Leap year

Leap year
You are encouraged to solve this task according to the task description, using any language you may know.

Task

Determine whether a given year is a leap year in the Gregorian calendar.

See also

• Leap year (wiki)

11l

Translation of: Python

F is_leap_year(year)
   I year % 100 == 0
      R year % 400 == 0
   R year % 4 == 0

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.  

 
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 CCYY BY 100: R0 = YY, R1 = CC            
     LTR  0,0           YY = 0? IF CCYY DIV BY 100, LY IFF DIV BY 400                                        
     BZ   A               YES: R0|R1 = CC; CCYY DIV BY 100, TEST CC                           
     SRDL 0,32            NO: R0|R1 = YY; CCYY NOT DIV BY 100, TEST YY                           
A    LA   2,4           DIVISOR = 4; DIVIDEND = YY, OR DIV BY 100 CC                                    
     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

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'.

68000 Assembly

;Example
        move.l  #2018,d0
        bsr     leap_year
        addi.l  #28,d1      ; # days in February 2018
        rts
; Leap Year
;
; Input
;   d0=year
;
; Output
;   d1=1 if leap year, 0 if not leap year
;   zero flag clear if leap year, set if not
;
leap_year:
        cmpi.l  #1752,d0
        ble.s   not_leap_year

        move.l  d0,d1
        lsr.l   #1,d1
        bcs.s   not_leap_year
        lsr.l   #1,d1
        bcs.s   not_leap_year

; If we got here, year is divisible by 4.
        move.l  d0,d1
        divu    #100,d1
        swap    d1
        tst.w   d1
        bne.s   is_leap_year

; If we got here, year is divisible by 100.
        move.l  d0,d1
        divu    #400,d1
        swap    d1
        tst.w   d1
        bne.s   not_leap_year

is_leap_year:
        moveq.l #1,d1
        rts
not_leap_year:
        moveq.l #0,d1
        rts

8080 Assembly

		org	100h
		jmp	test
		;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
		;; Check if a year is a leap year.
		;; Input: HL = year.
		;; Output: Carry flag set if HL is a leap year.
		;; Registers used: all.
leap:		mvi	a,3		; Divisible by 4?
		ana	l		; If not, not a leap year,
		rnz			; Return w/carry cleared
		mvi	b,2		; Divide by 4 (shift right 2)
leapdiv4:	mov	a,h
		rar
		mov	h,a
		mov 	a,l
		rar
		mov	l,a
		dcr 	b
		jnz	leapdiv4
		lxi	b,-1		; Divide by 25 using subtraction
		lxi	d,-25
leapdiv25:	inx	b		; Keep quotient in BC
		dad	d
		jc	leapdiv25
		mov	a,e		; If so, L==E afterwards.
		xra	l 		; (High byte is always FF.)
		stc			; Set carry, and
		rnz			; return if not divisible.
		mvi	a,3		; Is this divisble by 4?
		ana	c
		sui	1		; Set carry if so.
		ret
		;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
		;; Test code: get year from CP/M command line and see
		;; if it is a leap year.
test:		lxi	b,5Dh		; First char of "file name"
		lxi	h,0		; Accumulator
digit:		ldax	b		; Get character
		sui	'0'		; ASCII digit
		jc	getleap		; Not valid digit = done
		cpi	10
		jnc	getleap		; Not valid digit = done
		dad	h		; HL *= 10
		mov	d,h
		mov	e,l
		dad	h
		dad	h
		dad	d
		mvi	d,0		; Plus digit
		mov	e,a
		dad	d
		inx	b		; Next character
		jmp	digit
getleap:	call	leap		; Is HL a leap year?
		lxi	d,no		; No, 
		jnc	out		; unless carry is set,
		lxi	d,yes		; then it is a leap year.
out:		mvi	c,9
		jmp	5 
no:		db	'NOT '
yes:		db	'LEAP YEAR.$'

Action!

BYTE FUNC IsLeapYear(CARD year)
  IF year MOD 100=0 THEN
    IF year MOD 400=0 THEN
      RETURN (1)
    ELSE
      RETURN (0)
    FI
  FI
    
  IF year MOD 4=0 THEN
    RETURN (1)
  FI
RETURN (0)

PROC Main()
  CARD ARRAY t=[1900 1901 2000 2001 2004 2020 2021]
  BYTE i,leap
  CARD year

  FOR i=0 TO 6
  DO 
    year=t(i)
    leap=IsLeapYear(year)
    IF leap=0 THEN
      PrintF("%U is not a leap year%E",year)
    ELSE
      PrintF("%U is a leap year%E",year)
    FI
  OD
RETURN

Output:

Screenshot from Atari 8-bit computer

1900 is not a leap year
1901 is not a leap year
2000 is a leap year
2001 is not a leap year
2004 is a leap year
2020 is a leap year
2021 is not a leap year

ActionScript

public function isLeapYear(year:int):Boolean {
    if (year % 100 == 0) {
        return (year % 400 == 0);
    }
    return (year % 4 == 0);
}

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);

ALGOL 60

Works with: A60

begin
	integer year;

	integer procedure mod(i,j); value i,j; integer i,j;
	mod:=i-(i div j)*j;

	boolean procedure isLeapYear(year); value year; integer year;
	isLeapYear:=mod(year,400)=0 or (mod(year,4)=0 and mod(year,100) notequal 0);

	for year := 1899, 1900, 1901, 1902, 1903, 1904, 1905, 1999, 2000, 2001, 2002, 2003, 2004 do begin
		outinteger(1,year);
		if isLeapYear(year) then outstring(1,"True\n") else outstring(1, "False\n")
	end for year
end

Output:

 1899 False
 1900 False
 1901 False
 1902 False
 1903 False
 1904 True
 1905 False
 1999 False
 2000 True
 2001 False
 2002 False
 2003 False
 2004 True

ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

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
)

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

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.

ALGOL-M

BEGIN

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

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

% EXERCISE THE FUNCTION %
INTEGER Y;
WRITE("TEST OF CENTURY YEARS");
FOR Y := 1600 STEP 100 UNTIL 2000 DO
  BEGIN
    IF ISLEAP(Y) <> 0 THEN
      WRITE(Y, " IS A LEAP YEAR")
    ELSE
      WRITE(Y, " IS NOT A LEAP YEAR");
  END;
WRITE("TEST OF CURRENT DECADE");
FOR Y := 2010 STEP 1 UNTIL 2020 DO
  BEGIN
    IF ISLEAP(Y) <> 0 THEN
      WRITE(Y, " IS A LEAP YEAR")
    ELSE
      WRITE(Y, " IS NOT A LEAP YEAR");
  END;

END

Output:

TEST OF CENTURY YEARS
  1600 IS A LEAP YEAR
  1700 IS NOT A LEAP YEAR
  1800 IS NOT A LEAP YEAR
  1900 IS NOT A LEAP YEAR
  2000 IS A LEAP YEAR
TEST OF CURRENT DECADE
  2010 IS NOT A LEAP YEAR
  2011 IS NOT A LEAP YEAR
  2012 IS A LEAP YEAR
  2013 IS NOT A LEAP YEAR
  2014 IS NOT A LEAP YEAR
  2015 IS NOT A LEAP YEAR
  2016 IS A LEAP YEAR
  2017 IS NOT A LEAP YEAR
  2018 IS NOT A LEAP YEAR
  2019 IS NOT A LEAP YEAR
  2020 IS A LEAP YEAR

APL

Returns 1 if leap year, 0 otherwise:

∇ z←Leap year
Z←(0=4|year)∧(0=400|year)∨~0=100|year
∇

A much neater version of the above relies on the fact that every rule is an exception the the previous one:

∇ z←Leap year
  z←0≠.=400 100 4∘.|year
∇

This essentially works by running an XOR reduction over the divisibility by 4, 100, and 400. Some APL implementations support tacit (a.k.a. points-free) programming:

Leap←0≠.=400 100 4∘.|⊢

Dyalog APL version 18.0 added a built-in date-time function:

Leap←0⎕DT,∘2 29¨

This works by extending the year to February 29 of that year, and then checking if the date is valid.

