Dan Chisholm's Topic Exam Q.5

[Maria Garcia]

 
class A {
  private int[] val;
  private int hash;
  public int hashCode() {
    int h = hash;
    if (h == 0) {
      int off = 0;
      int len = val.length;
      for (int i = 0; i < len; i++) {
        h = 31*h + val[off++];
      }
      hash = h;
    }
    return h;
  }
  // The equals method has been omitted for clarity.
  A (int[] val) {this.val = val;}
  public static void main (String[] args) {
    A a = new A(new int[]{1,2,3});
    System.out.print(a.hashCode());
  }
}
 

The answer is 1026. How did you come up with this answer Dan ?

[Maria Garcia]

oops!! i made a boo-boo... I got it now.

[Ray Chang]

Can you explain? thank you.

[Dan Chisholm]

Originally posted by Ray Chang:
Can you explain? thank you.

If you would like to know how the hash code is calculated then just move the print statement into the hashCode method.

class A {
  private int[] val;
  private int hash;
  public int hashCode() {
    int h = hash;
    if (h == 0) {
      int off = 0;
      int len = val.length;
      for (int i = 0; i < len; i++) {
        h = 31*h + val[off++];
        System.out.print(h + " ");
      }
      hash = h;
    }
    return h;
  }
  A (int[] val) {this.val = val;}
  public static void main (String[] args) {
    A a = new A(new int[]{1,2,3});
    a.hashCode();
  }
}

A good hashcode calculation for an array will use the value of each element of the array. An even better hashcode method for an array will also depend on the position of each value in the array. The following is an example of a hashcode method that does not produce unique hashcodes for arrays that contain the same set of elements in different orders.

class A {
  private int[] val;
  private int hash;
  public int hashCode() {
    int h = hash;
    if (h == 0) {
      int off = 0;
      int len = val.length;
      for (int i = 0; i < len; i++) {
        h += val[off++];
      }
      hash = h;
    }
    return h;
  }
  A (int[] val) {this.val = val;}
  public static void main (String[] args) {
    A a = new A(new int[]{1,2,3});
    A b = new A(new int[]{2,3,1});
    A c = new A(new int[]{3,2,1});
    System.out.print(a.hashCode() + ",");
    System.out.print(b.hashCode() + ",");
    System.out.print(c.hashCode());
  }
}

The result is "6,6,6". A better hashCode method produces unique values when the position of the array elements is changed.

class A {
  private int[] val;
  private int hash;
  public int hashCode() {
    int h = hash;
    if (h == 0) {
      int off = 0;
      int len = val.length;
      for (int i = 0; i < len; i++) {
        h = 31*h + val[off++];
      }
      hash = h;
    }
    return h;
  }
  A (int[] val) {this.val = val;}
  public static void main (String[] args) {
    A a = new A(new int[]{1,2,3});
    A b = new A(new int[]{2,3,1});
    A c = new A(new int[]{3,2,1});
    System.out.print(a.hashCode() + ",");
    System.out.print(b.hashCode() + ",");
    System.out.print(c.hashCode());
  }
}

The result is "1026,2016,2946".

[Barkat Mardhani]

Hi Dan:
I am trying to find an easy read on HashCode() method. Can you suggest one?

[Dan Chisholm]

Take a look at item 8 of Chapter 3 of Effective Java.

[Barkat Mardhani]

Thanks Dan...