)
= (yr% MOD 4 = 0) AND ((yr% MOD 400 = 0) OR (yr% MOD 100 <> 0))</lang>
Befunge
<lang befunge>0"2("*:3-:1-:2-:"^"-v< v*%"d"\!%4::,,"is".:<| >\45*:*%!+#v_ "ton"vv< v"ear."+550<,,,,*84<$# >"y pael a ">:#,_$:#@^</lang>
- Output:
1900 is not a leap year. 1994 is not a leap year. 1996 is a leap year. 1997 is not a leap year. 2000 is a leap year.
Bracmat
<lang bracmat> ( leap-year
=
. mod$(!arg.100):0
& `(mod$(!arg.400):0) { The backtick skips the remainder of the OR operation,
even if the tested condition fails. }
| mod$(!arg.4):0
)
& 1600 1700 1899 1900 2000 2006 2012:?tests & whl
' ( !tests:%?test ?tests
& ( leap-year$!test&out$(!test " is a leap year")
| out$(!test " is not a leap year")
)
)
& ;</lang>
- Output:
1600 is a leap year 1700 is not a leap year 1899 is not a leap year 1900 is not a leap year 2000 is a leap year 2006 is not a leap year 2012 is a leap year
C
<lang c>#include <stdio.h>
int is_leap_year(int year) {
return (!(year % 4) && year % 100 || !(year % 400)) ? 1 : 0;
}
int main() {
int test_case[] = {1900, 1994, 1996, 1997, 2000}, key, end, year;
for (key = 0, end = sizeof(test_case)/sizeof(test_case[0]); key < end; ++key) {
year = test_case[key];
printf("%d is %sa leap year.\n", year, (is_leap_year(year) == 1 ? "" : "not "));
}
}</lang>
- Output:
1900 is not a leap year. 1994 is not a leap year. 1996 is a leap year. 1997 is not a leap year. 2000 is a leap year.
C++
Uses C++11. Compile with
g++ -std=c++11 leap_year.cpp
<lang cpp>#include <iostream>
bool is_leap_year(int year) {
return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
}
int main() {
for (auto year : {1900, 1994, 1996, 1997, 2000}) {
std::cout << year << (is_leap_year(year) ? " is" : " is not") << " a leap year.\n";
}
}</lang>
- Output:
1900 is not a leap year. 1994 is not a leap year. 1996 is a leap year. 1997 is not a leap year. 2000 is a leap year.
C#
<lang csharp>using System;
class Program {
static void Main()
{
foreach (var year in new[] { 1900, 1994, 1996, DateTime.Now.Year })
{
Console.WriteLine("{0} is {1}a leap year.",
year,
DateTime.IsLeapYear(year) ? string.Empty : "not ");
}
}
}</lang>
- Output:
1900 is not a leap year. 1994 is not a leap year. 1996 is a leap year. 2010 is not a leap year.
Clipper
<lang Clipper>Function IsLeapYear( nYear ) Return Iif( nYear%100 == 0, (nYear%400 == 0), (nYear%4 == 0) )</lang>
Clojure
<lang clojure>(defn leap-year? [y]
(and (zero? (mod y 4)) (or (pos? (mod y 100)) (zero? (mod y 400)))))</lang>
COBOL
<lang cobol> IDENTIFICATION DIVISION.
PROGRAM-ID. leap-year.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 examples VALUE "19001994199619972000".
03 year PIC 9(4) OCCURS 5 TIMES
INDEXED BY year-index.
01 remainders.
03 400-rem PIC 9(4).
03 100-rem PIC 9(4).
03 4-rem PIC 9(4).
PROCEDURE DIVISION.
PERFORM VARYING year-index FROM 1 BY 1 UNTIL 5 < year-index
MOVE FUNCTION MOD(year (year-index), 400) TO 400-rem
MOVE FUNCTION MOD(year (year-index), 100) TO 100-rem
MOVE FUNCTION MOD(year (year-index), 4) TO 4-rem
IF 400-rem = 0 OR ((100-rem NOT = 0) AND 4-rem = 0)
DISPLAY year (year-index) " is a leap year."
ELSE
DISPLAY year (year-index) " is not a leap year."
END-IF
END-PERFORM
GOBACK
.</lang>
Using Date Intrinsic Functions <lang COBOL>
program-id. leap-yr.
*> Given a year, where 1601 <= year <= 9999
*> Determine if the year is a leap year
data division.
working-storage section.
1 input-year pic 9999.
1 binary.
2 int-date pic 9(8).
2 cal-mo-day pic 9(4).
procedure division.
display "Enter calendar year (1601 thru 9999): "
with no advancing
accept input-year
if input-year >= 1601 and <= 9999
then
*> if the 60th day of a year is Feb 29
*> then the year is a leap year
compute int-date = function integer-of-day
( input-year * 1000 + 60 )
compute cal-mo-day = function mod (
(function date-of-integer ( int-date )) 10000 )
display "Year " input-year space with no advancing
if cal-mo-day = 229
display "is a leap year"
else
display "is NOT a leap year"
end-if
else
display "Input date is not within range"
end-if
stop run
.
end program leap-yr.
</lang>
- Output:
Enter calendar year (1601 thru 9999): 2016 Year 2016 is a leap year Enter calendar year (1601 thru 9999): 2017 Year 2017 is NOT a leap year Enter calendar year (1601 thru 9999): 2100 Year 2100 is NOT a leap year Enter calendar year (1601 thru 9999): 2400 Year 2400 is a leap year Enter calendar year (1601 thru 9999): 3000 Year 3000 is NOT a leap year Enter calendar year (1601 thru 9999): 4000 Year 4000 is a leap year
Common Lisp
<lang lisp>(defun leap-year-p (year)
(destructuring-bind (fh h f)
(mapcar #'(lambda (n) (zerop (mod year n))) '(400 100 4))
(or fh (and (not h) f))))</lang>
Component Pascal
BlackBox Component Builder <lang oberon2> MODULE LeapYear; IMPORT StdLog, Strings, Args;
PROCEDURE IsLeapYear(year: INTEGER): BOOLEAN; BEGIN IF year MOD 4 # 0 THEN
RETURN FALSE
ELSE IF year MOD 100 = 0 THEN IF year MOD 400 = 0 THEN RETURN TRUE ELSE RETURN FALSE END ELSE RETURN TRUE END END END IsLeapYear;
PROCEDURE Do*; VAR p: Args.Params; year,done,i: INTEGER; BEGIN Args.Get(p); FOR i := 0 TO p.argc - 1 DO Strings.StringToInt(p.args[i],year,done); StdLog.Int(year);StdLog.String(":>");StdLog.Bool(IsLeapYear(year));StdLog.Ln END;
END Do;
END LeapYear.
