How does the return value works in Hashcode override? [duplicate]

Question

In the below code, the overridden method hashCode returns the result. Why are we assigning so many values to result variable before returning it? I got the below code from some tutorials:

public class User {
private String name;
private int age;
private String passport;

//getters and setters, constructor

@Override
public boolean equals(Object o) {

    if (o == this) return true;
    if (!(o instanceof User)) {
        return false;
    }

    User user = (User) o;

    return user.name.equals(name) &&
            user.age == age &&
            user.passport.equals(passport);
}

//Idea from effective Java : Item 9
@Override
public int hashCode() {
    int result = 17;
    result = 31 * result + name.hashCode();
    result = 31 * result + age;
    result = 31 * result + passport.hashCode();
    return result;
}

}

If you're wondering why it's in a variable that gets modified instead of a single return statement, try writing it both ways and see which is easier to read, write and extend. — that other guy
– that other guy, Commented Sep 28, 2018 at 19:41
You aren't really assigning many values. You are just using the variable as a temporary holder. You could write this as return 31 * (31 * (31 * 17 + name.hashCode()) + age) + passport.hashCode(); but it just isn't as clear like that, it's harder to modify etc. — Andy Turner
– Andy Turner, Commented Sep 28, 2018 at 19:48
It's not very clear what you're confused by. The style? The logic? Be more specific. And the attitude is totally unnecessary and unhelpful. — shmosel
– shmosel, Commented Sep 28, 2018 at 19:48

Bill the Lizard · Accepted Answer · 2018-09-28 19:51:10Z

1

The successive assignments to result are to "scramble" the information contained in it, and to add more information from other sources. Note that result is on both sides of the assignment. Starting with a base value of 17, we then do this a few times:

result = 31 * result + someValue;

By multiplying result by 31, we move the existing value to the left a few bits, then add more bits by adding another value to it.

This is a very fast way to come up with a fairly unique hash code for every User object that you can define.

edited Sep 28, 2018 at 19:51

answered Sep 28, 2018 at 19:48

Bill the Lizard

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1 Comment

ikaerom Over a year ago

People claim that there is no more advantage to using this 17 + 31 * approach to hashCode() implementations, and using Objects.hashCode() is equivalent since JDK 1.7. Is that a statement that holds past the 21st year of the 21st century?

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