With any of the above definitions, no loop is necessary to check each year of an array:

      Leap 1899 1900 1901 1902 1903 1904 1905 1999 2000 2001 2002 2003 2004

Output:

0 0 0 0 0 1 0 0 1 0 0 0 1

AppleScript

on leap_year(y)
    return y mod 4 is equal to 0 and (y mod 100 is not equal to 0 or y mod 400 is equal to 0)
end leap_year

leap_year(1900)

Arc

(= leap? (fn (year)
  (if (and (is 0 (mod year 4)) (isnt 0 (mod year 100))) year
      (unless (< 0 (+ (mod year 100) (mod year 400))) year))))

Output:

(map [leap? _] '(1900 1904 2000 2019 2020 2100)) 
;; =>          '(     1904 2000      2020     )

Arturo

years: [
    1600 1660 1724 1788 1848 1912 1972 
    2032 2092 2156 2220 2280 2344 2348
    1698 1699 1700 1750 1800 1810 1900 
    1901 1973 2100 2107 2200 2203 2289
]

print select years => leap?

Output:

1600 1660 1724 1788 1848 1912 1972 2032 2092 2156 2220 2280 2344 2348

AutoHotkey

leapyear(year)
{
    if (Mod(year, 100) = 0)
        return (Mod(year, 400) = 0)
    return (Mod(year, 4) = 0)
}

MsgBox, % leapyear(1604)

Output:

Returns 1 if year is a leap year

or

IsLeapYear(Year)
{
    return !Mod(Year, 4) && Mod(Year, 100) || !Mod(Year, 400)
}

MsgBox % "The year 1604 was " (IsLeapYear(1604) ? "" : "not ") "a leap year"

Output:

The year 1600 was a leap year
The year 1601 was not a leap year
The year 1604 was a leap year

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)"

Output:

2012 is a leap year.

--BugFix (talk) 16:18, 16 November 2013 (UTC)

AWK

function leapyear( year )
{
    if ( year % 100 == 0 )
        return ( year % 400 == 0 )
    else
        return ( year % 4 == 0 )            
}

Bash

#!/bin/bash

is_leap_year ()  # Define function named is_leap_year
{

declare -i year=$1      # declare integer variable "year" and set it to function parm 1

echo -n "$year ($2)-> " # print the year passed in, but do not go to the next line

if (( $year % 4 == 0 )) # if year not dividable by 4, then not a leap year, % is the modulus operator 
then
        if (( $year % 400 == 0 ))       # if century dividable by 400, is a leap year
        then
                echo "This is a leap year"
        else
                if (( $year % 100 == 0 )) # if century not divisible by 400, not a leap year
                then
                        echo "This is not a leap year"
                else
                        echo "This is a leap year" # not a century boundary, but dividable by 4, is a leap year
                fi
        fi
else
        echo "This is not a leap year"
fi


}

# test all cases
# call the function is_leap_year several times with two parameters... year and test's expectation for 'is/not leap year.
is_leap_year 1900 not # a leap year
is_leap_year 2000 is  # a leap year
is_leap_year 2001 not # a leap year
is_leap_year 2003 not # a leap year
is_leap_year 2004 is  # a leap year

# Save the above to a file named is_leap_year.sh, then issue the following command to run the 5 tests of the function
# bash is_leap_year.sh

BASIC

Applesoft BASIC

A one-liner combination from the Commodore BASIC and GW-BASIC solutions.

FOR Y = 1750 TO 2021: PRINT  MID$ ( STR$ (Y) + " ",1,5 * (Y / 4 =  INT (Y / 4)) * ((Y / 100 <  >  INT (Y / 100)) + (Y / 400 =  INT (Y / 400))));: NEXT

BASIC256

Translation of: FreeBASIC

# year is a BASIC-256 keyword
function leapyear(year_)
	if (year_ mod 4) <> 0 then return FALSE
	if (year_ mod 100) = 0 and (year_ mod 400) <> 0 then return FALSE
	return TRUE
end function

for year_ = 1800 to 2900 step 100
	print year_;
	if leapyear(year_) then print " is a leap year" else print " is not a leap year"
next year_

print

for year_ = 2012 to 2031
	print year_;
	if leapyear(year_) = TRUE then print " = leap   "; else print " = no     ";
	if (year_ mod 4) = 3 then print ""
next year_
end

BaCon

From the Ada shortcut calculation

' 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

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

BBC BASIC

      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))

Much quicker without full evaluation:

DEFFNleap(yr%)
 IF yr% MOD 4   THEN =FALSE
 IF yr% MOD 400 ELSE =TRUE
 IF yr% MOD 100 ELSE =FALSE
=TRUE

Chipmunk Basic

Translation of: GW-BASIC

10 rem Leap year
20 for i% = 1 to 5
30   read year%
40   print year%;"is ";
50   if isleapyear(year%) = 0 then print "not "; else print "";
60   print "a leap year."
70 next i%
80 end

200 data 1900,1994,1996,1997,2000

400 sub isleapyear(y%)
410   isleapyear = ((y% mod 4 = 0) and (y% mod 100 <> 0)) or (y% mod 400 = 0)
420 end sub

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.

Commodore BASIC

An old-timey solution:

10 DEF FNLY(Y)=(Y/4=INT(Y/4))*((Y/100<>INT(Y/100))+(Y/400=INT(Y/400)))

Or, using Simons' BASIC's MOD function:

Simons' BASIC

10 DEF FNLY(Y)=(0=MOD(Y,4))*((0<MOD(Y,100))+(0=MOD(Y,400)))

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

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

window 1

// 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 BOOL
  BOOL 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

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

print

// 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

HandleEvents

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

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

Output:

2016 is a leap year.

GW-BASIC

With a function

Works with: BASICA

10  ' Leap year
20  DEF FN ISLEAPYEAR(Y%) = ((Y% MOD 4 = 0) AND (Y% MOD 100 <> 0)) OR (Y% MOD 400 = 0)
95  ' *** Test ***
100 FOR I% = 1 TO 5
110  READ YEAR%
120  PRINT YEAR%; "is ";
130  IF FN ISLEAPYEAR(YEAR%) = 0 THEN PRINT "not "; ELSE PRINT "";
140  PRINT "a leap year."
150 NEXT I%
160 END
200 DATA 1900, 1994, 1996, 1997, 2000

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.  

With a subroutine

Prints all the leap years from 1750 to 2021. Note the correct behaviour of 1800, 1900, and 2000.

Works with: BASICA

10 FOR Y = 1750 TO 2021
20 GOSUB 1000
30 IF L = 1 THEN PRINT Y;" ";
40 NEXT Y
50 END
1000 L = 0
1010 IF Y MOD 4 <> 0 THEN RETURN
1020 IF Y MOD 100 = 0 AND Y MOD 400 <> 0 THEN RETURN
1030 L = 1
1040 RETURN

Output:

1752   1756   1760   1764   1768   1772   1776   1780   1784   1788   1792   1796   1804   1808   1812   1816   1820   1824   1828
1832   1836   1840   1844   1848   1852   1856   1860   1864   1868   1872   1876   1880   1884   1888   1892   1896   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   2016   2020

IS-BASIC

100 PROGRAM "Leapyear.bas"
110 FOR I=1990 TO 2020
120   IF LEAPY(I) THEN
130     PRINT I;"is a leap year."
140   ELSE
150     PRINT I;"is not a leap year."
160   END IF
170 NEXT
180 DEF LEAPY(Y)=MOD(Y,4)=0 AND MOD(Y,100) OR MOD(Y,400)=0

Liberty BASIC

Simple method

if leap(1996)then
    print "leap"
else
    print "ordinary"
end if
wait

function leap(n)
    leap=date$("2/29/";n)
end function

Calculated method

    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

NS-HUBASIC

10 INPUT "ENTER A NUMBER, AND I'LL DETECT IF IT'S A LEAP YEAR OR NOT. ",A
20 IF A-(A/100)*100=0 AND A-(A/400)*400<>0 THEN RESULT$="NOT "
30 PRINT "THAT'S "RESULT$"A LEAP YEAR."

