Leap year
You are encouraged to solve this task according to the task description, using any language you may know.
Determine whether a given year is a leap year.
See Also
ActionScript
<lang actionscript>public function isLeapYear(year:int):Boolean {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
ALGOL 68
<lang algol68>MODE YEAR = INT, MONTH = INT, DAY = INT;
PROC year days = (YEAR year)DAY: # Ignore 1752 CE for the moment #
( month days(year, 2) = 28 | 365 | 366 );
PROC month days = (YEAR year, MONTH month) DAY:
( month | 31,
28 + ABS(year MOD 4 NE 0 OR year MOD 400 EQ 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, 1996, 1997, 2000);
FOR i TO UPB test cases DO
YEAR year = test cases[i];
printf(($g(0)" is "b("","not ")"a leap year."l$, year, is leap year(year)))
OD
)</lang> Output:
1900 is not a leap year. 1996 is not a leap year. 1997 is a leap year. 2000 is a leap year.
Haskell
Simple version <lang haskell> import Data.List import Control.Monad import Control.Arrow
leaptext x b | b = show x ++ " is a leap year" | otherwise = show x ++ " is not a leap year"
isleapsf j | 0==j`mod`100 = 0 == j`mod`400 | otherwise = 0 == j`mod`4 </lang> Algorithmic <lang haskell> isleap = foldl1 ((&&).not).flip map [400, 100, 4]. ((0==).).mod </lang> Example using isleap <lang haskell>
- Main> mapM_ (putStrLn. (ap leaptext isleap)) [1900,1994,1996,1997,2000]
1900 is not a leap year 1994 is not a leap year 1996 is a leap year 1997 is not a leap year 2000 is a leap year </lang>
Icon and Unicon
Gives leap year status for 2000,1900,2012 and any arguments you give <lang Icon>procedure main(arglist) every y := !([2000,1900,2012]|||arglist) do
write("The year ",y," is ", leapyear(y) | "not ","a leap year.")
end
procedure leapyear(year) #: determine if year is leap
if (numeric(year) % 4 = 0 & year % 100 ~= 0) | (numeric(year) % 400 = 0) then return
end </lang>
Icon
Unicon
The Icon solution works in Unicon.
J
<lang j>isLeap=: 0 -/@:= 4 100 400 |/ ]</lang>
Example use:
<lang> isLeap 1900 1996 1997 2000 0 1 0 1</lang>
Java
<lang java>new java.util.GregorianCalendar().isLeapYear(year)</lang>
or <lang java>public static boolean isLeapYear(int year){
if(year % 100 == 0) return year % 400 == 0; return year % 4 == 0;
}</lang>
JavaScript
<lang javascript>function isLeapYear(year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
Perl
<lang Perl>sub leap {
my $yr = $_[0];
if ($yr % 100 == 0) {
return ($yr % 400 == 0);
}
return ($yr % 4 == 0)
}</lang>
Or more concisely:
<lang Perl>sub leap {!($_[0] % 100) ? !($_[0] % 400) : !($_[0] % 4)}</lang>
PHP
<lang php><?php function isLeapYear(year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}</lang>
With date('L'):
<lang php><?php function isLeapYear(year) {
return (date('L', mktime(0, 0, 0, 2, 1, year)) === '1')
}</lang>
PicoLisp
<lang PicoLisp>(de isLeapYear (Y)
(bool (date Y 2 29)) )</lang>
Output:
: (isLeapYear 2010) -> NIL : (isLeapYear 2008) -> T : (isLeapYear 1600) -> T : (isLeapYear 1700) -> NIL
PureBasic
<lang PureBasic>Procedure isLeapYear(Year)
If (Year%4=0 And Year%100) Or Year%400=0 ProcedureReturn #True Else ProcedureReturn #False EndIf
EndProcedure</lang>
Python
<lang python>import calendar calendar.isleap(year)</lang>
or <lang python>def is_leap_year(year):
if year % 100 == 0:
return year % 400 == 0
return year % 4 == 0</lang>
Asking for forgiveness instead of permission:
<lang python>import datetime
def is_leap_year(year):
try:
datetime.date(year, 2, 29)
except ValueError:
return False
return True</lang>
Ruby
<lang ruby>require 'date' def is_leap_year(year)
Date.new(year).leap?
end</lang>
SNOBOL4
Predicate leap( ) succeeds/fails, returns nil.
<lang SNOBOL4> define('leap(yr)') :(end_leap) leap eq(remdr(yr,400),0) :s(return)
eq(remdr(yr,100),0) :s(freturn) eq(remdr(yr,4),0) :s(return)f(freturn)
end_leap
- # Test and display (with ?: kluge)
test = "output = ('10' ? (*leap(yr) 1 | 0)) ': ' yr"
yr = '1066'; eval(test)
yr = '1492'; eval(test)
yr = '1900'; eval(test)
yr = '2000'; eval(test)
end</lang>
Output:
0: 1066 1: 1492 0: 1900 1: 2000
Tcl
The "classic" modulo comparison <lang tcl>proc isleap1 {year} {
return [expr {($year % 4 == 0) && (($year % 100 != 0) || ($year % 400 == 0))}]
} isleap1 1988 ;# => 1 isleap1 1989 ;# => 0 isleap1 1900 ;# => 0 isleap1 2000 ;# => 1</lang>
Does Feb 29 exist in the given year? If not a leap year, the clock command will return "03-01" <lang tcl>proc isleap2 year {
return [expr {[clock format [clock scan "$year-02-29" -format "%Y-%m-%d"] -format "%m-%d"] eq "02-29"}]
} isleap2 1988 ;# => 1 isleap2 1989 ;# => 0 isleap2 1900 ;# => 0 isleap2 2000 ;# => 1</lang>