何时覆盖hashCode()和equals()方法

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

equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, and transitive). 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.

无论何时覆盖equals方法都需覆盖hashcode方法，并保持一致（equals为true, hashcode也一致），当需要把自己定义的类加到以hash的集合中时（HashMap会先比较hashcode, code一致时比较equals方法），如果要保持只有一个相同的实例时，需要覆盖这两方法。