Java Code for calculating Leap Year

Question

I am following  The Art and Science of Java  book and it shows how to calculate a leap year  The book uses ACM Java Task Force s library   Here is the code the books uses   import acm program     public class LeapYear extends ConsoleProgram       public void run                  println  This program calculates leap year             int year   readInt  Enter the year                    boolean isLeapYear     year   4    0   amp  amp   year   100    0      year   400    0             if  isLeapYear                        println year     is a leap year               else             println year     is not a leap year                Now  this is how I calculated the leap year    import acm program     public class LeapYear extends ConsoleProgram       public void run                  println  This program calculates leap year             int year   readInt  Enter the year               if   year   4    0   amp  amp  year   100    0                        println year     is a leap year                       else if   year   4    0   amp  amp   year   100    0   amp  amp   year   400    0                         println year     is a leap year                       else                       println year     is not a leap year                         Is there anything wrong with my code or should i use the one provided by the book    EDIT    Both of the above code works fine  What i want to ask is which code is the best way to calculate the leap year

User · Answer

I suggest you put this code into a method and create a unit test.

public static boolean isLeapYear(int year) {
    assert year >= 1583; // not valid before this date.
    return ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);
}

In the unit test

assertTrue(isLeapYear(2000));
assertTrue(isLeapYear(1904));
assertFalse(isLeapYear(1900));
assertFalse(isLeapYear(1901));

User · Answer

public static void main String   args     String strDate  Feb 2013           String   strArray strDate split    s                      Calendar cal   Calendar getInstance            cal setTime new SimpleDateFormat  MMM   parse strArray 0  toString              int monthInt   cal get Calendar MONTH           monthInt            cal set Calendar YEAR  Integer parseInt strArray 1                       strDate strArray 1  toString    -  monthInt  -  cal getActualMaximum Calendar DAY OF MONTH            System out println strDate

User · Answer

From JAVA s GregorianCalendar sourcecode          Returns true if   code year  is a leap year      public boolean isLeapYear int year        if  year  gt  changeYear            return year   4    0  amp  amp   year   100    0    year   400    0              return year   4    0      Where changeYear is the year the Julian Calendar becomes the Gregorian Calendar  1582       The Julian calendar specifies leap years every four years  whereas the   Gregorian calendar omits century years which are not divisible by 400    In the Gregorian Calendar documentation you can found more information about it

User · Answer

this answer is great but it won t work for years before Christ  using a proleptic Gregorian calendar   If you want it to work for B C  years  then use the following adaptation   public static boolean isLeapYear final int year        final Calendar cal   Calendar getInstance        if  year lt 0            cal set Calendar ERA  GregorianCalendar BC           cal set Calendar YEAR  -year         else         cal set Calendar YEAR  year       return cal getActualMaximum Calendar DAY OF YEAR   gt  365      You can verify that for yourself by considering that year -5  i e  4 BC  should be pronounced as a leap year assuming a proleptic Gregorian calendar  Same with year -1  the year before 1 AD   The linked to answer does not handle that case whereas the above adapted code does

User · Answer

As wikipedia states algorithm for the leap year should be       year 4    0   amp  amp   year 100   0       year 400  0       Here is a sample program how to check for leap year

User · Answer

The correct implementation is   public static boolean isLeapYear int year      Calendar cal   Calendar getInstance      cal set Calendar YEAR  year     return cal getActualMaximum Calendar DAY OF YEAR   gt  365      But if you are going to reinvent this wheel then   public static boolean isLeapYear int year      if  year   4    0        return false      else if  year   400    0        return true      else if  year   100    0        return false      else       return true

User · Answer

You can ask the GregorianCalendar class for this   boolean isLeapyear   new GregorianCalendar   isLeapYear year

User · Answer

boolean leapYear       year   4      0

User · Answer

If you are using java8    java time Year of year  isLeap     Java implementation of above method    public static boolean isLeap long year            return   year  amp  3     0   amp  amp    year   100     0     year   400     0

