This is one the most frequent questions I was asked in interview.

You can override the default implementation of the equals() method defined in java.lang.Object. If you override the equals(), you MUST also override hashCode(). Otherwise a violation of the general contract for Object.hashCode will occur, which can have unexpected repercussions when your class is in conjunction with all hash-based collections.

The default implementation of equals() method checks to see if the two objects have the same identity. Similarly, the default implementation of the hashCode() method returns an integer based on the object’s identity and is not based on the values of instance (and class) variables of the object. No matter how many times the values of its instance variables (data fields) change, the hash code calculated by the default hashCode implementation does not change during the life of the object.

Consider the following code, we have overridden equals() method to check if two objects are equal based on the values of their instance variables. Two objects may be stored at different memory addresses but may still be equal base on their instance variable.

public class CustomerID {
private long crmID;
private int nameSpace;

public CustomerID(long crmID, int nameSpace) {
super();
this.crmID = crmID;
this.nameSpace = nameSpace;
}

public boolean equals(Object obj) {
//null instanceof Object will always return false
if (!(obj instanceof CustomerID))
return false;
if (obj == this)
return true;
return this.crmID == ((CustomerID) obj).crmID &&
this.nameSpace == ((CustomerID) obj).nameSpace;
}

public static void main(String[] args) {
Map m = new HashMap();
m.put(new CustomerID(2345891234L,0),”Jeff Smith”);
System.out.println(m.get(new CustomerID(2345891234L,0)));
}

}

Compile and run the above code, the output result is

null
What is wrong? The two instances of CustomerID are logically equal according to the class’s equals method. Because the hashCode() method is not overridden, these two instances’ identities are not in common to the default hashCode implementation. Therefore, the Object.hashCode returns two seemingly random numbers instead of two equal numbers. Such behavior violates “Equal objects must have equal hash codes” rule defined in the hashCode contract.

Let’s provide a simple hashCode() method to fix this problem:

public class CustomerID {
private long crmID;
private int nameSpace;

public CustomerID(long crmID, int nameSpace) {
super();
this.crmID = crmID;
this.nameSpace = nameSpace;
}

public boolean equals(Object obj) {
//null instanceof Object will always return false
if (!(obj instanceof CustomerID))
return false;
if (obj == this)
return true;
return this.crmID == ((CustomerID) obj).crmID &&
this.nameSpace == ((CustomerID) obj).nameSpace;
}

public int hashCode() {
int result = 0;
result = (int)(crmID/12) + nameSpace;
return result;
}

public static void main(String[] args) {
Map m = new HashMap();
m.put(new CustomerID(2345891234L,0),”Jeff Smith”);
System.out.println(m.get(new CustomerID(2345891234L,0)));
}

}
Compile and run the above code, the output result is

Jeff Smith

Advertisements