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
11l
F is_leap_year(year)
I year % 100 == 0
R year % 400 == 0
R year % 4 == 0360 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
ENDSample 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
rts8080 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
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
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
# 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_
endBaCon
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
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
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
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.
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)=0Liberty BASIC
Simple method
if leap(1996)then
print "leap"
else
print "ordinary"
end if
wait
function leap(n)
leap=date$("2/29/";n)
end functionCalculated 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 ifNS-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
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
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
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
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
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
endZX 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")
$)- Output:
1899 is not a leap year. 1900 is not a leap year. 1901 is not a leap year. 1902 is not a leap year. 1903 is not a leap year. 1904 is a leap year. 1905 is not a leap year. 1999 is not a leap year. 2000 is a leap year. 2001 is not a leap year. 2002 is not a leap year. 2003 is not a leap year. 2004 is a leap year. 2021 is not a leap year. 2022 is not a leap year.
Befunge
0"2("*:3-:1-:2-:"^"-v<
v*%"d"\!%4::,,"is".:<|
>\45*:*%!+#v_ "ton"vv<
v"ear."+550<,,,,*84<$#
>"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:> $FALSE
Crystal
p Time.leap_year?(2020)
p Time.leap_year?(2021)
p Time.leap_year?(2022)
true false false
D
import std.algorithm;
bool leapYear(in uint y) pure nothrow {
return (y % 4) == 0 && (y % 100 || (y % 400) == 0);
}
void main() {
auto good = [1600, 1660, 1724, 1788, 1848, 1912, 1972, 2032,
2092, 2156, 2220, 2280, 2344, 2348];
auto bad = [1698, 1699, 1700, 1750, 1800, 1810, 1900, 1901,
1973, 2100, 2107, 2200, 2203, 2289];
assert(filter!leapYear(bad ~ good).equal(good));
}
Using the datetime library:
import std.datetime;
void main() {
assert(yearIsLeapYear(1724));
assert(!yearIsLeapYear(1973));
assert(!Date(1900, 1, 1).isLeapYear);
assert(DateTime(2000, 1, 1).isLeapYear);
}
Dart
class Leap {
bool leapYear(num year) {
return (year % 400 == 0) || (( year % 100 != 0) && (year % 4 == 0));
bool isLeapYear(int year) =>
(year % 4 == 0) && ((year % 100 != 0) || (year % 400 == 0));
// Source: https://api.flutter.dev/flutter/quiver.time/isLeapYear.html
}
}
Dc
Directly taken from Wikipedia.
[0q]s0
[1q]s1
[ S. [ l. 4% 0!=0 ## if y % 4: return 0
l. 100% 0!=1 ## if y % 100: return 1
l. 400% 0!=0 ## if y % 400: return 0
1 ## return 1
]x s.L.
]sL ## L = isleapYear()
[ Sy
lyn [ is ]P
ly lLx
[not ] 0:y
[ ] 1:y
;yP
[a leap year]P AP
OsyLyo
]sT ## T = testYear()
1988 lTx
1989 lTx
1900 lTx
2000 lTx- Output:
1988 is a leap year 1989 is not a leap year 1900 is not a leap year 2000 is a leap year
Delphi /Pascal
Delphi has standard function IsLeapYear in SysUtils unit.
program TestLeapYear;
{$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.
DuckDB
create or replace function is_leap_year(y) as
(y%4) == 0 and (y < 1582 or (y%400) == 0 or (y%100) != 0);
Examples In the following transcript, "D " signifies the DuckDB prompt:
D .header off
D .mode list
# Not leap years:
D select histogram( is_leap_year(y) )
from (select unnest([2400, 2012, 2000, 1600, 1500, 1400]) as y);
{true=6}
# Not leap years:
D select histogram( not is_leap_year(y) )
from (select unnest([2100, 2014, 1900, 1800, 1700, 1499]) as y);
{true=6}
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 == 0Elixir
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
(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 PROGRAMEuphoria
function isLeapYear(integer year)
return remainder(year,4)=0 and remainder(year,100)!=0 or remainder(year,400)=0
end functionExcel
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)
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:
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
HandleEventsOutput (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.
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
}
Golfscript
~.4%!\.100%!!\400%!|&
- Output — for input
1500:
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
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.
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
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
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
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
endLiveCode
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 trueLLVM
; 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]
endLogtalk
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)
endMaple
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);
endUsing 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
(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 functionMIPS 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 $raМК-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/ПModula-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
def isLeapYear(year)
if (year % 100 = 0)
return (year % 400 = 0)
else
return (year % 4 = 0)
end
endNeko
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;Oberon-07
First, a minimal Dates module which exports an isLeapYear method.
