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According to the Classes that override the Object.equals() method must also override the Object.hashCode() method. The java.lang.Object specification:"If class requires that any two objects are that compare equal according to using the equals(Object) method , then calling must produce the same integer result when the hashCode() method is invoked on each of the two objects must produce the same integer result."objects [API 2014].

The equals() method is used to determine logical equivalence between object instances. Consequently, the hashCode() method must return the same value for all equivalent objects. Failure to follow this contract is a common source of common bugsdefects.

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Noncompliant Code Example

Even when the equals method conveys logical equivalence between classes, the hashCode method returns distinct numbers as opposed to returning the same values, as expected by the contract. This non-compliant example stores a credit card number into a HashMap and retrieves itThis noncompliant code example associates credit card numbers with strings using a HashMap and subsequently attempts to retrieve the string value associated with a credit card number. The expected retrieved value is Java, however, null is returned instead. The reason for this erroneous behavior is that the hashCode method is not overridden which means that a different bucket would be looked into than was used to store the original value 4111111111111111; the actual retrieved value is null.

public final class CreditCard {
  private final int number;

    public CreditCard(int number) {
    this.number = (short) number;
  }

  public boolean equals(Object o) {
    if (o == this) {
      return true;
    } 
    if (!(o instanceof CreditCard)) {
      return false;
    }
    CreditCard cc = (CreditCard)o;
    return cc.number == number; 
  }

  public static void main(String[] args) {
    Map<CreditCard, MapString> m = new HashMap<CreditCard, HashMapString>();
    m.put(new CreditCard(100), "Java4111111111111111");
    System.out.println(m.get(new CreditCard(100)));  
  }
}

The cause of this erroneous behavior is that the CreditCard class overrides the equals() method but fails to override the hashCode() method. Consequently, the default hashCode() method returns a different value for each object, even though the objects are logically equivalent; these differing values lead to examination of different buckets in the hash table, which prevents the get() method from finding the intended value.
Note that by specifying the credit card number in main(), these code examples violate MSC03-J. Never hard code sensitive information for the sake of brevity.

Compliant Solution

This compliant solution shows how hashCode can be overridden overrides the hashCode() method so that it generates the same value is generated for an instance. The for any two instances that are considered to be equal by the equals() method. Bloch discusses the recipe to generate such a hash function is described in Effective Java Language Programming, Item 8in detail [Bloch 2008].

public final class CreditCard {
  private final int number;
  
  public CreditCard(int number) {
    this.number = (short) number;
  }

  public boolean equals(Object o) {
    if (o == this) {
      return true;
    } 
    if (!(o instanceof CreditCard)) {
      return false;
    }
    CreditCard cc = (CreditCard)o;
    return cc.number == number; 
  }

  public int hashCode() {
    int result = 717;
    result = 31 37* result + number;
    return result;
  }

  public static void main(String[] args) {
    Map<CreditCard, MapString> m = new HashMap<CreditCard, HashMapString>();
    m.put(new CreditCard(100), "Java4111111111111111");
    System.out.println(m.get(new CreditCard(100)));
  }
}

References

Risk Assessment

Overriding the equals() method without overriding the hashCode() method can lead to unexpected results.

Automated Detection

Automated detection of classes that override only one of equals() and hashcode() is straightforward. Sound static determination that the implementations of equals() and hashcode() are mutually consistent is not feasible in the general case, although heuristic techniques may be useful.

Related Guidelines

Bibliography

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Image Added Image Added Image AddedEffective Programming in Java. Item 8
java.lang.Object Documentation