</lang>
Execute: ^Q LeapYear.Do 2000 2004 2013~
- Output:
2000:> $TRUE 2004:> $TRUE 2013:> $FALSE
D
<lang d>import std.algorithm;
bool leapYear(in uint y) pure nothrow {
return (y % 4) == 0 && (y % 100 || (y % 400) == 0);
}
void main() {
auto good = [1600, 1660, 1724, 1788, 1848, 1912, 1972, 2032,
2092, 2156, 2220, 2280, 2344, 2348];
auto bad = [1698, 1699, 1700, 1750, 1800, 1810, 1900, 1901,
1973, 2100, 2107, 2200, 2203, 2289];
assert(filter!leapYear(bad ~ good).equal(good));
}</lang>
Using the datetime library:
<lang d>import std.datetime;
void main() {
assert(yearIsLeapYear(1724)); assert(!yearIsLeapYear(1973)); assert(!Date(1900, 1, 1).isLeapYear); assert(DateTime(2000, 1, 1).isLeapYear);
} </lang>
Delphi/Pascal
Delphi has standard function IsLeapYear in SysUtils unit. <lang Delphi>program TestLeapYear;
{$APPTYPE CONSOLE}
uses
SysUtils;
var
Year: Integer;
begin
Write('Enter the year: ');
Readln(Year);
if IsLeapYear(Year) then
Writeln(Year, ' is a Leap year')
else
Writeln(Year, ' is not a Leap year');
Readln;
end.</lang>
DWScript
<lang Delphi>function IsLeapYear(y : Integer) : Boolean; begin
Result:= (y mod 4 = 0)
and ( ((y mod 100) <> 0)
or ((y mod 400) = 0) );
end;
const good : array [0..13] of Integer =
[1600,1660,1724,1788,1848,1912,1972,2032,2092,2156,2220,2280,2344,2348];
const bad : array [0..13] of Integer =
[1698,1699,1700,1750,1800,1810,1900,1901,1973,2100,2107,2200,2203,2289];
var i : Integer;
PrintLn('Checking leap years'); for i in good do
if not IsLeapYear(i) then PrintLn(i);
PrintLn('Checking non-leap years'); for i in bad do
if IsLeapYear(i) then PrintLn(i);</lang>
Ela
<lang ela>isLeap y | y % 100 == 0 = y % 400 == 0
| else = y % 4 == 0</lang>
Elixir
<lang elixir>leap_year? = fn(year) -> :calendar.is_leap_year(year) end IO.inspect for y <- 2000..2020, leap_year?.(y), do: y</lang>
- Output:
[2000, 2004, 2008, 2012, 2016, 2020]
Erlang
<lang erlang> -module(gregorian). -export([leap/1]).
leap( Year ) -> calendar:is_leap_year( Year ). </lang>
ERRE
<lang ERRE>PROGRAM LEAP_YEAR
FUNCTION LEAP(YR%)
LEAP=(YR% MOD 4=0) AND ((YR% MOD 400=0) OR (YR% MOD 100<>0))
END FUNCTION
BEGIN
LOOP
INPUT("Enter a year: ",year%)
EXIT IF YEAR%=0
IF LEAP(year%) THEN
PRINT(year%;" is a leap year")
ELSE
PRINT(year%;" is not a leap year")
END IF
END LOOP
END PROGRAM</lang>
Euphoria
<lang euphoria>function isLeapYear(integer year)
return remainder(year,4)=0 and remainder(year,100)!=0 or remainder(year,400)=0
end function</lang>
Excel
Take two cells, say A1 and B1, in B1 type in :
<lang Excel> =IF(OR(NOT(MOD(A1,400)),AND(NOT(MOD(A1,4)),MOD(A1,100))),"Leap Year","Not a Leap Year") </lang>
- Output:
1900 Not a Leap Year 1954 Not a Leap Year 1996 Leap Year 2003 Not a Leap Year 2012 Leap Year
F#
<lang fsharp>let isLeapYear = System.DateTime.IsLeapYear assert isLeapYear 1996 assert isLeapYear 2000 assert not (isLeapYear 2001) assert not (isLeapYear 1900)</lang>
Factor
Call leap-year? word from calendars vocabulary. For example: <lang factor>USING: calendar prettyprint ; 2011 leap-year? .</lang> Factor uses proleptic Gregorian calendar.
Forth
<lang forth>: leap-year? ( y -- ? )
dup 400 mod 0= if drop true exit then
dup 100 mod 0= if drop false exit then
4 mod 0= ;</lang>
Or more simply (but always computing three "mod"): <lang forth>: leap-year? dup 4 mod 0= over 16 mod 0= rot 25 mod 0= not or and ;</lang>
Fortran
<lang fortran>program leap
implicit none
write(*,*) leap_year([1900, 1996, 1997, 2000])
contains
pure elemental function leap_year(y) result(is_leap) implicit none logical :: is_leap integer,intent(in) :: y
is_leap = (mod(y,4)==0 .and. .not. mod(y,100)==0) .or. (mod(y,400)==0)
end function leap_year
end program leap</lang>
- Output:
F T F T
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
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
' ------=< MAIN >=------
' year is a FreeBASIC keyword Dim As Integer Year_
For Year_ = 1800 To 2900 Step 100
Print Year_; IIf(leapyear(Year_), " is a leap year", " is not a leap year")
Next
Print : Print
For Year_ = 2012 To 2031
Print Year_;
If leapyear(Year_) = TRUE Then
Print " = leap",
Else
Print " = no",
End If
If year_ Mod 4 = 3 Then Print ' lf/cr
Next
' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
1800 is not a leap year 1900 is not a leap year 2000 is a leap year 2100 is not a leap year 2200 is not a leap year 2300 is not a leap year 2400 is a leap year 2500 is not a leap year 2600 is not a leap year 2700 is not a leap year 2800 is a leap year 2900 is not a leap year 2012 = leap 2013 = no 2014 = no 2015 = no 2016 = leap 2017 = no 2018 = no 2019 = no 2020 = leap 2021 = no 2022 = no 2023 = no 2024 = leap 2025 = no 2026 = no 2027 = no 2028 = leap 2029 = no 2030 = no 2031 = no
FutureBasic
<lang futurebasic> include "ConsoleWindow"
// In-line C function to generate random number in range BeginCFunction long randomInRange( long min, long max ) { int i = (arc4random()%(max-min+1))+min; return (long)i; } EndC toolbox fn randomInRange( long min, long max ) = long
// Leap year test function local fn LeapYear( year as long ) as Boolean dim as Boolean result : result = _false
if year mod 400 == 0 then result = _true : exit fn if year mod 100 == 0 then result = _false : exit fn if year mod 4 == 0 then result = _true : exit fn if year mod 4 != 0 then result = _false : exit fn end fn = result
dim as long i, y, knownLeapYear(10)
// Array of known leap years from 1980 through 2020 for control knownLeapYear(0) = 1980 : knownLeapYear(1) = 1984 : knownLeapYear(2) = 1988 knownLeapYear(3) = 1992 : knownLeapYear(4) = 1996 : knownLeapYear(5) = 2000 knownLeapYear(6) = 2004 : knownLeapYear(7) = 2008 : knownLeapYear(8) = 2012 knownLeapYear(9) = 2016 : knownLeapYear(10) = 2020
print "Known leap years:" for i = 0 to 9
if ( fn LeapYear( knownLeapYear(i) ) == _true ) print knownLeapYear(i); " is a leap year."