User · Answer

This is what I came up with  There is an added function to check to see if the int is past the date on which the exceptions were imposed year 100  year  400   Before 1582 those exceptions weren t around    import java util Scanner   public class lecture    public static void main String   args        boolean loop true      Scanner console   new Scanner  System in        while  loop           System out print   Enter the year                int year  console nextInt            System out println   The year is a leap year     leapYear year             System out print   again                int again   console nextInt            if  again    1               loop false             if         public static boolean leapYear   int year       boolean leaped   false      if  year 4  0           leaped   true          if year gt 1582               if  year 100  0 amp  amp year 400  0                   leaped false                                 1st if     return leaped

User · Answer

java time Year  isLeap  I d like to add the new java time way of doing this with the Year class and isLeap method   java time Year of year  isLeap

User · Answer

Most Efficient Leap Year Test   if   year  amp  3     0  amp  amp    year   25     0     year  amp  15     0            leap year        This is an excerpt from my detailed answer at https   stackoverflow com a 11595914 733805

User · Answer

new GregorianCalendar   isLeapYear year

User · Answer

Your code  as it is  without an additional class  does not appear to work for universal java  Here is a simplified version that works anywhere  leaning more towards your code   import java util    public class LeapYear       public static void main String   args            int year                        Scanner scan   new Scanner System in               System out println  Enter year                  year   scan nextInt                 if   year   4    0   amp  amp  year   100    0                    System out println year     is a leap year                   else if   year   4    0   amp  amp   year   100    0                       amp  amp   year   400    0                     System out println year     is a leap year                   else                   System out println year     is not a leap year                                       Your code  in context  works just as well  but note that book code always works  and is tested thoroughly  Not to say yours isn t

User · Answer

It s almost always wrong to have repetition in software  In any engineering discipline  form should follow function  and you have three branches for something which has two possible paths - it s either a leap year or not   The mechanism which has the test on one line doesn t have that issue  but generally it would be better to separate the test into a function which takes an int representing a year and returns a boolean representing whether or not the year is a leap year  That way you can do something with it other that print to standard output on the console  and can more easily test it   In code which is known to exceed its performance budget  it s usual to arrange the tests so that they are not redundant and perform the tests in an order which returns early  The wikipedia example does this - for most years you have to calculate modulo 400 100 and 4  but for a few you only need modulo 400 or 400 and 100  This is a small optimisation in terms of performance   at best  only one in a hundred inputs are effected    but it also means the code has less repetition  and there s less for the programmer to type in

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With the course of    TestMyCode Programming assignment evaluator  that one of the exercises was this kind of issue  I wrote this answer    import java util Scanner   public class LeapYear        public static void main String   args            Scanner reader   new Scanner System in           System out println  Type a year               int year   Integer parseInt reader nextLine              if  year   400    0  amp  amp  year   100    0  amp  amp  year   4    0                System out println  The year is a leap year              else          if  year   4    0  amp  amp  year 100  0                 System out println  The year is a leap year              else                        System out println  The year is not a leap year

User · Answer

easiest way ta make java leap year and more clear to understandenter code here  import  java util Scanner    class que19   public static void main String   args         Scanner input new Scanner System in        double a       System out println  enter the year here         a input nextDouble        if   a   4   0    amp  amp   a 100  0      a 400  0             System out println  leep year               else           System out println  not a leap year

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import java util Scanner       public class LeapYear       public static void main String   args               TODO Auto-generated method stub         Scanner input   new Scanner System in           System out print  Enter the year then press Enter               int year   input nextInt             if   year  lt  1580   amp  amp   year   4    0                 System out println  Leap year      year             else               if   year   4    0   amp  amp   year   100    0      year   400    0                     System out println  Leap year      year                 else                   System out println year     not a leap year

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import javax swing    public class LeapYear       public static void main String   args        int year  String yearStr   JOptionPane showInputDialog null   Enter radius        year   Integer parseInt  yearStr     boolean isLeapYear  isLeapYear    year   4    0  amp  amp  year   100    0      year   400    0       if isLeapYear    JOptionPane showMessageDialog null   Leap Year            else  JOptionPane showMessageDialog null   Not a Leap Year

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Pseudo code from Wikipedia translated into the most compact Java   year   400    0       year   4    0   amp  amp   year   100    0

[java] Java Code for calculating Leap Year

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