Palo Alto Tiny BASIC

Translation of: Tiny BASIC

    10 REM LEAP YEAR
    20 FOR Y=1750 TO 2022
    30 GOSUB 100
    40 IF L=1 PRINT Y
    50 NEXT Y
    60 STOP
   100 LET L=0
   110 IF Y-(Y/4)*4#0 RETURN
   120 IF Y-(Y/100)*100#0 LET L=1
   130 IF Y-(Y/400)*400=0 LET L=1
   140 RETURN

PureBasic

Procedure isLeapYear(Year)
  If (Year%4=0 And Year%100) Or Year%400=0
    ProcedureReturn #True
  Else
    ProcedureReturn #False
  EndIf
EndProcedure

QBasic

Note that the year% function is not needed for most modern BASICs.

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

QL SuperBASIC

AUTO  
   REM Is% a non-proleptic Gregorian year y$<=9999 leap (0) 0R ordinary (1)?
   DEF FN Is%(y$) 
     LOC l%,c%,y%
     LET c%=y$(1 TO 2)&"00" : y%=y$
     LET l%=c% MOD 16 AND y$(3 TO 4)="00" OR y% MOD 4 
     RET l%
   END DEF Is%
ctrl+space

using only power-of-2 divisions. N.B. the inverted logic brings home the BaCon code's flaw

Output:

1600  0
1700  1
1800  1
1900  1
2000  0
2100  1

Run BASIC

if date$("02/29/" + mid$(date$("mm/dd/yyyy"),7,4)) then print "leap year" else print "not"

S-BASIC

Since S-BASIC has no MOD operator or function, we have to supply one.

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

rem - return true (-1) if y is a leap year, otherwise 0
function isleapyear(y = integer) = integer
end = mod(y,4)=0 and mod(y,100)<>0 or mod(y,400)=0

rem - exercise the function
var y = integer

print "Test of century years"
for y = 1600 to 2000 step 100
   if isleapyear(y) then
     print y;" is a leap year"
   else
     print y;" is NOT a leap year"
next y

print "Test of current half-decade"
for y = 2015 to 2020
   if isleapyear(y) then
     print y; " is a leap year"
   else
     print y; " is NOT a leap year"
next y

end

Output:

Test of century years
 1600 is a leap year
 1700 is NOT a leap year
 1800 is NOT a leap year
 1900 is NOT a leap year
 2000 is a leap year
Test of current half-decade
 2015 is NOT a leap year
 2016 is a leap year
 2017 is NOT a leap year
 2018 is NOT a leap year
 2019 is NOT a leap year
 2020 is a leap year

Sinclair ZX81 BASIC

ZX81 BASIC does not support user-defined functions, even the single-expression functions that are provided by many contemporary dialects; so we have to fake it using a subroutine and pass everything in global variables.

5000 LET L=Y/4=INT (Y/4) AND (Y/100<>INT (Y/100) OR Y/400=INT (Y/400))
5010 RETURN

An example showing how to call it:

10 INPUT Y
20 GOSUB 5000
30 PRINT Y;" IS ";
40 IF NOT L THEN PRINT "NOT ";
50 PRINT "A LEAP YEAR."
60 STOP

Tiny BASIC

Works with: TinyBasic

REM Rosetta Code problem: https://rosettacode.org/wiki/Leap_year
REM by Jjuanhdez, 06/2022

10 REM Leap year
20 LET Y = 1750
30 IF Y = 2021 THEN GOTO 80
40 GOSUB 100
50 IF L = 1 THEN PRINT Y
60 LET Y = Y + 1
70 GOTO 30
80 END
100 LET L = 0
110 IF (Y - (Y / 4) * 4) <> 0 THEN RETURN
120 IF (Y - (Y / 100) * 100) = 0 THEN GOTO 140
130 LET L = 1
140 IF (Y - (Y / 400) * 400) <> 0 THEN GOTO 160
150 LET L = 1
160 RETURN

uBasic/4tH

Translation of: BBC BASIC

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)))

VBA

Public Function Leap_year(year As Integer) As Boolean
    Leap_year = (Month(DateSerial(year, 2, 29)) = 2)
End Function

VBScript

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

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.

Visual Basic

Works with: Visual Basic version VB6 Standard

Public Function IsLeapYear1(ByVal theYear As Integer) As Boolean
'this function utilizes documented behaviour of the built-in DateSerial function
IsLeapYear1 = (VBA.Day(VBA.DateSerial(theYear, 2, 29)) = 29)
End Function

Public Function IsLeapYear2(ByVal theYear As Integer) As Boolean
'this function uses the well-known formula
IsLeapYear2 = IIf(theYear Mod 100 = 0, theYear Mod 400 = 0, theYear Mod 4 = 0)
End Function

Testing:

Sub Main()
'testing the above functions
Dim i As Integer
  For i = 1750 To 2150
    Debug.Assert IsLeapYear1(i) Eqv IsLeapYear2(i)
  Next i
End Sub

Visual Basic .NET

Translation of: C#

Module Module1

    Sub Main()
        For Each y In {1900, 1994, 1996, Date.Now.Year}
            Console.WriteLine("{0} is {1}a leap year.", y, If(Date.IsLeapYear(y), String.Empty, "not "))
        Next
    End Sub

End Module

Output:

1900 is not a leap year.
1994 is not a leap year.
1996 is a leap year.
2019 is not a leap year.

Yabasic

sub leapyear(year)
    if mod(year, 4) <> 0 then return false : fi
    if mod(year, 100) = 0 and mod(year, 400) <> 0 then return false : fi
    return TRUE
end sub

for year = 1800 to 2900 step 100
    print year; 
	if leapyear(year) then print " is a leap year" else print " is not a leap year" : fi
next year
print
for year = 2012 to 2031
    print year;
    if leapyear(year) = TRUE then print " = leap "; else print " = no "; : fi
    if mod(year, 4) = 3 then print : fi
next year
end

ZX Spectrum Basic

10 DEF FN l(y)=y/4=INT (y/4) AND (y/100<>INT (y/100) OR y/400=INT (y/400))

Batch File

@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...

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 . . .

bc

define l(y) {
    if (y % 100 == 0) y /= 100
    if (y % 4 == 0) return(1)
    return(0)
}

BCPL

get "libhdr"

let leap(year) = year rem 400 = 0 | (year rem 4 = 0 & year rem 100 ~= 0)

let start() be 
$(  let years = table 1899, 1900, 1901, 1902, 1903, 1904, 1905, 1999,
                      2000, 2001, 2002, 2003, 2004, 2021, 2022
    for i = 0 to 14 do
        writef("%N %S a leap year.*N", 
               years!i, leap(years!i) -> "is", "is not")
$)

Output:

1899 is not a leap year.
1900 is not a leap year.
1901 is not a leap year.
1902 is not a leap year.
1903 is not a leap year.
1904 is a leap year.
1905 is not a leap year.
1999 is not a leap year.
2000 is a leap year.
2001 is not a leap year.
2002 is not a leap year.
2003 is not a leap year.
2004 is a leap year.
2021 is not a leap year.
2022 is not a leap year.

Befunge

Translation of: C

0"2("*:3-:1-:2-:"^"-v<
v*%"d"\!%4::,,"is".:<|
>\45*:*%!+#v_ "ton"vv<
v"ear."+550<,,,,*84<$#
>"y pael a ">:#,_$:#@^

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.

BQN

Leap ← 0=4|100÷˜⍟(0=|)¨⊢

Or:

Leap ← -˝0=4‿100‿400|⌜⊢

Test:

Leap 1900‿1996‿1998‿1999‿2000‿2024‿2100

Output:

⟨ 0 1 0 0 1 1 0 ⟩

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")
      )
    )
& ;

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

#include <stdio.h>

int is_leap_year(unsigned year)
{
    return !(year & (year % 100 ? 3 : 15));
}

int main(void)
{
    const unsigned test_case[] = {
        1900, 1994, 1996, 1997, 2000, 2024, 2025, 2026, 2100
    };
    const unsigned n = sizeof test_case / sizeof test_case[0];

    for (unsigned i = 0; i != n; ++i) {
        unsigned year = test_case[i];
        printf("%u is %sa leap year.\n", year, is_leap_year(year) ? "" : "not ");
    }
    return 0;
}

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.
2024 is a leap year.
2025 is not a leap year.
2026 is not a leap year.
2100 is not a leap year.