else
print knownLeapYear(i); " is a not leap year." end if
next
// Random years from 1980 to 2020 to test print "Check random years:" for i = 0 to 20
y = fn randomInRange( 1980, 2020 )
if ( fn LeapYear( y ) == _true )
print y; " is a leap year."
else
print y; " is a not leap year."
end if
next </lang>
Output (results will vary for random years):
Known leap years: 1980 is a leap year. 1984 is a leap year. 1988 is a leap year. 1992 is a leap year. 1996 is a leap year. 2000 is a leap year. 2004 is a leap year. 2008 is a leap year. 2012 is a leap year. 2016 is a leap year. Check random years: 1998 is a not leap year. 1987 is a not leap year. 2015 is a not leap year. 1998 is a not leap year. 2020 is a leap year. 2020 is a leap year. 2009 is a not leap year. 2020 is a leap year. 2018 is a not leap year. 2013 is a not leap year. 2003 is a not leap year. 1994 is a not leap year. 1989 is a not leap year. 1999 is a not leap year. 1984 is a leap year. 1980 is a leap year. 1998 is a not leap year. 2008 is a leap year. 1983 is a not leap year. 2007 is a not leap year. 2004 is a leap year.
Gambas
<lang gambas>Public Sub Form_Open() Dim dDate As Date Dim siYear As Short = InputBox("Enter a year", "Leap year test") Dim sMessage As String = " is a leap year."
Try dDate = Date(siYear, 02, 29) If Error Then sMessage = " is not a leap year."
Message(siYear & sMessage)
End</lang>
Output:
2016 is a leap year.
GAP
<lang gap>IsLeapYear := function(n)
return (n mod 4 = 0) and ((n mod 100 <> 0) or (n mod 400 = 0));
end;
- alternative using built-in function
IsLeapYear := function(n)
return DaysInYear(n) = 366;
end;</lang>
Go
<lang go>func isLeap(year int) bool {
return year%400 == 0 || year%4 == 0 && year%100 != 0
}</lang>
Groovy
Solution: <lang groovy>(1900..2012).findAll {new GregorianCalendar().isLeapYear(it)}.each {println it}</lang>
- Output:
1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012
Harbour
<lang visualfoxpro>FUNCTION IsLeapYear( nYear )
RETURN iif( nYear % 100 == 0, nYear % 400 == 0, nYear % 4 == 0 )</lang>
Haskell
Simple version <lang haskell>import Data.List import Control.Monad import Control.Arrow
leaptext x b | b = show x ++ " is a leap year" | otherwise = show x ++ " is not a leap year"
isleapsf j | 0==j`mod`100 = 0 == j`mod`400 | otherwise = 0 == j`mod`4</lang> Algorithmic <lang haskell>isleap = foldl1 ((&&).not).flip map [400, 100, 4]. ((0==).).mod</lang> Example using isleap <lang haskell>*Main> mapM_ (putStrLn. (ap leaptext isleap)) [1900,1994,1996,1997,2000] 1900 is not a leap year 1994 is not a leap year 1996 is a leap year 1997 is not a leap year 2000 is a leap year</lang>
TDD version <lang haskell>import Test.HUnit
isLeapYear::Int->Bool isLeapYear y
| mod y 400 == 0 = True | mod y 100 == 0 = False | mod y 4 == 0 = True | otherwise = False
tests = TestList[TestCase $ assertEqual "4 is a leap year" True $ isLeapYear 4
,TestCase $ assertEqual "1 is not a leap year" False $ isLeapYear 1
,TestCase $ assertEqual "64 is a leap year" True $ isLeapYear 64
,TestCase $ assertEqual "2000 is a leap year" True $ isLeapYear 2000
,TestCase $ assertEqual "1900 is not a leap year" False $ isLeapYear 1900]</lang>
Hy
<lang clojure>(defn leap? [y]
(and
(= (% y 4) 0)
(or
(!= (% y 100) 0)
(= (% y 400) 0))))</lang>
Icon and Unicon
Gives leap year status for 2000,1900,2012 and any arguments you give <lang Icon>procedure main(arglist) every y := !([2000,1900,2012]|||arglist) do
write("The year ",y," is ", leapyear(y) | "not ","a leap year.")
end
procedure leapyear(year) #: determine if year is leap
if (numeric(year) % 4 = 0 & year % 100 ~= 0) | (numeric(year) % 400 = 0) then return
end</lang>
J
<lang j>isLeap=: 0 -/@:= 4 100 400 |/ ]</lang> Example use: <lang j> isLeap 1900 1996 1997 2000 0 1 0 1</lang>
Java
By default, java.util.GregorianCalendar switches from Julian calendar to Gregorian calendar at 15 October 1582. The code below uses both the GregorianCalendar class and the algorithm from the wiki. Both values are printed in the output.
<lang java>import java.util.GregorianCalendar; import java.text.MessageFormat;
public class Leapyear{
public static void main(String[] argv){
int[] yrs = {1800,1900,1994,1998,1999,2000,2001,2004,2100};
GregorianCalendar cal = new GregorianCalendar();
for(int year : yrs){
System.err.println(MessageFormat.format("The year {0,number,#} is leaper: {1} / {2}.",
year, cal.isLeapYear(year), isLeapYear(year)));
}
}
public static boolean isLeapYear(int year){
return (year % 100 == 0) ? (year % 400 == 0) : (year % 4 == 0);
}
}
</lang>
- Output:
The year 1800 is leaper: false / false. The year 1900 is leaper: false / false. The year 1994 is leaper: false / false. The year 1998 is leaper: false / false. The year 1999 is leaper: false / false. The year 2000 is leaper: true / true. The year 2001 is leaper: false / false. The year 2004 is leaper: true / true. The year 2100 is leaper: false / false.
