![]() That’s all about why we need to override equals and hashcode methods in Java. It is perfectly valid to override hashCode() without overriding equals() as objects with equal hash codes need not be equal. In Java language the important contract is whenever you override one of the methods ( equals() and hashCode() ), then you must override the other method. Please note that even though equal objects must have equal hash codes, the reverse is not true. It only checks if two objects belongs to the same memory location. However, each time we override the equals method we should make sure that the hashCode method is changed in a compatible way. But since the equals() method is not overridden, when the set hashes e2 and iterates through the bucket looking if there is an Employee e such that e2.equals(e) is true, it won’t find any as e2.equals(e1) will be false. In simple words the equals () method does not check what is stored inside the Object. If we only override the hashCode() method, both e1 and e2 will hash to the same bucket as they produce the same hash code. What could happen if I only override equals Suppose you only override equals but not hashCode This means that hashCode is inherited from Object Object. Why, you might ask, did the String class. The String class overrides the equals method it inherited from the Object class and implemented logic to compare the two String objects character by character. Notice the use of the equals method instead of the operator. Override only hashCode() without overriding equals() method These are the outcomes one would typically expect and want. It is worth noting that if the class instance is never used in any hash-based collections, then it doesn’t really matter if hashCode() is overridden or not. That's the equals method for HashMap.Entry, not HashMap itself - look at how it's trying to use the reference passed into it as a Map.Entry. Even though both e1 and e2 are equal, they don’t hash to the same bucket, and both reside in the collection as separate keys. Since the default hashCode implementation in the Object class return distinct integers for distinct objects, if only equals() method is overridden, e1 will be placed in some bucket and e2 will be placed in some other bucket as e1.hashCode() != e1.hashCode(). Overriding only equals() method without overriding hashCode() causes the two equal instances to have unequal hash codes, which violates the hashCode contract (mentioned in Javadoc) that clearly says, if two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result. Override only equals() without overriding hashCode() method ![]() Now let’s discuss the above program’s behavior if the equals() method is overridden without overriding hashCode(), or vice versa. In the following example we have a class Employee with two variables name, age and a parameterized constructor. This is because we have overridden both the equals() and hashCode() method in the Employee class, and both objects now point to the same bucket and hold the same location within the bucket. As evident from the generated output, the set contains only one Employee object even though two different Employee objects are added.
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