javascript to find leap year

How can I get the code below to work when I have a month of february? Currently it is getting to the day and then stopping before getting to the if to determine whether it is a leap year.

 if (month == 2) {
    if (day == 29) {
        if (year % 4 != 0 || year % 100 == 0 && year % 400 != 0) {
            field.focus();
             field.value = month +'/' +  '';
        }
    }
    else if (day > 28) {
        field.focus();
             field.value = month +'/' +  '';
    }
}

Solution 1:

It's safer to use Date objects for datetime stuff, e.g.

isLeap = new Date(year, 1, 29).getMonth() == 1

Since people keep asking about how exactly this works, it has to do with how JS calculates the date value from year-month-day (details here). Basically, it first calculates the first of the month and then adds N -1 days to it. So when we're asking for the 29th Feb on a non-leap year, the result will be the 1st Feb + 28 days = 1st March:

> new Date(2015, 1, 29)
< Sun Mar 01 2015 00:00:00 GMT+0100 (CET)

On a leap year, the 1st + 28 = 29th Feb:

> new Date(2016, 1, 29)
< Mon Feb 29 2016 00:00:00 GMT+0100 (CET)

In the code above, I set the date to 29th Feb and look if a roll-over took place. If not (the month is still 1, i.e. February), this is a leap year, otherwise a non-leap one.

Solution 2:

Compared to using new Date() this is is around 100 times faster!

Update:

This latest version uses a bit test of the bottom 3 bits (is it a multiple of 4), as well as a check for the year being a multiple of 16 (bottom 4 bits in binary is 15) and being a multiple of 25.

ily = function(y) {return !(y & 3 || !(y % 25) && y & 15);};

http://jsperf.com/ily/15

It is slightly faster again than my previous version (below):

ily = function(yr) {return !((yr % 4) || (!(yr % 100) && (yr % 400)));};

http://jsperf.com/ily/7

It is also 5% faster, compared to the already fast conditional operator version by broc.seib

Speed Test results: http://jsperf.com/ily/6

Expected logic test results:

alert(ily(1900)); // false
alert(ily(2000)); // true
alert(ily(2001)); // false
alert(ily(2002)); // false
alert(ily(2003)); // false
alert(ily(2004)); // true
alert(ily(2100)); // false
alert(ily(2400)); // true

Solution 3:

isLeap = !(new Date(year, 1, 29).getMonth()-1)

...subtraction by one should work even faster than compare on most CPU architectures.