JavaScript
<lang javascript>var isLeapYear = function (year) { return (year % 100 === 0) ? (year % 400 === 0) : (year % 4 === 0); };</lang> Or, by setting the day to the 29th and checking if the day remains <lang javascript>// Month values start at 0, so 1 is for February var isLeapYear = function (year) { return new Date(year, 1, 29).getDate() === 29; };</lang>
jq
<lang jq>def leap:
. as $y | ($y%4) == 0 and ($y < 1582 or ($y%400) == 0 or ($y%100) != 0);</lang>
Examples: <lang jq>def assert(value; f):
value as $value
| ($value|f) | if . then empty else error("assertion violation: \($value) => \(.)") end;
((2400, 2012, 2000, 1600, 1500, 1400) | assert(.; leap)),
((2100, 2014, 1900, 1800, 1700, 1499) | assert(.; leap|not)) </lang>
- Output:
$ jq -n -f Leap_year.jq
Julia
<lang julia>leap(y) = y%4==0 && (y<1582 || y%400==0 || y%100!=0)
- some tests
@assert all([leap(yr) for yr in [2400, 2012, 2000, 1600, 1500, 1400]]) @assert all([~leap(yr) for yr in [2100, 2014, 1900, 1800, 1700, 1499]])</lang>
K
<lang K> leapyear:{(+/~x!'4 100 400)!2}
a@&leapyear'a:1900,1994,1996,1997,2000
1996 2000</lang>
Kotlin
<lang kotlin>fun isLeapYear(year: Int) = year % 400 == 0 || (year % 100 != 0 && year % 4 == 0)</lang>
Lasso
<lang Lasso>define isLeapYear(y::integer) => { #y % 400 == 0 ? return true #y % 100 == 0 ? return false #y % 4 == 0 ? return true return false }
with test in array(2012,2016,1933,1900,1999,2000) do => {^ isLeapYear(#test) '\r' ^}</lang>
- Output:
true true false false false true
Liberty BASIC
Simple method
<lang lb>if leap(1996)then
print "leap"
else
print "ordinary"
end if wait
function leap(n)
leap=date$("2/29/";n)
end function</lang>
Calculated method
<lang lb> year = 1908
select case
case year mod 400 = 0
leapYear = 1
case year mod 4 = 0 and year mod 100 <> 0
leapYear = 1
case else
leapYear = 0
end select
if leapYear = 1 then
print year;" is a leap year."
else
print year;" is not a leap year."
end if</lang>
Lingo
<lang lingo>on isLeapYear (year)
return date(year, 2, 29).month=2
end</lang>
LiveCode
<lang LiveCode>function isLeapYear year
return (year MOD 4 is 0) AND ((year MOD 400 is 0) OR (year MOD 100 is not 0))
end isLeapYear
command testLeapYear
set itemDelimiter to comma
put "1900,1994,1996,1997,2000" into years
repeat for each item y in years
put y && "is" && isLeapYear(y) && return after tyears
end repeat
put tyears
end testLeapYear
1900 is false 1994 is false 1996 is true 1997 is false 2000 is true </lang>
Logo
<lang logo>to multiple? :n :d
output equal? 0 modulo :n :d
end to leapyear? :y
output ifelse multiple? :y 100 [multiple? :y 400] [multiple? :y 4]
end</lang>
Logtalk
<lang logtalk>leap_year(Year) :-
( mod(Year, 4) =:= 0, mod(Year, 100) =\= 0 ->
true
; mod(Year, 400) =:= 0
).</lang>
LOLCODE
<lang lolcode>BTW Determine if a Gregorian calendar year is leap HAI 1.3 HOW IZ I Leap YR Year
BOTH SAEM 0 AN MOD OF Year AN 4
O RLY?
YA RLY
BOTH SAEM 0 AN MOD OF Year AN 100
O RLY?
YA RLY
BOTH SAEM 0 AN MOD OF Year AN 400
O RLY?
YA RLY
FOUND YR WIN
NO WAI
FOUND YR FAIL
OIC
NO WAI
FOUND YR WIN
OIC
NO WAI
FOUND YR FAIL
OIC
IF U SAY SO
I HAS A Yearz ITZ A BUKKIT Yearz HAS A SRS 0 ITZ 1900 Yearz HAS A SRS 1 ITZ 1904 Yearz HAS A SRS 2 ITZ 1994 Yearz HAS A SRS 3 ITZ 1996 Yearz HAS A SRS 4 ITZ 1997 Yearz HAS A SRS 5 ITZ 2000
IM IN YR Loop UPPIN YR Index WILE DIFFRINT Index AN 6
I HAS A Yr ITZ Yearz'Z SRS Index I HAS A Not I IZ Leap YR Yr MKAY O RLY? YA RLY Not R "" NO WAI Not R " NOT" OIC VISIBLE Yr " is" Not " a leap year"
IM OUTTA YR Loop
KTHXBYE </lang>
- Output:
1900 is NOT a leap year 1904 is a leap year 1994 is NOT a leap year 1996 is a leap year 1997 is NOT a leap year 2000 is a leap year
Lua
<lang Lua>function isLeapYear(year)
return year%4==0 and (year%100~=0 or year%400==0)
end</lang>
Maple
<lang maple>isLeapYear := proc(year) if not year mod 4 = 0 or (year mod 100 = 0 and not year mod 400 = 0) then return false; else return true; end if; end proc:</lang>
Mathematica
Dates are handled by built-in functions in the Wolfram Language <lang Mathematica>LeapYearQ[2002]</lang>
MATLAB / Octave
MATLAB, conveniently, provides a function that returns the last day of an arbitrary month of the calendar given the year. Using the fact that February is 29 days long during a leap year, we can write a one-liner that solves this task. <lang MATLAB>function TrueFalse = isLeapYear(year)
TrueFalse = (eomday(year,2) == 29);
end</lang>
Maxima
<lang maxima>leapyearp(year) := is(mod(year, 4) = 0 and
(mod(year, 100) # 0 or mod(year, 400) = 0))$</lang>
МК-61/52
<lang>П0 1 0 0 / {x} x=0 14 ИП0 4 0 0 ПП 18 ИП0 4 ПП 18 / {x} x=0 24 1 С/П 0 С/П</lang>
Mercury
<lang mercury>:- pred is_leap_year(int::in) is semidet.
is_leap_year(Year) :-
( if Year mod 100 = 0 then Year mod 400 = 0 else Year mod 4 = 0 ).</lang>
Usage:
<lang mercury>:- module leap_year.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module int, list, string.
main(!IO) :-
Years = [1600, 1700, 1899, 1900, 2000, 2006, 2012], io.write_list(Years, "", write_year_kind, !IO).