C#

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 ");
        }
    }
}

Output:

1900 is not a leap year.
1994 is not a leap year.
1996 is a leap year.
2010 is not a leap year.

C++

Uses C++11. Compile with

g++ -std=c++11 leap_year.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";
  }
}

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.

Clipper

Function IsLeapYear( nYear )
Return Iif( nYear%100 == 0, (nYear%400 == 0), (nYear%4 == 0) )

Clojure

A simple approach:

(defn leap-year? [y]
  (and (zero? (mod y 4))
       (or (pos?  (mod y 100))
           (zero? (mod y 400)))))

A slightly terser, if slightly less obvious approach:

(defn leap-year? [y]
  (condp #(zero? (mod %2 %1)) y
    400 true
    100 false
    4   true
    false))

CLU

is_leap_year = proc (year: int) returns (bool)
    return(year//400 =0 cor (year//4 = 0 cand year//100 ~= 0))
end is_leap_year

start_up = proc ()
    po: stream := stream$primary_output()
    years: sequence[int] := sequence[int]$
        [1899, 1900, 1901, 1902, 1903, 1904, 1905, 1999,
         2000, 2001, 2002, 2003, 2004, 2021, 2022]
    
    for year: int in sequence[int]$elements(years) do
        stream$puts(po, int$unparse(year) || " is ")
        if ~is_leap_year(year) then stream$puts(po, "not ") end
        stream$putl(po, "a leap year.")
    end
end start_up

Output:

1899 is not a leap year.
1900 is not a leap year.
1901 is not a leap year.
1902 is not a leap year.
1903 is not a leap year.
1904 is a leap year.
1905 is not a leap year.
1999 is not a leap year.
2000 is a leap year.
2001 is not a leap year.
2002 is not a leap year.
2003 is not a leap year.
2004 is a leap year.
2021 is not a leap year.
2022 is not a leap year.

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
           .

Using Date Intrinsic Functions

       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.

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

(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))))

Component Pascal

BlackBox Component Builder

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.

Execute: ^Q LeapYear.Do 2000 2004 2013~

Output:

 2000:> $TRUE
 2004:> $TRUE
 2013:> $FALSE

Crystal

p Time.leap_year?(2020)
p Time.leap_year?(2021)
p Time.leap_year?(2022)

true
false
false

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));
}

Using the datetime library:

import std.datetime;

void main() {
    assert(yearIsLeapYear(1724));
    assert(!yearIsLeapYear(1973));
    assert(!Date(1900, 1, 1).isLeapYear);
    assert(DateTime(2000, 1, 1).isLeapYear);
}

Dart

class Leap {
  bool leapYear(num year) {
    return (year % 400 == 0) || (( year % 100 != 0) && (year % 4 == 0));

  bool isLeapYear(int year) =>
    (year % 4 == 0) && ((year % 100 != 0) || (year % 400 == 0));
  // Source: https://api.flutter.dev/flutter/quiver.time/isLeapYear.html
  }
}

Dc

Directly taken from Wikipedia.

Works with: GNU dc

[0q]s0
[1q]s1

[ S. [  l.   4% 0!=0    ## if y %   4:  return 0
        l. 100% 0!=1    ## if y % 100:  return 1
        l. 400% 0!=0    ## if y % 400:  return 0
        1               ##              return 1
     ]x s.L.
]sL                     ## L = isleapYear()

[   Sy
    lyn [ is ]P
    ly lLx
    [not ] 0:y
    [    ] 1:y
    ;yP
    [a leap year]P AP
    OsyLyo
]sT                     ## T = testYear()

1988 lTx
1989 lTx
1900 lTx
2000 lTx

Output:

1988 is     a leap year
1989 is not a leap year
1900 is not a leap year
2000 is     a leap year

Delphi/Pascal

Delphi has standard function IsLeapYear in SysUtils unit.

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.

Draco

proc nonrec leap_year(word year) bool:
    year%400=0 or (year%4=0 and year%100/=0)
corp

proc nonrec main() void:
    [15]word years = (1899, 1900, 1901, 1902, 1903, 1904, 1905, 1999,
                      2000, 2001, 2002, 2003, 2004, 2021, 2022);
    word i;
    
    for i from 0 upto 14 do
        writeln(years[i],
                if leap_year(years[i]) then " is " else " is not " fi,
                "a leap year.")
    od
corp

Output:

1899 is not a leap year.
1900 is not a leap year.
1901 is not a leap year.
1902 is not a leap year.
1903 is not a leap year.
1904 is a leap year.
1905 is not a leap year.
1999 is not a leap year.
2000 is a leap year.
2001 is not a leap year.
2002 is not a leap year.
2003 is not a leap year.
2004 is a leap year.
2021 is not a leap year.
2022 is not a leap year.

DWScript

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);

Dyalect

func isLeap(y) {
    if y % 100 == 0 {
        y % 400 == 0
    } else {
        y % 4 == 0
    }
}

print(isLeap(1984))

Output:

true

EasyLang

func leapyear y .
   if y mod 4 = 0 and (y mod 100 <> 0 or y mod 400 = 0)
      return 1
   .
   return 0
.
print leapyear 2000

Ela

isLeap y | y % 100 == 0 = y % 400 == 0
         | else         = y % 4 == 0

Elixir

leap_year? = fn(year) -> :calendar.is_leap_year(year) end
IO.inspect for y <- 2000..2020, leap_year?.(y), do: y

Output:

[2000, 2004, 2008, 2012, 2016, 2020]

Emacs Lisp

Translation of: Scheme

(defun leap-year-p (year)
  (apply (lambda (a b c) (or a (and (not b) c)))
	 (mapcar (lambda (n) (zerop (mod year n)))
		 '(400 100 4))))

Erlang

-module(gregorian).
-export([leap/1]).

leap( Year ) -> calendar:is_leap_year( Year ).

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

Euphoria

function isLeapYear(integer year)
    return remainder(year,4)=0 and remainder(year,100)!=0 or remainder(year,400)=0
end function

Excel

Take two cells, say A1 and B1, in B1 type in :

=IF(OR(NOT(MOD(A1,400)),AND(NOT(MOD(A1,4)),MOD(A1,100))),"Leap Year","Not a Leap Year")

Output:

1900	Not a Leap Year
1954	Not a Leap Year
1996	Leap Year
2003	Not a Leap Year
2012	Leap Year

LAMBDA

Binding the name ISLEAPYEAR to the following lambda expression in the Name Manager of the Excel WorkBook,

as a reusable custom function:

(See LAMBDA: The ultimate Excel worksheet function)

Works with: Office 365 betas 2021

ISLEAPYEAR
=LAMBDA(y,
    OR(
        0 = MOD(y, 400),
        AND(
            0 = MOD(y, 4),
            0 <> MOD(y, 100)
        )
    )
)

Output:

	fx	=ISLEAPYEAR(A2)
	A	B
1	Year	Verdict
2	1900	FALSE
3	1954	FALSE
4	1996	TRUE
5	2003	FALSE
6	2012	TRUE

F#

let isLeapYear = System.DateTime.IsLeapYear
assert isLeapYear 1996
assert isLeapYear 2000
assert not (isLeapYear 2001)
assert not (isLeapYear 1900)

Factor

Call leap-year? word from calendars vocabulary. For example:

USING: calendar prettyprint ;
2011 leap-year? .

Factor uses proleptic Gregorian calendar.

Fermat

Function IsLeap(y) = if y|4>0 then 0 else if y|100=0 and y|400>0 then 0 else 1 fi fi.

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= ;

Or more simply (but always computing three "mod"):

: leap-year? dup 4 mod 0= over 16 mod 0= rot 25 mod 0= not or and ;

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

Output:

  F T F T 

Fōrmulæ

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

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

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

Solution

In a more concise way:

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;

Genie

Dialect conversion from Vala entry.