MODULE Dates;
(* returns TRUE if year is a leap year, FALSE otherwise.
assumes year is in the Gregorian Calendar
*)
PROCEDURE isLeapYear* ( year : INTEGER ) : BOOLEAN ;
BEGIN
RETURN ( year MOD 400 = 0 )
OR ( ( year MOD 4 = 0 ) & ( year MOD 100 # 0 ) )
END isLeapYear ;
END Dates.Use the Dates module to test for leap years.
MODULE LeapYear;
IMPORT Dates, Out;
PROCEDURE test ( year : INTEGER ) ;
BEGIN
Out.Int( year, 4 );
Out.String( " is " );
IF ~ Dates.isLeapYear( year ) THEN Out.String( "not " ) END;
Out.String( "a leap year" );
Out.Ln
END test ;
BEGIN
test( 1899 ); test( 1900 ); test( 1901 ); test( 1902 ); test( 1903 );
test( 1904 ); test( 1905 ); test( 1999 ); test( 2000 ); test( 2001 );
test( 2002 ); test( 2003 ); test( 2004 )
END LeapYear.- 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
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) = 0Using 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~isLeapYearOpenEdge/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))isLeap(n)={
if(n%4,0,
n%100,1,
n%400,0,1
)
};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
PascalABC.NET
##
function IsLeapYear(year: word): boolean :=
if year mod 100 = 0 then year mod 400 = 0
else year mod 4 = 0;
IsLeapYear(2000).println;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, $year)) === '1')
}Picat
go =>
foreach(Y in [1600,1700,1899,1900,2000,2006,2012])
println(Y=cond(leap_year(Y),leap_year,not_leap_year))
end,
nl.
leap_year(Year) =>
(Year mod 4 == 0, Year mod 100 != 0)
;
Year mod 400 == 0.- Output:
1600 = leap_year 1700 = not_leap_year 1899 = not_leap_year 1900 = not_leap_year 2000 = leap_year 2006 = not_leap_year 2012 = leap_year
PicoLisp
(de isLeapYear (Y)
(bool (date Y 2 29)) )- Output:
: (isLeapYear 2010) -> NIL : (isLeapYear 2008) -> T : (isLeapYear 1600) -> T : (isLeapYear 1700) -> NIL
PL/0
var isleap, year;
procedure checkifleap;
begin
isleap := 0;
if (year / 4) * 4 = year then
begin
if year - (year / 100) * 100 <> 0 then isleap := 1;
if year - (year / 400) * 400 = 0 then isleap := 1
end;
end;
begin
year := 1759;
while year <= 2022 do
begin
call checkifleap;
if isleap = 1 then ! year;
year := year + 1
end
end.- Output:
1760
1764
1768
1772
1776
1780
1784
1788
1792
1796
1804
1808
1812
1816
1820
1824
1828
1832
1836
1840
1844
1848
1852
1856
1860
1864
1868
1872
1876
1880
1884
1888
1892
1896
1904
1908
1912
1916
1920
1924
1928
1932
1936
1940
1944
1948
1952
1956
1960
1964
1968
1972
1976
1980
1984
1988
1992
1996
2000
2004
2008
2012
2016
2020
PL/I
dcl mod builtin;
dcl year fixed bin (31);
do year = 1900, 1996 to 2001;
if mod(year, 4) = 0 &
(mod(year, 100) ^= 0 |
mod(year, 400) = 0) then
put skip edit(year, 'is a leap year') (p'9999b', a);
else
put skip edit(year, 'is not a leap year') (p'9999b', a);
end;- Output:
1900 is not a leap year 1996 is a leap year 1997 is not a leap year 1998 is not a leap year 1999 is not a leap year 2000 is a leap year 2001 is not a leap year
PL/M
100H: /* DETERMINE WHETHER SOME YEARS ARE LEAP YEARS OR NOT */
/* CP/M BDOS SYSTEM CALL */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5;END;
/* CONSOLE OUTPUT ROUTINES */
PR$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
} defPowerShell
$Year = 2016
[System.DateTime]::IsLeapYear( $Year )Prolog
leap_year(L) :-
partition(is_leap_year, L, LIn, LOut),
format('leap years : ~w~n', [LIn]),
format('not leap years : ~w~n', [LOut]).
is_leap_year(Year) :-
R4 is Year mod 4,
R100 is Year mod 100,
R400 is Year mod 400,
( (R4 = 0, R100 \= 0); R400 = 0).- Output:
?- leap_year([1900,1994,1996,1997,2000 ]).
leap years : [1996,2000]
not leap years : [1900,1994,1997]
L = [1900,1994,1996,1997,2000].There is an handy builtin that simplifies a lot, ending up in a simple query:
?- findall(Y, (between(1990,2030,Y),day_of_the_year(date(Y,12,31),366)), L).
