Check if year is leap year in javascript

Question

 function leapYear(year){
    var result; 
    year = parseInt(document.getElementById("isYear").value);
    if (years/400){
      result = true
    }
    else if(years/100){
      result = false
    }
    else if(years/4){
      result= true
    }
    else{
      result= false
    }
    return result
 }

This is what I have so far (the entry is on a from thus stored in "isYear"), I basically followed this here, so using what I already have, how can I check if the entry is a leap year based on these conditions(note I may have done it wrong when implementing the pseudocode, please correct me if I have) Edit: Note this needs to use an integer not a date function

check @ http://stackoverflow.com/questions/8175521/javascript-to-find-leap-year — swetha, May 03 '13 at 06:51
The thing is I need to deal with this as if I am working with only the year say 2014 as an integer not a date value — BigBob, May 03 '13 at 06:56
A very detailed information on calculation is here - https://www.mathsisfun.com/leap-years.html — Manohar Reddy Poreddy, Sep 11 '22 at 04:56

score 222 · Accepted Answer · answered May 03 '13 at 06:51

222

function leapYear(year)
{
  return ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
}

answered May 03 '13 at 06:51

MMeersseman

2,541
1
13
8

Glad I could help. You can just use ´((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)´ where you need it if you don't want a function. I find functions helpful for having good human readable code though. – MMeersseman May 03 '13 at 07:04
I already did as instructed, but thanks for the advice. I would love to use more functions, however I am not allowed to add any extra functions. Fairly restricting and making some other calculations annoying, but managing. – BigBob May 03 '13 at 07:06
2

FYI: This seems to be much more performant than `isLeap` (below) -- http://jsfiddle.net/atwright147/7dqzvzpr/ – atwright147 Jul 10 '15 at 14:39
@atwright147, +1 for the performance check. It's interesting to note that nearly 96% of the time spent in "ily + defining function" goes to defining the function when compared to "ily".... (referring to your example jsfiddle.net/atwright147/7dqzvzpr) – KSK Jul 22 '15 at 17:26
5

the fastest ever!!! can be reduced to that ... which speeds up a bit function isLeap (y) { return !((y % 4) || (!(y % 100) && y % 400)); } – fedeghe Aug 26 '15 at 14:23
2

@fedeghe And I thought I had made the most concise one `!(y%4)&&(!(y%400)||!!(y%100));` until I read your comment, but you did the inverse logic and saved two characters yet! Haha. Very well done. – Modular Jun 30 '17 at 16:26
1

@fedeghe What about `!(y&3||y&15&&!(y%25));`? See https://stackoverflow.com/q/9852837/3167040 for details! – ReinstateMonica3167040 Jul 26 '17 at 23:59
sounds definitely interesting! I want to check it... thx! – fedeghe Jul 27 '17 at 05:20
@Userthatisnotauser I ran a benchmark and `!((y % 4) || (!(y % 100) && y % 400))` is a way faster, although less concise....anyway thx for sharing ;) – fedeghe Aug 09 '17 at 13:17
@fedeghe Very interesting. How much faster was it? – ReinstateMonica3167040 Aug 09 '17 at 22:40
@Userthatisnotauser ~twice... curious! – fedeghe Aug 10 '17 at 00:01
@Userthatisnotauser on ff, in chrome e.g. are similar, safari seems unstable in times...anyway ... jmvc.org/test_leap – fedeghe Aug 10 '17 at 07:27
@Userthatisnotauser sometimes your shortest one is even faster (some mobile bro) – fedeghe Aug 11 '17 at 10:45
1

minor nitpick but it is, for the most part, correct to use the strict (in)equality operator "===". – Neil Meyer Nov 13 '17 at 09:19

score 72 · Answer 2 · edited Sep 29 '20 at 16:53

72

The function checks if February has 29 days. If it does, then we have a leap year.

