Implementing stock methods

[Steve Butcher]

"They" say that any "serious" class should override or implement toString(), equals(), hashCode(), and the Cloneable and Serializable interfaces.
Some of these are easy enough:

public class Point2D implements Cloneable, Serializable {
   private int x;
   private int y;
   /*
     constructors and incredibly useful methods omitted
     for reasons of national security
   */
   public String toString() {
      return new String( "(" + x + ", " + y + ")" );
   }
   public boolean equals( Object o ) {
      boolean result = false;
      if ( !( o == null ) && ( o instanceof Point2D ) ) {
         Point2D p = (Point2D) o;
         if ( ( this.x == p.x ) && ( this.y == p.y ) ) {
            result = true;
         }
      }
      return result;
   }
}

How do you implement hashCode()? It is a native method of the class java.lang.Object; do you just "get" it through inheritance (then why are they saying to implement it) or do you have to do something and what do you do?
Steve
[e-mail]

[This message has been edited by Steve Butcher (edited March 22, 2000).]

[Thandapani Saravanan]

One thing about objects are:
Objects have state and indentity.
Is the method hashCode() provided to give identity to an object?
Coming to the question, we are supposed to implement hashCode() if we implement equals(). The method hashCode() has to return same value for two objects if calling obj1.equals(obj2) returns true. It also has to return same value when called on same object many times.
I wonder how we can do this.

[Pawel Veselov]

the hashCode() method can be just overriden without any problems.
The standart implementation of equals() method does the following:
boolean equals(a) {
return ((a.hashCode() == this.hashCode()) && (a == this));
}

So, the idea of hashCoding is to be able to tell some comparing stuff that object DO differ. So, if the hashCode()s are different and the implementation of hashCode() method is supposed to be very fast, at least faster than byte-to-byte comparison, we can know that the object we search is wrong.
Anyhow, there can be definetely be two objects with the same hashCodes. For example, Long classs.

------------------
With best of best regards, Pawel S. Veselov ( aka Black Angel )

[Steve Butcher]

Pawel,
I am now beginning to understand more fully the ramifications of equals() and hashCode(). However, clearly Object.hashCode() does not meet your criteria for subclasses--it would cause your equals() method to fail for objects that meet the criteria of being equals(). For example,

public class Test {
   private int i;
   
   public Test() {
      this( 0 );
   }
   
   public Test( int i ) {
      this.i = i;
   }
   
   public void setI( int i ) { 
      this.i = i; 
   }
   
   public int getI() {
      return this.i;
   }
   
   public String toString() {
      return new String( " { " + i + " } " );
   }
   
   public static void main( String[] args ) {
      Test[] t = new Test[ 4 ];
      
      t[ 0 ] = new Test( 1 );
      t[ 1 ] = new Test( 2 );
      t[ 2 ] = new Test( 1 );
      t[ 3 ] = new Test( 4 );
      
      for ( int j = 0; j < 2; j++ ) {
         for ( int i = 0; i < 4; i++ ) {
            System.out.println( "Test object " + i + " Test.toString(): " + t[ i ].toString() +
                               "with Object.hashCode(): " + t[ i ].hashCode() );
         }
         System.out.println();
      }
   }
}

Results in the following output:

Test object 0 Test.toString(): { 1 } with Object.hashCode(): 3179625
Test object 1 Test.toString(): { 2 } with Object.hashCode(): 3180238
Test object 2 Test.toString(): { 1 } with Object.hashCode(): 3180240
Test object 3 Test.toString(): { 4 } with Object.hashCode(): 3180242
Test object 0 Test.toString(): { 1 } with Object.hashCode(): 3179625
Test object 1 Test.toString(): { 2 } with Object.hashCode(): 3180238
Test object 2 Test.toString(): { 1 } with Object.hashCode(): 3180240
Test object 3 Test.toString(): { 4 } with Object.hashCode(): 3180242

Clearly the first condition, that for every hashCode() on the same object, hashCode() returns the same result during a single run of the application BUT for two objects that are equals(), in this case t[0] and t[2], Object.hashCode() does not return the same hashCode() so this violates the general contract of the hashCode() API.
Obviously, this is because hashCode() has not been overridden to provide appropriate hashcodes for Test objects. I have, using the greatest of all teachers, javac, discovered that you can override a method in base class that was declared native (that was one basic question I had). Now, my basic question is: How do you compute a hashcode? On what should the hashcode be computed?
Thank you all for your assistance; I should have been more precise in my questions.
Steve
[e-mail]
[PS I'm sorry Thandapani but I really don't know what you're getting at, I have read and understood the API, it has not helped. I just don't know how to implement a proper hashCode() method based on the general contract or any contract for that matter. That is the help I am looking for. ]