- - pred write_year_kind(int::in, io::di, io::uo) is det.
write_year_kind(Year, !IO) :-
io.format("%d %s a leap year.\n",
[i(Year), s(if is_leap_year(Year) then "is" else "is not" )], !IO).</lang>
MIPS Assembly
Pass year in a0, returns boolean in v0. <lang mips> IsLeap: andi $a1, $a0, 3 #a0 is year to test bnez $a1 NotLeap li $a1, 100 div $a0, $a1 mfhi $a1 bnez $a1, Leap mflo $a1 andi $a1, $a1, 3 bnez $a1, NotLeap Leap: li $v0, 1 jr $ra NotLeap:li $v0, 0 jr $ra </lang>
MUMPS
<lang MUMPS>ILY(X) ;IS IT A LEAP YEAR?
QUIT ((X#4=0)&(X#100'=0))!((X#100=0)&(X#400=0))</lang>
Usage:
USER>W $SELECT($$ILY^ROSETTA(1900):"Yes",1:"No") No USER>W $SELECT($$ILY^ROSETTA(2000):"Yes",1:"No") Yes USER>W $SELECT($$ILY^ROSETTA(1999):"Yes",1:"No") No
Nemerle
Demonstrating implementation as well as use of standard library function. <lang Nemerle>using System; using System.Console; using Nemerle.Assertions; using Nemerle.Imperative;
module LeapYear {
IsLeapYear(year : int) : bool
requires year >= 1582 otherwise throw ArgumentOutOfRangeException("year must be in Gregorian calendar.")
// without the contract enforcement would work for proleptic Gregorian Calendar
// in that case we might still want to require year > 0
{
when (year % 400 == 0) return true;
when (year % 100 == 0) return false;
when (year % 4 == 0) return true;
false
}
Main() : void
{
WriteLine("2000 is a leap year: {0}", IsLeapYear(2000));
WriteLine("2100 is a leap year: {0}", IsLeapYear(2100));
try {
WriteLine("1500 is a leap year: {0}", IsLeapYear(1500));
}
catch {
|e is ArgumentOutOfRangeException => WriteLine(e.Message)
}
WriteLine("1500 is a leap year: {0}", DateTime.IsLeapYear(1500)); // is false, indicating use of proleptic
// Gregorian calendar rather than reverting to
// Julian calendar
WriteLine("{0} is a leap year: {1}", DateTime.Now.Year,
DateTime.IsLeapYear(DateTime.Now.Year));
}
}</lang>
- Output:
2000 is a leap year: True 2100 is a leap year: False Specified argument was out of the range of valid values. Parameter name: year must be in Gregorian calendar. 1500 is a leap year: False 2013 is a leap year: False
NetRexx
Demonstrates both a Gregorian/proleptic Gregorian calendar leap-year algorithm and use of the Java library's GregorianCalendar object to determine which years are leap-years.
Note that the Java library indicates that the year 1500 is a leap-year as the Gregorian calendar wasn't established until 1582. The Java library implements the Julian calendar for dates prior to the Gregorian cut-over and leap-year rules in the Julian calendar are different to those for the Gregorian calendar. <lang NetRexx>/* NetRexx */
options replace format comments java crossref savelog symbols nobinary
years = '1500 1580 1581 1582 1583 1584 1600 1700 1800 1900 1994 1996 1997 2000 2004 2008 2009 2010 2011 2012 2100 2200 2300 2400 2500 2600' years['l-a'] = years['n-a'] = years['l-j'] = years['n-j'] =
loop y_ = 1 to years.words
year = years.word(y_)
if isLeapyear(year) then years['l-a'] = years['l-a'] year
else years['n-a'] = years['n-a'] year
if GregorianCalendar().isLeapYear(year) then years['l-j'] = years['l-j'] year
else years['n-j'] = years['n-j'] year
end y_
years['l-a'] = years['l-a'].strip years['n-a'] = years['n-a'].strip years['l-j'] = years['l-j'].strip years['n-j'] = years['n-j'].strip
say ' Sample years:' years['all'].changestr(' ', ',') say ' Leap years (algorithmically):' years['l-a'].changestr(' ', ',') say ' Leap years (Java library) :' years['l-j'].changestr(' ', ',') say ' Non-leap years (algorithmically):' years['n-a'].changestr(' ', ',') say ' Non-leap years (Java library) :' years['n-j'].changestr(' ', ',')
return
-- algorithmically method isLeapyear(year = int) public constant binary returns boolean
select when year // 400 = 0 then ly = isTrue when year // 100 \= 0 & year // 4 = 0 then ly = isTrue otherwise ly = isFalse end return ly
method isTrue public constant binary returns boolean
return 1 == 1
method isFalse public constant binary returns boolean
return \isTrue</lang>
- Output:
Sample years: 1500,1580,1581,1582,1583,1584,1600,1700,1800,1900,1994,1996,1997,2000,2004,2008,2009,2010,2011,2012,2100,2200,2300,2400,2500,2600
Leap years (algorithmically): 1580,1584,1600,1996,2000,2004,2008,2012,2400
Leap years (Java library) : 1500,1580,1584,1600,1996,2000,2004,2008,2012,2400
Non-leap years (algorithmically): 1500,1581,1582,1583,1700,1800,1900,1994,1997,2009,2010,2011,2100,2200,2300,2500,2600
Non-leap years (Java library) : 1581,1582,1583,1700,1800,1900,1994,1997,2009,2010,2011,2100,2200,2300,2500,2600
Nim
<lang nim>import times let year = 1980 echo isLeapYear(year)
- or
proc isLeapYear2(year): bool =
if year mod 100 == 0: year mod 400 == 0 else: year mod 4 == 0
echo isLeapYear2(year)</lang>
- Output:
true true
Oberon-2
<lang oberon2> PROCEDURE IsLeapYear(year: INTEGER): BOOLEAN; BEGIN
IF year MOD 4 # 0 THEN
RETURN FALSE
ELSE
IF year MOD 100 = 0 THEN
IF year MOD 400 = 0 THEN
RETURN TRUE
ELSE
RETURN FALSE
END
ELSE
RETURN TRUE
END
END
END IsLeapYear; </lang>
Objeck
<lang objeck>bundle Default {
class LeapYear {
function : Main(args : String[]) ~ Nil {
test_case := [1900, 1994, 1996, 1997, 2000];
each(i : test_case) {
test_case[i]->Print();
if(IsLeapYear(test_case[i])) {
" is a leap year."->PrintLine();
}
else {
" is not a leap year."->PrintLine();
};
};
}
function : native : IsLeapYear(year : Int) ~ Bool {
if(year % 4 = 0 & year % 100 <> 0) {
return true;
}
else if(year % 400 = 0) {
return true;
};
return false; } }
}</lang>
OCaml
<lang ocaml>let is_leap_year ~year =
if (year mod 100) = 0 then (year mod 400) = 0 else (year mod 4) = 0</lang>
Using Unix Time functions: <lang ocaml>let is_leap_year ~year =
let tm =
Unix.mktime {
(Unix.gmtime (Unix.time())) with
Unix.tm_year = (year - 1900);
tm_mon = 1 (* feb *);
tm_mday = 29
}
in
(tm.Unix.tm_mday = 29)</lang>
Oforth
<lang Oforth>Date.IsLeapYear(2000)</lang>
ooRexx
<lang ooRexx>
- routine isLeapYear
use arg year d = .datetime~new(year, 1, 1) return d~isLeapYear
</lang>
OpenEdge/Progress
The DATE function converts month, day, year integers to a date data type and will set the error status if invalid values are passed. <lang progress>FUNCTION isLeapYear RETURNS LOGICAL (
i_iyear AS INTEGER
):
DATE( 2, 29, i_iyear ) NO-ERROR. RETURN NOT ERROR-STATUS:ERROR.
