Cautions while overriding equals and hashcode - Java

The theory (for the language lawyers and the mathematically inclined):

equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, andtransitive). In addition, it must be consistent (if the objects are not modified, then it must keep returning the same value). Furthermore, o.equals(null) must always return false.

hashCode() (javadoc) must also be consistent (if the object is not modified in terms of equals(), it must keep returning the same value).

The relation between the two methods is:

Whenever a.equals(b), then a.hashCode() must be same as b.hashCode().

In practice:

If you override one, then you should override the other.

Use the same set of fields that you use to compute equals() to compute hashCode().

Use the excellent helper classes EqualsBuilder and HashCodeBuilder from the Apache Commons Lang library. An example:

public class Person {
    private String name;
    private int age;
    // ...

    @Override
    public int hashCode() {
        return new HashCodeBuilder(17, 31). // two randomly chosen prime numbers
            // if deriving: appendSuper(super.hashCode()).
            append(name).
            append(age).
            toHashCode();
    }

    @Override
    public boolean equals(Object obj) {
       if (!(obj instanceof Person))
            return false;
        if (obj == this)
            return true;

        Person rhs = (Person) obj;
        return new EqualsBuilder().
            // if deriving: appendSuper(super.equals(obj)).
            append(name, rhs.name).
            append(age, rhs.age).
            isEquals();
    }
}

Also remember:

When using a hash-based Collection or Map such as HashSet, LinkedHashSet, HashMap,Hashtable, or WeakHashMap, make sure that the hashCode() of the key objects that you put into the collection never changes while the object is in the collection. The bulletproof way to ensure this is to make your keys immutable, which has also other benefits.

Source: Stackoverflow

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Yes, it is important if your item will be used as a key in a dictionary, or HashSet<T>, etc - since this is used (in the absence of a custom IEqualityComparer<T>) to group items into buckets. If the hash-code for two items does not match, they may never be considered equal (Equals will simply never be called).

The GetHashCode() method should reflect the Equals logic; the rules are:

• if two things are equal (Equals(...) == true) then they must return the same value for GetHashCode()
• if the GetHashCode() is equal, it is not necessary for them to be the same; this is a collision, and Equals will be called to see if it is a real equality or not.

In this case, it looks like "return FooId;" is a suitable GetHashCode() implementation. If you are testing multiple properties, it is common to combine them using code like below, to reduce diagonal collisions (i.e. so that new Foo(3,5) has a different hash-code to new Foo(5,3)):

int hash = 13;
hash = (hash * 7) + field1.GetHashCode();
hash = (hash * 7) + field2.GetHashCode();
...
return hash;

Oh - for convenience, you might also consider providing == and != operators when overriding Equals and GetHashCode.