Leap year
From Rosetta Code
You are encouraged to solve this task according to the task description, using any language you may know.
See Also
[edit] ActionScript
public function isLeapYear(year:int):Boolean {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}
[edit] 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.
[edit] AppleScript
on leap_year(y)
if (y mod 100 is equal to 0) then
return (y mod 400 is equal to 0)
end if
return (y mod 4 is equal to 0)
end leap_year
leap_year(1900)
[edit] C/C++
#include <stdio.h>
#include <stdbool.h>
bool is_leap_year(int year)
{
return !(year%4) && year%100 || !(year%400);
}
int main(){
int test_case[] = {1900, 1994, 1996, 1997, 2000}, key;
for(key=0; key<sizeof test_case/sizeof test_case[0]; key++){
int year = test_case[key];
printf("%d is %sa leap year.\n", year, is_leap_year(year)?"":"not ");
}
}
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.
[edit] C#
using System;
class Program {
public static bool IsLeapYear(int year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}
public static bool IsLeapYear(DateTime dt) {
return IsLeapYear(dt.Year);
}
static void Main(string[] args) { //test the func
Console.WriteLine(IsLeapYear(1900).ToString());
Console.WriteLine(IsLeapYear(1994).ToString());
Console.WriteLine(IsLeapYear(1996).ToString());
Console.WriteLine(IsLeapYear(DateTime.Now).ToString());
}
}
[edit] Clojure
(defn leap-year? [y]
(and (zero? (mod y 4)) (or (pos? (mod y 100)) (zero? (mod y 400)))))
[edit] 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))))
[edit] 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= ;
[edit] 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
[edit] 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
[edit] 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
[edit] Icon
[edit] Unicon
The Icon solution works in Unicon.
[edit] J
isLeap=: 0 -/@:= 4 100 400 |/ ]
Example use:
isLeap 1900 1996 1997 2000
0 1 0 1
[edit] Java
new java.util.GregorianCalendar().isLeapYear(year)
or
public static boolean isLeapYear(int year){
if(year % 100 == 0) return year % 400 == 0;
return year % 4 == 0;
}
[edit] JavaScript
function isLeapYear(year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}
[edit] Liberty BASIC
[edit] Simple method
if leap(1996)then
print "leap"
else
print "ordinary"
end if
wait
function leap(n)
leap=date$("2/29/";n)
end function
[edit] 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
[edit] 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
[edit] Lua
function isLeapYear(year)
return year%4==0 and (year%100~=0 or year%400==0)
end
[edit] MATLAB
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
[edit] MUMPS
ILY(X) ;IS IT A LEAP YEAR?Usage:
QUIT ((X#4=0)&(X#100'=0))!((X#100=0)&(X#400=0))
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
[edit] OCaml
let is_leap_year ~year =
if (year mod 100) = 0
then (year mod 400) = 0
else (year mod 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)
[edit] 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.
[edit] Perl
sub leap {
my $yr = $_[0];
if ($yr % 100 == 0) {
return ($yr % 400 == 0);
}
return ($yr % 4 == 0);
}
Or more concisely:
sub leap {!($_[0] % 100) ? !($_[0] % 400) : !($_[0] % 4)}
Using library function/method:
use Date::Manip;
Date_LeapYear($year);
use Date::Manip::Base;
my $dmb = new Date::Manip::Base;
$dmb->leapyear($year);
[edit] 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 $y %% 400
}
[edit] PHP
<?php
function isLeapYear(year) {
if (year % 100 == 0) {
return (year % 400 == 0);
}
return (year % 4 == 0);
}
With date('L'):
<?php
function isLeapYear(year) {
return (date('L', mktime(0, 0, 0, 2, 1, year)) === '1')
}
[edit] PicoLisp
(de isLeapYear (Y)
(bool (date Y 2 29)) )
Output:
: (isLeapYear 2010) -> NIL : (isLeapYear 2008) -> T : (isLeapYear 1600) -> T : (isLeapYear 1700) -> NIL
[edit] PL/I
declare year picture '9999'; /* 16 August 2010 */
get edit (year) (f(4));
if days(year || '1231', 'YYYYMMDD') - days(year || '0101', 'YYYYMMDD') = 365 then
put ( year || ' is a leap year');
else
put (year || ' is not a leap year');
[edit] PureBasic
Procedure isLeapYear(Year)
If (Year%4=0 And Year%100) Or Year%400=0
ProcedureReturn #True
Else
ProcedureReturn #False
EndIf
EndProcedure
[edit] Python
import calendar
calendar.isleap(year)
or
def is_leap_year(year):
if year % 100 == 0:
return year % 400 == 0
return year % 4 == 0
Asking for forgiveness instead of permission:
import datetime
def is_leap_year(year):
try:
datetime.date(year, 2, 29)
except ValueError:
return False
return True
[edit] REXX
Standard solution:
leapyear:
arg year
return (year // 400 = 0 | (year // 100 <> 0 & year // 4 = 0))
Tricky solution: Check day number of 31st december
leapyear:
arg year
return date("D",year"1231","S") > 365
[edit] Ruby
require 'date'
def is_leap_year(year)
Date.new(year).leap?
end
[edit] SNOBOL4
Predicate leap( ) succeeds/fails, returns nil.
define('leap(yr)') :(end_leap)
leap eq(remdr(yr,400),0) :s(return)
eq(remdr(yr,100),0) :s(freturn)
eq(remdr(yr,4),0) :s(return)f(freturn)
end_leap
* # Test and display (with ?: kluge)
test = "output = ('10' ? (*leap(yr) 1 | 0)) ': ' yr"
yr = '1066'; eval(test)
yr = '1492'; eval(test)
yr = '1900'; eval(test)
yr = '2000'; eval(test)
end
Output:
0: 1066 1: 1492 0: 1900 1: 2000
[edit] Tcl
The "classic" modulo comparison
proc isleap1 {year} {
return [expr {($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"
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
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