Grokking The Java Developer Interview: More Than 200 Questions To Crack The Java, Spring, SpringBoot & Hibernate Interview

Scenario 2: when only equals() method is implemented by
Employee class

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Now, can you predict the output of our TestHashMap class?

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Let’s see the output:
Well, nothing’s changed here. Because even though e1 and e2 are
equal according to our newly implemented equals() method, they still
have different hashCode as the Object’s class hashCode() is used.
So the equals and hashCode contract is not followed and both e1,
e2 got inserted in HashMap.
Scenario 3: when only hashCode() method is implemented:

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Let’s run our TestHashMap class again and see the output:

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Well, now we have same hashCode for e1 and e2, but Object’s
equals method still checks the references and as references are
different, both are not equal and are inserted in the hashmap.
Scenario 4: When both equals and hashCode are implemented
properly:

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Output:

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Here, both e1 and e2 are equals as we are comparing the contents
of them in our equals() method, so their hashCodes must be same,
which they are. So value of e1 which was 100 got replaced by 200,
and size of hashmap is 2.
You can be asked to write the equals() and hashCode() methods
implementation by hand also, so you should pay attention to how
these are implemented.
Question 106: How to make a HashMap
synchronized?
Answer: Collections.synchronizedMap(map);

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