END FUNCTION. /* isLeapYear */
MESSAGE
1900 isLeapYear( 1900 ) SKIP 1994 isLeapYear( 1994 ) SKIP 1996 isLeapYear( 1996 ) SKIP 1997 isLeapYear( 1997 ) SKIP 2000 isLeapYear( 2000 )
VIEW-AS ALERT-BOX.</lang>
Oz
<lang oz>declare
fun {IsLeapYear Year}
case Year mod 100 of 0 then
Year mod 400 == 0
else
Year mod 4 == 0
end end
in
for Y in [1900 1996 1997 2000] do
if {IsLeapYear Y} then
{System.showInfo Y#" is a leap year."}
else
{System.showInfo Y#" is NOT a leap year."}
end end</lang>
- Output:
1900 is NOT a leap year. 1996 is a leap year. 1997 is NOT a leap year. 2000 is a leap year.
PARI/GP
<lang parigp>isLeap(n)={
if(n%400==0, return(1)); if(n%100==0, return(0)); n%4==0
};</lang>
Alternate version: <lang parigp>isLeap(n)=!(n%if(n%100,4,400))</lang>
<lang parigp>isLeap(n)={
if(n%4,0,
n%100,1,
n%400,0,1
)
};</lang>
Pascal
<lang pascal>program LeapYear; uses
sysutils;//includes isLeapYear
procedure TestYear(y: word); begin
if IsLeapYear(y) then writeln(y,' is a leap year') else writeln(y,' is NO leap year');
end; Begin
TestYear(1900); TestYear(2000); TestYear(2100); TestYear(1904);
end.</lang> Output:
1900 is NO leap year 2000 is a leap year 2100 is NO leap year 1904 is a leap year
Perl
<lang Perl>sub isleap {
my $year = shift;
if ($year % 100 == 0) {
return ($year % 400 == 0);
}
return ($year % 4 == 0);
}</lang>
Or more concisely:
<lang Perl>sub isleap { !($_[0] % 100) ? !($_[0] % 400) : !($_[0] % 4) }</lang>
Alternatively, using functions/methods from CPAN modules:
<lang Perl>use Date::Manip; print Date_LeapYear(2000);
use Date::Manip::Base; my $dmb = new Date::Manip::Base; print $dmb->leapyear(2000);
use DateTime; my $date = DateTime->new(year => 2000); print $date->is_leap_year();</lang>
Perl 6
<lang perl6>say "$year is a {Date.is-leap-year($year) ?? 'leap' !! 'common'} year."</lang>
In Rakudo 2010.07, Date.is-leap-year is implemented as
<lang perl6>multi method is-leap-year($y = $!year) {
$y %% 4 and not $y %% 100 or $y %% 400
}</lang>
Phix
Available as an auto-include, implemented as: <lang Phix>global function is_leap_year(integer y)
return remainder(y,4)=0 and (remainder(y,100)!=0 or remainder(y,400)=0)
end function</lang>
PHP
<lang php><?php function isLeapYear($year) {
if ($year % 100 == 0) {
return ($year % 400 == 0);
}
return ($year % 4 == 0);
}</lang>
With date('L'):
<lang php><?php
function isLeapYear($year) {
return (date('L', mktime(0, 0, 0, 2, 1, $year)) === '1')
}</lang>
PicoLisp
<lang PicoLisp>(de isLeapYear (Y)
(bool (date Y 2 29)) )</lang>
- Output:
: (isLeapYear 2010) -> NIL : (isLeapYear 2008) -> T : (isLeapYear 1600) -> T : (isLeapYear 1700) -> NIL
PL/I
<lang pli>dcl mod builtin; dcl year fixed bin (31);
do year = 1900, 1996 to 2001;
if mod(year, 4) = 0 &
(mod(year, 100) ^= 0 |
mod(year, 400) = 0) then
put skip edit(year, 'is a leap year') (p'9999b', a);
else
put skip edit(year, 'is not a leap year') (p'9999b', a);
end;</lang>
- Output:
1900 is not a leap year 1996 is a leap year 1997 is not a leap year 1998 is not a leap year 1999 is not a leap year 2000 is a leap year 2001 is not a leap year
PostScript
<lang postscript>/isleapyear {
dup dup 4 mod 0 eq % needs to be divisible by 4 exch 100 mod 0 ne % but not by 100 and exch 400 mod 0 eq % or by 400 or
} def</lang>
PowerShell
<lang powershell>$Year = 2016 [System.DateTime]::IsLeapYear( $Year )</lang>
Prolog
<lang Prolog>leap_year(L) :- partition(is_leap_year, L, LIn, LOut), format('leap years : ~w~n', [LIn]), format('not leap years : ~w~n', [LOut]).
is_leap_year(Year) :- R4 is Year mod 4, R100 is Year mod 100, R400 is Year mod 400, ( (R4 = 0, R100 \= 0); R400 = 0).</lang>
- Output:
<lang Prolog> ?- leap_year([1900,1994,1996,1997,2000 ]). leap years : [1996,2000] not leap years : [1900,1994,1997] L = [1900,1994,1996,1997,2000].</lang>
There is an handy builtin that simplifies a lot, ending up in a simple query:
<lang Prolog> ?- findall(Y, (between(1990,2030,Y),day_of_the_year(date(Y,12,31),366)), L). L = [1992, 1996, 2000, 2004, 2008, 2012, 2016, 2020, 2024, 2028]. </lang>
PureBasic
<lang PureBasic>Procedure isLeapYear(Year)
If (Year%4=0 And Year%100) Or Year%400=0 ProcedureReturn #True Else ProcedureReturn #False EndIf
EndProcedure</lang>
Python
<lang python>import calendar calendar.isleap(year)</lang> or <lang python>def is_leap_year(year):
if year % 100 == 0:
return year % 400 == 0
return year % 4 == 0</lang>
Asking for forgiveness instead of permission: <lang python>import datetime
def is_leap_year(year):
try:
datetime.date(year, 2, 29)
except ValueError:
return False
return True</lang>
Scala
Ad hoc solution as REPL scripts:
Straightforward
<lang scala>def isLeapYear(year: Int) = { if (year % 100 == 0) year % 400 == 0 else year % 4 == 0 }</lang>
JDK 7 (not recommended)
<lang scala>new java.util.GregorianCalendar().isLeapYear(year)</lang>
JDK 8
Using JSR-310 java.time. <lang scala>java.time.LocalDate.ofYearDay(year, 1).isLeapYear()</lang>
R
<lang R>isLeapYear <- function(year) {
ifelse(year%%100==0, year%%400==0, year%%4==0)
}
for (y in c(1900, 1994, 1996, 1997, 2000)) {
print(paste(y, " is ", ifelse(isLeapYear(y), "", "not "), "a leap year.", sep=""))
}</lang>
- Output:
1900 is not a leap year. 1994 is not a leap year. 1996 is a leap year. 1997 is not a leap year. 2000 is a leap year.