[indent=4]
/*
   Leap year, in Genie

   valac leapYear.gs
   ./leapYear
*/
init
    years:array of DateYear = {1900, 1994, 1996, 1997, 2000, 2100}

    for year in years
        status:string = year.is_leap_year() ? "" : "not "
        stdout.printf("%d is %sa leap year.\n", year, status)

Output:

prompt$ valac leapYear.gs
prompt$ ./leapYear
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.
2100 is not a leap year.

Go

func isLeap(year int) bool {
    return year%400 == 0 || year%4 == 0 && year%100 != 0
}

Groovy

Solution:

(1900..2012).findAll {new GregorianCalendar().isLeapYear(it)}.each {println it}

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

FUNCTION IsLeapYear( nYear )
   RETURN iif( nYear % 100 == 0, nYear % 400 == 0, nYear % 4 == 0 )

Haskell

Simple version

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

Algorithmic

isleap = foldl1 ((&&).not).flip map [400, 100, 4]. ((0==).).mod

Example using isleap

*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

TDD version

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]

Hy

(defn leap? [y]
    (and
        (= (% y 4) 0)
        (or
            (!= (% y 100) 0)
            (= (% y 400) 0))))

Icon and Unicon

Gives leap year status for 2000,1900,2012 and any arguments you give

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

J

isLeap=: 0 -/@:= 4 100 400 |/ ]

Example use:

   isLeap 1900 1996 1997 2000
0 1 0 1

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.

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);
        }
}

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.

Works with: Java version 8

import java.time.Year;

public class IsLeap {

    public static void main(String[] args) {
        System.out.println(Year.isLeap(2004));
    }
}

JavaScript

var isLeapYear = function (year) { return (year % 100 === 0) ? (year % 400 === 0) : (year % 4 === 0); };

Or, by setting the day to the 29th and checking if the day remains

// Month values start at 0, so 1 is for February
var isLeapYear = function (year) { return new Date(year, 1, 29).getDate() === 29; };

Joy

DEFINE leapyear == dup 100 div null rotate choice 4 rem null.

jq

Translation of: Julia

def leap:
  . as $y | ($y%4) == 0 and ($y < 1582 or ($y%400) == 0 or ($y%100) != 0);

Examples:

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))

Output:

$ jq -n -f Leap_year.jq

Julia

Works with: Julia version 0.6

isleap(yr::Integer) = yr % 4 == 0 && (yr < 1582 || yr % 400 == 0 || yr % 100 != 0)

@assert all(isleap, [2400, 2012, 2000, 1600, 1500, 1400])
@assert !any(isleap, [2100, 2014, 1900, 1800, 1700, 1499])

K

K3

Works with: Kona

Leap year predicate:

  lyp:{(+/~x!'4 100 400)!2}

  lyp'1996+!6
1 0 0 0 1 0

Leap year selection:

  lys:{a@&lyp'a:x}

  lys@1900,1994,1996,1997,2000
1996 2000

Koka

Chain of boolean expressions

 pub fun is-leap-year(year: int)
   year % 4 == 0 && (year % 100 != 0 || year % 400 == 0)

If-Then-Else

 pub fun is-leap-year'(year: int)
   year % (if year % 100 == 0 then 400 else 4) == 0

This approach use the buit-in libraries to create the february 28th date and the adds a day to it, which if it's in a leap year the next day wil be the 29th.

 import std/time
 import std/time/date
 import std/time/time

 pub fun is-leap-year''(year: int)
   Date(year, 2, 28).time.add-days(1).day == 29

Kotlin

fun isLeapYear(year: Int) = year % 400 == 0 || (year % 100 != 0 && year % 4 == 0)

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'
^}

Output:

true
true
false
false
false
true

Lingo

on isLeapYear (year)
  return date(year, 2, 29).month=2
end

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

LLVM

; This is not strictly LLVM, as it uses the C library function "printf".
; LLVM does not provide a way to print values, so the alternative would be
; to just load the string into memory, and that would be boring.

$"EMPTY_STR" = comdat any
$"NOT_STR" = comdat any
$"IS_A_LEAP_YEAR" = comdat any

@main.test_case = private unnamed_addr constant [5 x i32] [i32 1900, i32 1994, i32 1996, i32 1997, i32 2000], align 16
@"EMPTY_STR" = linkonce_odr unnamed_addr constant [1 x i8] zeroinitializer, comdat, align 1
@"NOT_STR" = linkonce_odr unnamed_addr constant [5 x i8] c"not \00", comdat, align 1
@"IS_A_LEAP_YEAR" = linkonce_odr unnamed_addr constant [22 x i8] c"%d is %sa leap year.\0A\00", comdat, align 1

;--- The declaration for the external C printf function.
declare i32 @printf(i8*, ...)

; Function Attrs: noinline nounwind optnone uwtable
define i32 @is_leap_year(i32) #0 {
  %2 = alloca i32, align 4              ;-- allocate a local copy of year
  store i32 %0, i32* %2, align 4        ;-- store a copy of year

  %3 = load i32, i32* %2, align 4       ;-- load the year
  %4 = srem i32 %3, 4                   ;-- year % 4
  %5 = icmp ne i32 %4, 0                ;-- (year % 4) != 0
  br i1 %5, label %c1false, label %c1true

c1true:
  %6 = load i32, i32* %2, align 4       ;-- load the year
  %7 = srem i32 %6, 100                 ;-- year % 100
  %8 = icmp ne i32 %7, 0                ;-- (year % 100) != 0
  br i1 %8, label %c2true, label %c1false

c1false:
  %9 = load i32, i32* %2, align 4       ;-- load the year
  %10 = srem i32 %9, 400                ;-- year % 400
  %11 = icmp ne i32 %10, 0              ;-- (year % 400) != 0
  %12 = xor i1 %11, true
  br label %c2true

c2true:
  %13 = phi i1 [ true, %c1true ], [ %12, %c1false ]
  %14 = zext i1 %13 to i64
  %15 = select i1 %13, i32 1, i32 0
  ret i32 %15
}

; Function Attrs: noinline nounwind optnone uwtable
define i32 @main() #0 {
  %1 = alloca [5 x i32], align 16       ;-- allocate test_case
  %2 = alloca i32, align 4              ;-- allocate key
  %3 = alloca i32, align 4              ;-- allocate end
  %4 = alloca i32, align 4              ;-- allocate year
  %5 = bitcast [5 x i32]* %1 to i8*
  call void @llvm.memcpy.p0i8.p0i8.i64(i8* %5, i8* bitcast ([5 x i32]* @main.test_case to i8*), i64 20, i32 16, i1 false)
  store i32 0, i32* %2, align 4         ;-- store 0 in key
  store i32 5, i32* %3, align 4         ;-- store 5 in end
  br label %loop

loop:
  %6 = load i32, i32* %2, align 4       ;-- load key
  %7 = load i32, i32* %3, align 4       ;-- load end
  %8 = icmp slt i32 %6, %7              ;-- key < end
  br i1 %8, label %loop_body, label %exit

loop_body:
  %9 = load i32, i32* %2, align 4       ;-- load key
  %10 = sext i32 %9 to i64              ;-- sign extend key
  %11 = getelementptr inbounds [5 x i32], [5 x i32]* %1, i64 0, i64 %10
  %12 = load i32, i32* %11, align 4     ;-- load test_case[key]
  store i32 %12, i32* %4, align 4       ;-- store test_case[key] as year

  %13 = load i32, i32* %4, align 4      ;-- load year
  %14 = call i32 @is_leap_year(i32 %13) ;-- is_leap_year(year)
  %15 = icmp eq i32 %14, 1              ;-- is_leap_year(year) == 1
  %16 = zext i1 %15 to i64              ;-- zero extend
  %17 = select i1 %15, i8* getelementptr inbounds ([1 x i8], [1 x i8]* @"EMPTY_STR", i32 0, i32 0), i8* getelementptr inbounds ([5 x i8], [5 x i8]* @"NOT_STR", i32 0, i32 0)

  %18 = load i32, i32* %4, align 4      ;-- load year
  %19 = call i32 (i8*, ...) @printf(i8* getelementptr inbounds ([22 x i8], [22 x i8]* @"IS_A_LEAP_YEAR", i32 0, i32 0), i32 %18, i8* %17)

  %20 = load i32, i32* %2, align 4      ;-- load key
  %21 = add nsw i32 %20, 1              ;-- increment key
  store i32 %21, i32* %2, align 4       ;-- store key
  br label %loop

exit:
  ret i32 0
}

; Function Attrs: argmemonly nounwind
declare void @llvm.memcpy.p0i8.p0i8.i64(i8* nocapture writeonly, i8* nocapture readonly, i64, i32, i1) #1

attributes #0 = { noinline nounwind optnone uwtable "correctly-rounded-divide-sqrt-fp-math"="false" "disable-tail-calls"="false" "less-precise-fpmad"="false" "no-frame-pointer-elim"="false" "no-infs-fp-math"="false" "no-jump-tables"="false" "no-nans-fp-math"="false" "no-signed-zeros-fp-math"="false" "no-trapping-math"="false" "stack-protector-buffer-size"="8" "target-cpu"="x86-64" "target-features"="+fxsr,+mmx,+sse,+sse2,+x87" "unsafe-fp-math"="false" "use-soft-float"="false" }
attributes #1 = { argmemonly nounwind }

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.

