Checking whether a year is leap or not

This tutorial is going to teach you the coding for checking whether a year is a leap year or not. Here, we have taken the year 2000. So define an integer n=2000 in the class "Leapyear" and now apply "if else" condition. As we know leap year is divided by the integer 4 and so applying if condition as n/4=0, then "n" is a leap year. Now in the System.out.println write the message that the year is a leap year. Again applying "else" condition the output will be that the year is not a leap year.

Here is the code of program:


class  Leapyear

{

  public static void main(String[] args) 

  {

    int n=2000;

    if (n%4==0){

    System.out.println("The given year is a leap year");

      }

    else{

    System.out.println("This is not a leap year");

  }

}

}

Download the program:

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Current Comments

4 comments so far (post your own) View All Comments Latest 10 Comments:

This Program Dose Not Meet All The Conditions For A Leap Year...

Posted by Amar on Monday, 09.29.08 @ 13:39pm | #80769

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Your leap year example gives false results for the years 1900, 2100 and others.
The leap year rules are:
1. Years that are multiples of 4 are leap years,
except for the following special years.
2. Years that are mutliples of 100 are not leap years, except for the following special years.
3. Years that are multiples 400 are leap years.
Your program does not obey to rule 2.

Posted by Hubert Partl on Wednesday, 01.2.08 @ 16:52pm | #44367

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Its very helpful for beginners.

Posted by Monisha agarwal on Sunday, 03.25.07 @ 11:21am | #12652

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good example for the biginners

Posted by visakh m g on Friday, 12.1.06 @ 14:12pm | #316

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