Racket
<lang racket>(define (leap-year? y)
(and (zero? (modulo y 4)) (or (positive? (modulo y 100)) (zero? (modulo y 400)))))</lang>
Raven
<lang Raven>define is_leap_year use $year
$year 100 % 0 = if
$year 400 % 0 =
$year 4 % 0 =</lang>
REBOL
<lang rebol>leap-year?: func [
{Returns true if the specified year is a leap year; false otherwise.}
year [date! integer!]
/local div?
][
either date? year [year: year/year] [
if negative? year [throw make error! join [script invalid-arg] year]
]
; The key numbers are 4, 100, and 400, combined as follows:
; 1) If the year is divisible by 4, it’s a leap year.
; 2) But, if the year is also divisible by 100, it’s not a leap year.
; 3) Double but, if the year is also divisible by 400, it is a leap year.
div?: func [n] [zero? year // n]
to logic! any [all [div? 4 not div? 100] div? 400]
]</lang>
Retro
<lang Retro> : isLeapYear? ( y-f )
dup 400 mod 0 = [ drop -1 0 ] [ 1 ] if 0; drop dup 100 mod 0 = [ drop 0 0 ] [ 1 ] if 0; drop 4 mod 0 = ;</lang>
This is provided by the standard calendar library.
REXX
local variables
<lang rexx>leapyear: procedure; parse arg yr return yr//400==0 | (yr//100\==0 & yr//4==0)</lang>
with short-circuit
The REXX language doesn't support short-circuits, so here is a version that does a short-circuit. <lang rexx>leapyear: procedure; parse arg yr if y//4\==0 then return 0 /*Not ÷ by 4? Not a leap year.*/ return yr//400==0 | yr//100\==0</lang>
no local variables
This version doesn't need a PROCEDURE to hide local variable(s) [because there aren't any local variables]. <lang rexx>leapyear: if arg(1)//4\==0 then return 0 return arg(1)//400==0 | arg(1)//100\==0</lang>
handles 2 digit year
This REXX version has the proviso that if the year is exactly two digits,
the current century is assumed (i.e., no year windowing).
If a year below 100 is to be used, the year should have leading zeroes added (to make it four digits).
<lang rexx>leapyear: procedure; parse arg y /*year could be: Y, YY, YYY, YYYY*/
if y//4\==0 then return 0 /*Not ÷ by 4? Not a leap year.*/
if length(y)==2 then y=left(date('S'),2)y /*adjust for a 2─digit YY year.*/
return y//100\==0 | y//400==0 /*apply 100 and 400 year rule. */
- * * End of File * * *</lang>
Ring
<lang ring> give year leap = isLeapYear(year) if leap true see year + " is leap year." else see year + " is not leap year." ok
Func isLeapYear year
if (year % 400) = 0 return true
but (year % 100) = 0 return false
but (year % 4) = 0 return true
else return false ok
</lang>
Ruby
<lang ruby>require 'date'
Date.leap?(year)</lang>
The leap? method is aliased as gregorian_leap? And yes, there is a julian_leap? method.
Run BASIC
<lang runbasic>if date$("02/29/" + mid$(date$("mm/dd/yyyy"),7,4)) then print "leap year" else print "not"</lang>
Rust
<lang rust>fn is_leap(year: i32) -> bool {
let factor = |x| year % x == 0; factor(4) && (!factor(100) || factor(400)) }
}</lang>
Scala
By default, java.util.GregorianCalendar switches from Julian calendar to Gregorian calendar at 15 October 1582.
<lang scala>//use Java's calendar class new java.util.GregorianCalendar().isLeapYear(year)</lang>
For proleptic Gregorian calendar:
<lang scala>def isLeapYear(year:Int)=if (year%100==0) year%400==0 else year%4==0;
//or use Java's calendar class def isLeapYear(year:Int):Boolean = {
val c = new java.util.GregorianCalendar c.setGregorianChange(new java.util.Date(Long.MinValue)) c.isLeapYear(year)
}</lang>
Scheme
<lang scheme>(define (leap-year? n) (apply (lambda (a b c) (or a (and (not b) c)))
(map (lambda (m) (zero? (remainder n m)))
'(400 100 4))))</lang>
Seed7
This function is part of the "time.s7i" library. It returns TRUE if the year is a leap year in the Gregorian calendar. <lang seed7>const func boolean: isLeapYear (in integer: year) is
return (year rem 4 = 0 and year rem 100 <> 0) or year rem 400 = 0;</lang>
Original source: [1]
Sidef
<lang ruby>func isleap(year) {
if (year %% 100) {
return (year %% 400);
}
return (year %% 4);
}</lang>
or a little bit simpler: <lang ruby>func isleap(year) { year %% 100 ? (year %% 400) : (year %% 4) };</lang>
Smalltalk
Smalltalk has a built-in method named isLeapYear: <lang smalltalk> Date today isLeapYear. </lang>
SNOBOL4
Predicate leap( ) succeeds/fails, returns nil. <lang SNOBOL4> define('leap(yr)') :(end_leap) leap eq(remdr(yr,400),0) :s(return)
eq(remdr(yr,100),0) :s(freturn) eq(remdr(yr,4),0) :s(return)f(freturn)
end_leap
- # Test and display (with ?: kluge)
test = "output = ('10' ? (*leap(yr) 1 | 0)) ': ' yr"
yr = '1066'; eval(test)
yr = '1492'; eval(test)
yr = '1900'; eval(test)
yr = '2000'; eval(test)
end</lang>
- Output:
0: 1066 1: 1492 0: 1900 1: 2000
Swift
<lang Swift>func isLeapYear(year:Int) -> Bool {
return (year % 100 == 0) ? (year % 400 == 0) : (year % 4 == 0)
}
print(isLeapYear(2000)) print(isLeapYear(2011))</lang>
- Output:
true false
Tcl
The "classic" modulo comparison: <lang tcl>proc isleap1 {year} {