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

Logtalk

leap_year(Year) :-
    (   mod(Year, 4) =:= 0, mod(Year, 100) =\= 0 ->
        true
    ;   mod(Year, 400) =:= 0
    ).

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

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

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

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:

Mathematica/Wolfram Language

Dates are handled by built-in functions in the Wolfram Language

LeapYearQ[2002]

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.

function TrueFalse = isLeapYear(year)
    TrueFalse = (eomday(year,2) == 29);
end

Using Logical and modular functions

x = ~mod(YEAR, 4) & (mod(YEAR, 100) | ~mod(YEAR, 400))

Maxima

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

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 ).

Usage:

:- 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).

min

Works with: min version 0.19.6

(mod 0 ==) :divisor?
(((400 divisor?) (4 divisor?) (100 divisor? not)) cleave and or) :leap-year?

MiniScript

isLeapYear = function(year)
  return year%4==0 and (year % 100 or not year % 400)
end function

MIPS Assembly

Pass year in a0, returns boolean in v0.

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

МК-61/52

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

Modula-2

MODULE LeapYear;
FROM FormatString IMPORT FormatString;
FROM Terminal IMPORT WriteString,ReadChar;

PROCEDURE IsLeapYear(year : INTEGER) : BOOLEAN;
BEGIN
    IF year MOD 100 = 0 THEN
        RETURN year MOD 400 = 0;
    END;
    RETURN year MOD 4 = 0
END IsLeapYear;

PROCEDURE Print(year : INTEGER);
VAR
    buf : ARRAY[0..63] OF CHAR;
    leap : BOOLEAN;
BEGIN
    leap := IsLeapYear(year);
    FormatString("Is %i a leap year? %b\n", buf, year, leap);
    WriteString(buf)
END Print;

BEGIN
    Print(1900);
    Print(1994);
    Print(1996);
    Print(1997);
    Print(2000);
    ReadChar
END LeapYear.

MUMPS

ILY(X) ;IS IT A LEAP YEAR?
 QUIT ((X#4=0)&(X#100'=0))!((X#100=0)&(X#400=0))

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

Nanoquery

Translation of: Python

def isLeapYear(year)
	if (year % 100 = 0)
		return (year % 400 = 0)
	else
		return (year % 4 = 0)
	end
end

Neko

Translating from C

/**
 <doc><h2>Leap year, in Neko</h2></doc>
**/

var leapyear = function(y) return ($not(y % 4) && $istrue(y % 100) || $not(y % 400))

var tests = $array(2000, 1997, 1996, 1994, 1990, 1980, 1900)
var cnt = $asize(tests)
while (cnt -= 1) >= 0 $print(tests[cnt], if leapyear(tests[cnt]) " is" else " is not", " a leapyear", "\n")

Output:

prompt$ nekoc leapyear.neko
prompt$ neko leapyear.n
1900 is not a leapyear
1980 is a leapyear
1990 is not a leapyear
1994 is not a leapyear
1996 is a leapyear
1997 is not a leapyear
2000 is a leapyear

Nemerle

Demonstrating implementation as well as use of standard library function.

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));
    }
}

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.

/* 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

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

import times
let year = 1980
echo isLeapYear(year)

# or

proc isLeapYear2(year: Natural): bool =
  if year mod 100 == 0:
    year mod 400 == 0
  else: year mod 4 == 0

echo isLeapYear2(year)

Output:

true
true

Oberon-2

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;

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;
    }
  }
}

OCaml

let is_leap_year ~year =
  year mod (if year mod 100 = 0 then 400 else 4) = 0

Using Unix Time functions:

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)

Oforth

Date.IsLeapYear(2000)

ooRexx

::routine isLeapYear
  use arg year 
  d = .datetime~new(year, 1, 1) 
  return d~isLeapYear

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.

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.

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

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

isLeap(n)={
  if(n%400==0, return(1));
  if(n%100==0, return(0));
  n%4==0
};

Alternate version:

isLeap(n)=!(n%if(n%100,4,400))

Works with: PARI/GP version 2.6.0 and above

isLeap(n)={
  if(n%4,0,
    n%100,1,
      n%400,0,1
  )
};

Pascal

Works with: Free 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.

Output:

1900 is NO leap year
2000 is a leap year
2100 is NO leap year
1904 is a leap year

Perl

sub isleap {
    my $year = shift;
    if ($year % 100 == 0) {
        return ($year % 400 == 0);
    }
    return ($year % 4 == 0);
}

Or more concisely:

sub isleap { not $_[0] % ($_[0] % 100 ? 4 : 400) }

Alternatively, using functions/methods from CPAN modules:

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();

Phix

Available as an auto-include, implemented as:

global function is_leap_year(integer y)
    return remainder(y,4)=0 and (remainder(y,100)!=0 or remainder(y,400)=0)
end function

PHP

<?php
function isLeapYear($year) {
    if ($year % 100 == 0) {
        return ($year % 400 == 0);
    }
    return ($year % 4 == 0);
}

With date('L'):

<?php
function isLeapYear($year) {
    return (date('L', mktime(0, 0, 0, 2, 1, $year)) === '1')
}

Picat

go =>
  foreach(Y in [1600,1700,1899,1900,2000,2006,2012])
     println(Y=cond(leap_year(Y),leap_year,not_leap_year))
  end,
  nl.

leap_year(Year) => 
  (Year mod 4 == 0, Year mod 100 != 0) 
  ; 
  Year mod 400 == 0.

Output:

1600 = leap_year
1700 = not_leap_year
1899 = not_leap_year
1900 = not_leap_year
2000 = leap_year
2006 = not_leap_year
2012 = leap_year

PicoLisp

(de isLeapYear (Y)
   (bool (date Y 2 29)) )

Output:

: (isLeapYear 2010)
-> NIL

: (isLeapYear 2008)
-> T

: (isLeapYear 1600)
-> T

: (isLeapYear 1700)
-> NIL

PL/0

Translation of: Tiny BASIC

var isleap, year;

procedure checkifleap;
begin
  isleap := 0;
  if (year / 4) * 4 = year then
  begin
    if year - (year / 100) * 100 <> 0 then isleap := 1;
    if year - (year / 400) * 400 = 0 then isleap := 1
  end;
end;

begin
  year := 1759;
  while year <= 2022 do
  begin
    call checkifleap;
    if isleap = 1 then ! year;
    year := year + 1
  end
end.

