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.
This is clearly stated in hashCode() description.
Please refer its Object class API here for more clarity.
Harivenkatesh Polnati wrote:Why always override hashcode() if overriding equals()?
You don't have to do it, but it's basically good practise, because it ensures that your object will always work correctly in a hashed collection.
If you don't follow this rule - and that includes not just overriding hashCode(), but also ensuring that it's always in sync with equals() (ie, if you updateequals(), you also need to update hashCode()) - you run the risk of someone using it in a HashMap or HashSet and having it behave very strangely.
Isn't it funny how there's always time and money enough to do it WRONG?
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