return [expr {($year % 4 == 0) && (($year % 100 != 0) || ($year % 400 == 0))}]
} isleap1 1988 ;# => 1 isleap1 1989 ;# => 0 isleap1 1900 ;# => 0 isleap1 2000 ;# => 1</lang> Does Feb 29 exist in the given year? If not a leap year, the clock command will return "03-01". (This code will switch to the Julian calendar for years before 1582.) <lang tcl>proc isleap2 year {
return [expr {[clock format [clock scan "$year-02-29" -format "%Y-%m-%d"] -format "%m-%d"] eq "02-29"}]
} isleap2 1988 ;# => 1 isleap2 1989 ;# => 0 isleap2 1900 ;# => 0 isleap2 2000 ;# => 1</lang>
TUSCRIPT
<lang tuscript>$$ MODE TUSCRIPT LOOP year="1900'1994'1996'1997'2000",txt="" SET dayoftheweek=DATE(number,29,2,year,number) IF (dayoftheweek==0) SET txt="not " PRINT year," is ",txt,"a leap year" ENDLOOP</lang>
- Output:
1900 is not a leap year 1994 is not a leap year 1996 is a leap year 1997 is not a leap year 2000 is a leap year
uBasic/4tH
<lang>DO
INPUT "Enter a year: "; y IF FUNC(_FNleap(y)) THEN PRINT y; " is a leap year" ELSE PRINT y; " is not a leap year" ENDIF
LOOP END
_FNleap Param (1) RETURN ((a@ % 4 = 0) * ((a@ % 400 = 0) + (a@ % 100 # 0)))</lang>
UNIX Shell
Original Bourne: <lang sh>leap() {
if expr $1 % 4 >/dev/null; then return 1; fi if expr $1 % 100 >/dev/null; then return 0; fi if expr $1 % 400 >/dev/null; then return 1; fi return 0;
}</lang>
Using GNU date(1): <lang sh>leap() {
date -d "$1-02-29" >/dev/null 2>&1;
}</lang>
Defining a bash function is_leap which accepts a YEAR argument, and uses no IO redirection, nor any extra processes. <lang sh>is_leap() {
local year=$(( 10#${1:?'Missing year'} ))
(( year % 4 == 0 && ( year % 100 != 0 || year % 400 == 0 ) )) && return 0
return 1
}</lang>
Using the cal command: (note that this invokes two processes with IO piped between them and is relatively heavyweight compared to the above shell functions: leap and is_leap) <lang sh>leap() {
cal 02 $1 | grep -q 29
} </lang>
Ursa
This program takes a year as a command line argument. <lang ursa>decl int year set year (int args<1>) if (= (mod year 4) 0)
if (and (= (mod year 100) 0) (not (= (mod year 400) 0)))
out year " is not a leap year" endl console
else
out year " is a leap year" endl console
end if
else
out year " is not a leap year" endl console
end if</lang> Output in Bash:
$ ursa leapyear.u 1900 1900 is not a leap year $ ursa leapyear.u 2000 2000 is a leap year
Vala
<lang Vala>void main() {
DateYear[] years = {1900, 1994, 1996, 1997, 2000};
foreach (DateYear year in years) {
string status = year.is_leap_year() ? "" : "not ";
print("%d is %sa leap year.\n", (int) year, status);
}
}</lang>
VBScript
<lang vb> Function IsLeapYear(yr) IsLeapYear = False If yr Mod 4 = 0 And (yr Mod 400 = 0 Or yr Mod 100 <> 0) Then IsLeapYear = True End If End Function
'Testing the function. arr_yr = Array(1900,1972,1997,2000,2001,2004)
For Each yr In arr_yr If IsLeapYear(yr) Then WScript.StdOut.WriteLine yr & " is leap year." Else WScript.StdOut.WriteLine yr & " is NOT leap year." End If Next </lang>
- Output:
1900 is NOT leap year. 1972 is leap year. 1997 is NOT leap year. 2000 is leap year. 2001 is NOT leap year. 2004 is leap year.
Vedit macro language
<lang vedit>while (#1 = Get_Num("Year: ")) {
#2 = (#1 % 4 == 0) && ((#1 % 100 != 0) || (#1 % 400 == 0))
if (#2) {
Message(" is leap year\n")
} else {
Message(" is not leap year\n")
}
}</lang>
The following version requires Vedit 6.10 or later: <lang vedit>while (#1 = Get_Num("Year: ")) {
if (Is_Leap_Year(#1)) {
Message(" is leap year\n")
} else {
Message(" is not leap year\n")
}
}</lang>
Wortel
<lang wortel>@let {
isLeapYear !?{\~%%1H \~%%4H \~%%4}
!-isLeapYear @range[1900 2000]
}</lang> Returns:
[1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000]
X86 Assembly
Using FASM syntax. Leaf function fits nicely into your program. <lang asm> align 16
- Input year as signed dword in EAX
IsLeapYear:
test eax,11b jz .4 retn ; 75% : ZF=0, not a leap year
.4:
mov ecx,100 cdq idiv ecx test edx,edx jz .100 cmp edx,edx retn ; 24% : ZF=1, leap year
.100:
test eax,11b retn ; 1% : ZF=?, leap year if EAX%400=0</lang>
XLISP
<lang xlisp>(DEFUN LEAP-YEARP (YEAR)
(AND (= (MOD YEAR 4) 0) (OR (/= (MOD YEAR 100) 0) (= (MOD YEAR 400) 0))))
- Test the function
(DISPLAY (MAPCAR LEAP-YEARP '(1600 1640 1800 1928 1979 1990 2000 2004 2005 2016)))</lang>
- Output:
(#T #T () #T () () #T #T () #T)
XPL0
<lang XPL0>func LeapYear(Y); \Return 'true' if Y is a leap year int Y; [if rem(Y/100)=0 then return rem(Y/400)=0; return rem(Y/4)=0; ];</lang>
Yorick
This solution is vectorized and can be applied to scalar or array input. <lang yorick>func is_leap(y) {
return ((y % 4 == 0) & (y % 100 != 0)) | (y % 400 == 0);
}</lang> Interactive example usage:
> is_leap(1988) 1 > is_leap([1988,1989,1900,2000]) [1,0,0,1]
zkl
<lang zkl>Time.Date.isLeapYear(1988) //-->True T(1988,1989,1900,2000).apply(Time.Date.isLeapYear)
//-->L(True,False,False,True)</lang>
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