L = [1992, 1996, 2000, 2004, 2008, 2012, 2016, 2020, 2024, 2028].Python
import calendar
calendar.isleap(year)or
def is_leap_year(year):
return not year % (4 if year % 100 else 400)Asking for forgiveness instead of permission:
import datetime
def is_leap_year(year):
try:
datetime.date(year, 2, 29)
except ValueError:
return False
return TrueQ
ly:{((0<>x mod 100) | 0=x mod 400) & 0=x mod 4} / Return 1b if x is a leap year; 0b otherwiseQuackery
[ dup 400 mod 0 = iff [ drop true ] done
dup 100 mod 0 = iff [ drop false ] done
4 mod 0 = ] is leap? ( n --> b )R
isLeapYear <- function(year) {
ifelse(year%%100==0, year%%400==0, year%%4==0)
}
for (y in c(1900, 1994, 1996, 1997, 2000)) {
cat(y, ifelse(isLeapYear(y), "is", "isn't"), "a leap year.\n")
}- Output:
1900 isn't a leap year. 1994 isn't a leap year. 1996 is a leap year. 1997 isn't a leap year. 2000 is a leap year.
Racket
(define (leap-year? y)
(and (zero? (modulo y 4)) (or (positive? (modulo y 100)) (zero? (modulo y 400)))))Raku
(formerly Perl 6)
say "$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
endRaven
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]
]Red
leapyear?: function [
"returns true for a leap year."
date [date!]
][
year: date/year
if zero? year // 400 and year / 100 [return true]
if zero? year // 4 [return true]
false
]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\==0no 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\==0handles 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 okRPG
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 LRRPL
≪ 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) = 0Stata
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)!=0See 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) || ($year % 400 == 0))}]
}
isleap1 1988 ;# => 1
isleap1 1989 ;# => 0
isleap1 1900 ;# => 0
isleap1 2000 ;# => 1Does Feb 29 exist in the given year? If not a leap year, the clock command will return "03-01". (This code will switch to the Julian calendar for years before 1582.)
proc isleap2 year {
return [expr {[clock format [clock scan "$year-02-29" -format "%Y-%m-%d"] -format "%m-%d"] eq "02-29"}]
}
isleap2 1988 ;# => 1
isleap2 1989 ;# => 0
isleap2 1900 ;# => 0
isleap2 2000 ;# => 1TUSCRIPT
$$ 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
Uiua
Leap ← ◿2/+=0◿100_400_4UNIX 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 ifOutput 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 $isLeapYear))
)Wortel
@let {
isLeapYear !?{\~%%1H \~%%4H \~%%4}
!-isLeapYear @range[1900 2000]
}Returns:
[1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000]
Wren
var isLeapYear = Fn.new { |y|
return ((y % 4 == 0) && (y % 100!= 0)) || (y % 400 == 0)
}
System.print("Leap years between 1900 and 2020 inclusive:")
var c = 0
for (i in 1900..2020) {
if (isLeapYear.call(i)) {
System.write("%(i) ")
c = c + 1
if (c % 15 == 0) System.print()
}
}- Output:
Leap years between 1900 and 2020 inclusive: 1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1976 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 2020
X86 Assembly
Using FASM syntax. Leaf function fits nicely into your program.
align 16
; Input year as signed dword in EAX
IsLeapYear:
test eax,11b
jz .4
retn ; 75% : ZF=0, not a leap year
.4:
mov ecx,100
cdq
idiv ecx
test edx,edx
jz .100
cmp edx,edx
retn ; 24% : ZF=1, leap year
.100:
test eax,11b
retn ; 1% : ZF=?, leap year if EAX%400=0XLISP
(DEFUN LEAP-YEARP (YEAR)
(AND (= (MOD YEAR 4) 0) (OR (/= (MOD YEAR 100) 0) (= (MOD YEAR 400) 0))))
; Test the function
(DISPLAY (MAPCAR LEAP-YEARP '(1600 1640 1800 1928 1979 1990 2000 2004 2005 2016)))- Output:
(#T #T () #T () () #T #T () #T)
XPL0
func LeapYear(Y); \Return 'true' if Y is a leap year
int Y;
[if rem(Y/100)=0 then return rem(Y/400)=0;
return rem(Y/4)=0;
];YAMLScript
!YS-v0
defn main(year=2024):
say: "$year is $when-not(
leap-year(year) 'not ')a
leap year."
defn leap-year(year):
(year % 4) == 0 &&:
(year % 100):pos? ||:
(year % 400).eq(0)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)Categories:
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- Date and time
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