Overriding hashCode() in eclipse- Java

Question

Whenever I override hashcode() using eclipse 'source' menu it generates following code in the class

final int prime = 31;
int result = 1;
result = prime * result + ((fieldName1== null) ? 0 : fieldName1.hashCode());
result = prime * result + ((fieldName2== null) ? 0 : fieldName2.hashCode());

Could anyone please explain why it is doing all this calculation(multiplication and then addition), why it is not returning simply

fieldName.hashCode();
or
fieldName2.hashCode();

?

Try having two or more fields in your class and then generate and see what happens. — darijan, Jun 14 '13 at 12:02
@Zavior this is no duplicate as in the oter question it is asked why not XOR is used. Here it is asked why the calculation is used to create the `hashCode`. — Uwe Plonus, Jun 14 '13 at 12:02
fair enough, the first answer there does answer this one as well :P — Zavior, Jun 14 '13 at 12:05
@darijan I have 15 fields in my class I put just a section of code here coz of readability purpose, although I put Two fields here. — Just_another_developer, Jun 14 '13 at 12:06
See also http://stackoverflow.com/questions/11795104/is-the-hashcode-function-generated-by-eclipse-any-good — Raedwald, Jun 14 '13 at 12:27

AllTooSir · Answer 1 · 2013-06-14T12:06:30.303

0

Multiplying reduces collisions.

Please read Joshua Bloch

The value 31 was chosen because it is an odd prime. If it were even and the multiplication overflowed, information would be lost, as multiplication by 2 is equivalent to shifting. The advantage of using a prime is less clear, but it is traditional. A nice property of 31 is that the multiplication can be replaced by a shift and a subtraction for better performance: 31 * i == (i << 5) - i. Modern VMs do this sort of optimization automatically.

edited Jun 14 '13 at 12:06

answered Jun 14 '13 at 11:58

AllTooSir

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This doesnt answer his question though – Zavior Jun 14 '13 at 11:59
ok I get that y '31' is chosen but I am asking why we need to do(or y IDE generates) 'all' of this calculation. Please read question for reference. – Just_another_developer Jun 14 '13 at 12:01
By multiplying, bits are left shifted which uses more of the available space of hash codes, reducing collisions. – AllTooSir Jun 14 '13 at 12:02
As The New Idiot says: it reduces collisions – Puce Jun 14 '13 at 12:03

score 0 · Answer 2 · edited May 23 '17 at 11:56

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It is explained here.

Basically you use primes in multiplication to have a better distribution of the hash values. Then HashSet and HashMap work better because they distibute according to the hash value. And badly distributed hash values lead to many collisions.

edited May 23 '17 at 11:56

Community

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answered Jun 14 '13 at 12:01

Uwe Plonus

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score 0 · Answer 3 · answered Jun 14 '13 at 12:19

why it is not returning simply fieldName.hashCode(); or fieldName2.hashCode();

To understand why, you must also examine the implementation of equals() and understand the constraints placed on the hashCode() implementation.

A hashCode() implementation must return equal values for objects that compare as equal: if x.equals(y) then x.hashCode() == y.hashCode().
A good hashCode() implementation should rarely return the same hash code for objects that do not compare as equal: if !x.equals(y) then often x.hashCode() != y.hashCode().
The equals() implementation that Eclipse generated examines both fileName1 and filename2. If either are different for two objects, the method will consider them the two objects to be not equivalent.
Therefore a corresponding good hashCode() implementation should produce different hash code values if either fileName1 or fileName2 are different.
Therefore a corresponding good hashCode() implementation should use fileName1.hashCode() and fileName2.hashCode().

Overriding hashCode() in eclipse- Java

3 Answers3