Output:

    1760
    1764
    1768
    1772
    1776
    1780
    1784
    1788
    1792
    1796
    1804
    1808
    1812
    1816
    1820
    1824
    1828
    1832
    1836
    1840
    1844
    1848
    1852
    1856
    1860
    1864
    1868
    1872
    1876
    1880
    1884
    1888
    1892
    1896
    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
    2016
    2020

PL/I

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;

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 

PL/M

100H: /* DETERMINE WHETHER SOME YEARS ARE LEAP YEARS OR NOT */

   /* CP/M BDOS SYSTEM CALL */
   BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5;END;
   /* CONSOLE OUTPUT ROUTINES */
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C ); END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
   PR$NL:     PROCEDURE; CALL PR$STRING( .( 0DH, 0AH, '$' ) );      END;
   PR$NUMBER: PROCEDURE( N );
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR( 6 ) BYTE INITIAL( '.....$' ), W BYTE;
      N$STR( W := LAST( N$STR ) - 1 ) = '0' + ( ( V := N ) MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;

   /* TASK */
   /* RETURNS TRUE IF YEAR IS A LEAP YEAR, FALSE OTHERWISE  */
   /*         ASSUMES YEAR IS IN THE GREGORIAN CALENDAR     */
   IS$LEAP$YEAR: PROCEDURE( YEAR )BYTE;
      DECLARE YEAR ADDRESS;
      RETURN (  YEAR MOD 400 = 0
             OR ( YEAR MOD 4 = 0 AND YEAR MOD 100 <> 0 )
             );
   END IS$LEAPYEAR ;
   /* TEST CASES */
   DECLARE TEST$YEAR ( 15 )ADDRESS INITIAL( 1899, 1900, 1901, 1902, 1903
                                          , 1904, 1905, 1999, 2000, 2001
                                          , 2002, 2003, 2004, 2021, 2022
                                          );
   DECLARE Y$POS BYTE;
   DO Y$POS = 0 TO LAST( TEST$YEAR );
      CALL PR$NUMBER( TEST$YEAR( Y$POS ) );
      CALL PR$STRING( .' IS $' );
      IF NOT IS$LEAP$YEAR( TEST$YEAR( Y$POS ) ) THEN DO;
         CALL PR$STRING( .'NOT $' );
      END;
      CALL PR$STRING( .'A LEAP YEAR$' );
      CALL PR$NL;
   END;

EOF

Output:

1899 IS NOT A LEAP YEAR
1900 IS NOT A LEAP YEAR
1901 IS NOT A LEAP YEAR
1902 IS NOT A LEAP YEAR
1903 IS NOT A LEAP YEAR
1904 IS A LEAP YEAR
1905 IS NOT A LEAP YEAR
1999 IS NOT A LEAP YEAR
2000 IS A LEAP YEAR
2001 IS NOT A LEAP YEAR
2002 IS NOT A LEAP YEAR
2003 IS NOT A LEAP YEAR
2004 IS A LEAP YEAR
2021 IS NOT A LEAP YEAR
2022 IS NOT A LEAP YEAR

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

PowerShell

$Year = 2016
[System.DateTime]::IsLeapYear( $Year )

Prolog

Works with: SWI-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).

Output:

 ?- leap_year([1900,1994,1996,1997,2000 ]).
leap years : [1996,2000]
not leap years : [1900,1994,1997]
L = [1900,1994,1996,1997,2000].

There is an handy builtin that simplifies a lot, ending up in a simple query:

?- 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].

Python

import calendar
calendar.isleap(year)

or

def is_leap_year(year):
    return not year % (4 if year % 100 else 400)

Asking for forgiveness instead of permission:

import datetime

def is_leap_year(year):
    try:
        datetime.date(year, 2, 29)
    except ValueError:
        return False
    return True

Q

ly:{((0<>x mod 100) | 0=x mod 400) & 0=x mod 4}    / Return 1b if x is a leap year; 0b otherwise

Quackery

Translation of: Forth

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

R

isLeapYear <- function(year) {
    ifelse(year%%100==0, year%%400==0, year%%4==0)
}

for (y in c(1900, 1994, 1996, 1997, 2000)) {
  cat(y, ifelse(isLeapYear(y), "is", "isn't"), "a leap year.\n")
}

Output:

1900 isn't a leap year.
1994 isn't a leap year.
1996 is a leap year.
1997 isn't a leap year.
2000 is a leap year.

Racket

(define (leap-year? y)
  (and (zero? (modulo y 4)) (or (positive? (modulo y 100)) (zero? (modulo y 400)))))

Raku

(formerly Perl 6)

Works with: Rakudo version 2010.07

say "$year is a {Date.is-leap-year($year) ?? 'leap' !! 'common'} year."

In Rakudo 2010.07, Date.is-leap-year is implemented as

multi method is-leap-year($y = $!year) {
    $y %% 4 and not $y %% 100 or $y %% 400
}

Rapira

fun is_leap_year(year)
  if (year /% 100) = 0 then
    return (year /% 400) = 0
  fi
  return (year /% 4) = 0
end

Raven

define is_leap_year use $year
    $year 100 % 0 = if
        $year 400 % 0 =
    $year 4 % 0 =

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]
]

Retro

:isLeapYear? (y-f)
    dup #400 mod n:zero? [ drop #-1 #0 ] [ #1 ] choose 0; drop
    dup #100 mod n:zero? [ drop  #0 #0 ] [ #1 ] choose 0; drop
    #4 mod n:zero? ;

REXX

local variables

leapyear:  procedure;    parse arg yr
return  yr//400==0  |  (yr//100\==0  &  yr//4==0)

with short-circuit

The REXX language doesn't support short-circuits, so here is a version that does a short-circuit.

leapyear:  procedure;   parse arg yr
if yr//4\==0  then return 0                 /*Not ÷ by 4?    Not a leap year.*/
return  yr//400==0  |  yr//100\==0

no local variables

This version doesn't need a PROCEDURE to hide local variable(s)   [because there aren't any local variables],
but it does invoke the   ARG   BIF multiple times.

leapyear: if arg(1)//4\==0  then return 0
          return arg(1)//400==0  |  arg(1)//100\==0

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).

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. */

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

RPG

Works with: RPGIII

     C*0N01N02N03Factor1+++OpcdeFactor2+++ResultLenDHHiLoEqComments+++++++
     C           *ENTRY    PLIST
     C                     PARM           YEAR    40       input (year)
     C                     PARM           ISLEAP  1        output (Y/N)
     C*
     C                     MOVE 'N'       ISLEAP
     C           YEAR      CABLE1752      DONE             not Gregorian
     C*
     C           YEAR      DIV  4         RESULT  40
     C                     MVR            REMAIN  40
     C           REMAIN    CABNE0         DONE
     C*
     C* If we got here, year is divisible by 4.
     C           YEAR      DIV  100       RESULT
     C                     MVR            REMAIN
     C           REMAIN    CABNE0         LEAPYR
     C*
     C* If we got here, year is divisible by 100.
     C           YEAR      DIV  400       RESULT
     C                     MVR            REMAIN
     C           REMAIN    CABNE0         DONE
     C*
     C           LEAPYR    TAG
     C                     MOVE 'Y'       ISLEAP
     C*
     C           DONE      TAG
     C                     SETON                     LR

RPL

Works with: Halcyon Calc version 4.2.7

≪ DUP 100 MOD 4 400 IFTE MOD NOT
≫ 'LEAP?' STO

2000 LEAP?
2001 LEAP?
2020 LEAP?
2100 LEAP?

Output:

4: 1
3: 0
2: 1
1: 0

Ruby

require 'date'

Date.leap?(year)

The leap? method is aliased as gregorian_leap? And yes, there is a julian_leap? method.

Rust

fn is_leap(year: i32) -> bool {
    let factor = |x| year % x == 0;
    factor(4) && (!factor(100) || factor(400))
}

Scala

JDK 7 (not recommended)

By default, java.util.GregorianCalendar switches from Julian calendar to Gregorian calendar at 15 October 1582.

//use Java's calendar class
new java.util.GregorianCalendar().isLeapYear(year)

JDK 8

Using JSR-310 java.time.

java.time.LocalDate.ofYearDay(year, 1).isLeapYear()

Implementation

For proleptic Gregorian calendar:

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)
}

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))))

sed

h
s/00$//
/[02468][048]$/!{
/[13579][26]$/!d
}
g
s/$/ is a leap year/

Test:

$ seq 1900 2100 | sed -f leap.sed
1904 is a leap year
1908 is a leap year
1912 is a leap year
...

Seed7

This function is part of the "time.s7i" library. It returns TRUE if the year is a leap year in the Gregorian calendar.

const func boolean: isLeapYear (in integer: year) is
  return (year rem 4 = 0 and year rem 100 <> 0) or year rem 400 = 0;

Original source: [1]

Sidef

func isleap(year) {
    if (year %% 100) {
        return (year %% 400);
    }
    return (year %% 4);
}

or a little bit simpler:

func isleap(year) { year %% 100 ? (year %% 400) : (year %% 4) };

Smalltalk

Smalltalk has a built-in method named isLeapYear:

Date today isLeapYear.