ES5

function isLeap(year) {
  return new Date(year, 1, 29).getDate() === 29;
}

ES6

const isLeap = year => new Date(year, 1, 29).getDate() === 29;

Result

isLeap(1004) // true
isLeap(1001) // false

edited Sep 29 '20 at 16:53

poolsideDev

390
4
13

answered May 06 '17 at 10:46

EugenSunic

13,162
13
64
86

12

I actually like this solution the best, because it doesn't require any underlying knowledge of the gregorian calendar. – Matt Korostoff Nov 04 '17 at 00:37

score 9 · Answer 3 · answered May 23 '17 at 20:19

9

A faster solution is provided by Kevin P. Rice here:https://stackoverflow.com/a/11595914/5535820 So here's the code:

function leapYear(year)
{
    return (year & 3) == 0 && ((year % 25) != 0 || (year & 15) == 0);
}

answered May 23 '17 at 20:19

C-Y

335
4
11

7

This is more concise: `!(y&3||y&15&&!(y%25))` – ReinstateMonica3167040 Aug 11 '17 at 15:57
7

and less readable @ReinstateMonica3167040 – elad.chen Dec 15 '21 at 13:50
2

@elad.chen Agreed, I must have commented that because of the comments on the accepted answer. Looking back, it's kind of silly: there's not any real benefits except when trying to make it hard to read... – ReinstateMonica3167040 Dec 16 '21 at 18:39
@elad.chen or include a link to this post to make it more readable! – C-Y Dec 17 '21 at 07:31

score 1 · Answer 4 · answered Jul 07 '15 at 05:59

1

If you're doing this in an Node.js app, you can use the leap-year package:

npm install --save leap-year

Then from your app, use the following code to verify whether the provided year or date object is a leap year:

var leapYear = require('leap-year');

leapYear(2014);
//=> false 

leapYear(2016);
//=> true

Using a library like this has the advantage that you don't have to deal with the dirty details of getting all of the special cases right, since the library takes care of that.

answered Jul 07 '15 at 05:59

nwinkler

52,665
21
154
168

21

But you're adding yet another dependency to your project for such a simple function. That's a no-no for me. – Frondor Jul 23 '18 at 16:57
11

Pulling in a package to check for three division remainders is frankly ridiculous. – Kuba Szymanowski Aug 01 '18 at 09:56
3

Is it? If you take a look at the wrong answers in this question (people simply dividing by 4 and checking the remainder), using a library would probably be a wise decision for some people. The advantage of the library is that it includes test cases and has a better chance of working than people coming up with their own wrong implementation (see below for a couple of examples). Having said that, I understand that writing your own quick function (if you get it right) saves you from including one dependency - to each their own. – nwinkler Aug 01 '18 at 09:59
3

On the other hand, all of this might be overkill anyway. The next year where the simply `year % 4` does not work is 2100, which most of the software built today is not going to see anyway. Add tags if you like... – nwinkler Aug 01 '18 at 10:03
right, and while a decent amount of software needs to handle 2000, 2000 also is a leap year. so unless your software needs to handle 1900 you're good – MalcolmOcean Oct 17 '20 at 00:03

score -1 · Answer 5 · edited Mar 30 '15 at 17:48

-1

You can use the following code to check if it's a leap year:

ily = function(yr) {
    return (yr % 400) ? ((yr % 100) ? ((yr % 4) ? false : true) : false) : true;
}

edited Mar 30 '15 at 17:48

Chris Forrence

10,042
11
48
64

answered Mar 30 '15 at 17:17

Brian Hayes

23
1

1

Could you elaborate a bit? – tsnorri Mar 30 '15 at 17:46
2

This seems like just a refactoring of the accepted answer. It's not any more compact or performing any different operations, so I'm not sure what value it would add. – Matt Johnson-Pint Mar 30 '15 at 17:55
12

this is pretty much molesting the ternary operator – Willem D'Haeseleer Feb 10 '16 at 01:39
2

This will be inefficient as it is always checking if a year is divisible by 400. You are getting 399 times `False` and one time `True` – Daut Feb 04 '19 at 13:49

score -11 · Answer 6 · answered May 03 '13 at 06:53

-11

You can try using JavaScript's Date Object

new Date(year,month).getFullYear()%4==0

This will return true or false.

answered May 03 '13 at 06:53

SrikanthManian

150
7

8

It doesn't handle special cases (divisibility by 100 and 400). – Pif Nov 20 '13 at 13:41

score -15 · Answer 7 · edited Aug 22 '16 at 04:45

-15

My Code Is Very Easy To Understand

var year = 2015;
var LeapYear = year % 4;

if (LeapYear==0) {
    alert("This is Leap Year");
} else {
    alert("This is not leap year");
}

edited Aug 22 '16 at 04:45

user3335966

2,673
4
30
33

answered Aug 22 '16 at 01:21

Maruf Sarkar

1
1
1

8

Your code is incorrect. 1900 was not a leap year, but your code will say that it was. – Aug 22 '16 at 01:38

Check if year is leap year in javascript

7 Answers7

Linked

Related