Override public int hashCode() using BigInteger

Question

A BigInteter is too big to be converted into an integer. But I have to store objects with an id (SHA 512) in a HashMap and need a hash function without many collisions.

I tried this. however, I am not sure if there is no clustering somewhere.

@Override
public boolean equals(Object obj) {
    if (this == obj) {
        return true;
    }
    if (obj == null) {
        return false;
    }
    if (getClass() != obj.getClass()) {
        return false;
    }
    Advertisement other = (Advertisement) obj;
    return this.getId().equals(other.getId());
}

@Override
public int hashCode() {
    return new BigInteger(getId(), 16).hashCode();
}

would be a cast to an integer (bi.intValue()) more efficient?

Bohemian · Accepted Answer · 2012-04-24 00:18:51Z

6

Don't try to reinvent the wheel - simply use getId().hashCode():

@Override
public int hashCode() {
    return getId().hashCode();
}

String.hashCode() uses an efficient, high-quality hashing algorithm, so it's the best choice. And it makes your code simpler, which is always a good thing.

edited Apr 24, 2012 at 0:18

answered Apr 23, 2012 at 23:38

Bohemian♦

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