SNOBOL4

Predicate leap( ) succeeds/fails, returns nil.

        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

Output:

0: 1066
1: 1492
0: 1900
1: 2000

Standard ML

fun isLeapYear y =
  y mod (if y mod 100 = 0 then 400 else 4) = 0

Stata

Given a dataset with a "year" variable, generate a variable "leap" which is 1 for a leap year, 0 otherwise.

gen leap = mod(year,400)==0 | mod(year,4)==0 & mod(year,100)!=0

See also the article How do I identify leap years in Stata? by Nicholas J. Cox in Stata FAQ.

Swift

func isLeapYear(year: Int) -> Bool {
    return year.isMultiple(of: 100) ? year.isMultiple(of: 400) : year.isMultiple(of: 4)
}

[1900, 1994, 1996, 1997, 2000].forEach { year in
    print("\(year): \(isLeapYear(year: year) ? "YES" : "NO")")
}

Output:

1900: NO
1994: NO
1996: YES
1997: NO
2000: YES

Tcl

The "classic" modulo comparison:

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

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.)

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

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

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 

UNIX Shell

POSIX compatible:

is_leap() {
  return $((${1%00} & 3))
}

Original Bourne:

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;
}

Using GNU date(1):

leap() {
  date -d "$1-02-29" >/dev/null 2>&1;
}

Defining a bash function is_leap which accepts a YEAR argument (defaulting to zero), and uses no IO redirection, nor any extra processes.

is_leap() (( year=${1-0}, year % 4 == 0 && ( year % 100 != 0 || year % 400 == 0 )))

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)

leap() {
  cal 02 $1 | grep -q 29
}

Ursa

This program takes a year as a command line argument.

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

Output in Bash:

$ ursa leapyear.u 1900
1900 is not a leap year
$ ursa leapyear.u 2000
2000 is a leap year

Vala

void main() {
  DateYear[] years = { 1900, 1994, 1996, 1997, 2000 };
  foreach ( DateYear year in years ) {
    string status = year.is_leap_year() ? "" : "not ";
    print (@"$year is $(status)a leap year.\n");
  }
}

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.

Vedit macro language

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")
    }
}

The following version requires Vedit 6.10 or later:

while (#1 = Get_Num("Year: ")) {
    if (Is_Leap_Year(#1)) {
        Message(" is leap year\n")
    } else {
	Message(" is not leap year\n")
    }
}

V (Vlang)

fn is_leap(year int) bool {
    return year %400 ==0 || (year%4 ==0 && year%100!=0)
}
 
fn main() {
    for y in 1950..2012 {
        if is_leap(y) {
            println(y)
        }
    }
}

Returns:

1952
1956
1960
1964
1968
1972
1976
1980
1984
1988
1992
1996
2000
2004
2008

WDTE

let str => import 'strings';

let multiple of n => == (% n of) 0;

let leap year => str.format '{} is{} a leap year.' year (switch year {
  multiple 400 => '';
  multiple 100 => ' not';
  multiple 4 => '';
  default => ' not';
}) -- io.writeln io.stdout;

WebAssembly

First, with syntactic sugar that allows us to put opcode arguments after the opcode itself:

(module
  ;; function isLeapYear: returns 1 if its argument (e.g. 2004) is a leap year, 0 otherwise.
  ;; Returns year%4==0 and (year%100!=0 or year%400==0)
  (func $isLeapYear (param $year i32) (result i32)
    (i32.and
      (i32.eqz (i32.rem_u (get_local $year) (i32.const 4)))  ;; year%4 == 0
      (i32.or
        (i32.ne (i32.rem_u (get_local $year) (i32.const 100)) (i32.const 0))   ;; year%100 != 0
        (i32.eqz (i32.rem_u (get_local $year) (i32.const 400)))  ;; yaer%400 == 0
      )
    )
  )
  (export "isLeapYear" (func $isLeapYear))
)

And then the same code, without the syntactic sugar:

(module
  ;; function isLeapYear: returns 1 if its argument (e.g. 2004) is a leap year, 0 otherwise.
  ;; Returns year%4==0 and (year%100!=0 or year%400==0)
  (func $isLeapYear (param $year i32) (result i32)
    get_local $year
    i32.const 4
    i32.rem_u
    i32.eqz           ;; year % 4 == 0
    get_local $year
    i32.const 100
    i32.rem_u
    i32.const 0
    i32.ne            ;; year % 100 != 0
    get_local $year
    i32.const 400
    i32.rem_u
    i32.eqz           ;; year % 400 == 0
    i32.or
    i32.and
  )
  (export "isLeapYear" (func $isLeapYear))
)

Wortel

@let {
  isLeapYear !?{\~%%1H \~%%4H \~%%4}
  !-isLeapYear @range[1900 2000]
}

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]

Wren

var isLeapYear = Fn.new { |y|
    return ((y % 4 == 0) && (y % 100!= 0)) || (y % 400 == 0)
}

System.print("Leap years between 1900 and 2020 inclusive:")
var c = 0
for (i in 1900..2020) {
    if (isLeapYear.call(i)) {
        System.write("%(i) ")
        c = c + 1
        if (c % 15 == 0) System.print()
    }
}

Output:

Leap years between 1900 and 2020 inclusive:
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 2016 2020 

X86 Assembly

Using FASM syntax. Leaf function fits nicely into your program.

    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

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)))

Output:

(#T #T () #T () () #T #T () #T)

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;
];

YAMLScript

!yamlscript/v0

defn main(year):
  say: "$year is $when-not(leap-year(year) 'not ')a leap year."

# Either one works:

defn leap-year(year):
  ((year % 4) == 0) && (((year % 100) > 0) || ((year % 100) == 0))

defn leap-year(year):
  and:
    zero?: (year % 4)
    or:
      pos?: (year % 100)
      zero?: (year % 400)

Yorick

This solution is vectorized and can be applied to scalar or array input.

func is_leap(y) {
  return ((y % 4 == 0) & (y % 100 != 0)) | (y % 400 == 0);
}

Interactive example usage:

> is_leap(1988)
1
> is_leap([1988,1989,1900,2000])
[1,0,0,1]

Zig

pub fn isLeapYear(year: anytype) bool {
  const inttype = @TypeOf(year);
  if (@typeInfo(inttype) != .Int) {
    @compileError("non-integer type used on leap year: " ++ @typeName(inttype));
  }
  return (if (@mod(year, @as(inttype, 100)) == 0)
    @mod(year, @as(inttype, 400)) == 0
  else
    @mod(year, @as(inttype, 4)) == 0);
}

Alternative (inspired by the C solution):

/// The type that holds the current year, i.e. 2016
pub const Year = u16;

/// Returns true for years with 366 days
/// and false for years with 365 days.
pub fn isLeapYear(year: Year) bool {
    // In the western Gregorian Calendar leap a year is
    // a multiple of 4, excluding multiples of 100, and
    // adding multiples of 400. In code:
    //
    // if (@mod(year, 4) != 0)
    //     return false;
    // if (@mod(year, 100) != 0)
    //     return true;
    // return (0 == @mod(year, 400));

    // The following is equivalent to the above
    // but uses bitwise operations when testing
    // for divisibility, masking with 3 as test
    // for multiples of 4 and with 15 as a test
    // for multiples of 16. Multiples of 16 and
    // 100 are, conveniently, multiples of 400.
    const mask: Year = switch (year % 100) {
        0 => 0b1111,
        else => 0b11,
    };
    return 0 == year & mask;
}

test "isLeapYear" {
    try testing.expectEqual(false, isLeapYear(2095));
    try testing.expectEqual(true, isLeapYear(2096));
    try testing.expectEqual(false, isLeapYear(2100));
    try testing.expectEqual(true, isLeapYear(2400));
}

zkl

Time.Date.isLeapYear(1988) //-->True
T(1988,1989,1900,2000).apply(Time.Date.isLeapYear) 
    //-->L(True,False,False,True)