java中为什么重写equals时必须重写hashCode方法？

在上一篇博文Java中equals和==的区别中介绍了Object类的equals方法，并且也介绍了我们可在重写equals方法，本章我们来说一下为什么重写equals方法的时候也要重写hashCode方法。

先让我们来看看Object类源码

/**
     * Returns a hash code value for the object. This method is
     * supported for the benefit of hash tables such as those provided by
     * {@link java.util.HashMap}.
     * <p>
     * The general contract of {@code hashCode} is:
     * <ul>
     * <li>Whenever it is invoked on the same object more than once during
     *     an execution of a Java application, the {@code hashCode} method
     *     must consistently return the same integer, provided no information
     *     used in {@code equals} comparisons on the object is modified.
     *     This integer need not remain consistent from one execution of an
     *     application to another execution of the same application.
     * <li>If two objects are equal according to the {@code equals(Object)}
     *     method, then calling the {@code hashCode} method on each of
     *     the two objects must produce the same integer result.
     * <li>It is <em>not</em> required that if two objects are unequal
     *     according to the {@link java.lang.Object#equals(java.lang.Object)}
     *     method, then calling the {@code hashCode} method on each of the
     *     two objects must produce distinct integer results.  However, the
     *     programmer should be aware that producing distinct integer results
     *     for unequal objects may improve the performance of hash tables.
     * </ul>
     * <p>
     * As much as is reasonably practical, the hashCode method defined by
     * class {@code Object} does return distinct integers for distinct
     * objects. (This is typically implemented by converting the internal
     * address of the object into an integer, but this implementation
     * technique is not required by the
     * Java&trade; programming language.)
     *
     * @return  a hash code value for this object.
     * @see     java.lang.Object#equals(java.lang.Object)
     * @see     java.lang.System#identityHashCode
     */
    public native int hashCode();

/**
     * Indicates whether some other object is "equal to" this one.
     * <p>
     * The {@code equals} method implements an equivalence relation
     * on non-null object references:
     * <ul>
     * <li>It is <i>reflexive</i>: for any non-null reference value
     *     {@code x}, {@code x.equals(x)} should return
     *     {@code true}.
     * <li>It is <i>symmetric</i>: for any non-null reference values
     *     {@code x} and {@code y}, {@code x.equals(y)}
     *     should return {@code true} if and only if
     *     {@code y.equals(x)} returns {@code true}.
     * <li>It is <i>transitive</i>: for any non-null reference values
     *     {@code x}, {@code y}, and {@code z}, if
     *     {@code x.equals(y)} returns {@code true} and
     *     {@code y.equals(z)} returns {@code true}, then
     *     {@code x.equals(z)} should return {@code true}.
     * <li>It is <i>consistent</i>: for any non-null reference values
     *     {@code x} and {@code y}, multiple invocations of
     *     {@code x.equals(y)} consistently return {@code true}
     *     or consistently return {@code false}, provided no
     *     information used in {@code equals} comparisons on the
     *     objects is modified.
     * <li>For any non-null reference value {@code x},
     *     {@code x.equals(null)} should return {@code false}.
     * </ul>
     * <p>
     * The {@code equals} method for class {@code Object} implements
     * the most discriminating possible equivalence relation on objects;
     * that is, for any non-null reference values {@code x} and
     * {@code y}, this method returns {@code true} if and only
     * if {@code x} and {@code y} refer to the same object
     * ({@code x == y} has the value {@code true}).
     * <p>
     * Note that it is generally necessary to override the {@code hashCode}
     * method whenever this method is overridden, so as to maintain the
     * general contract for the {@code hashCode} method, which states
     * that equal objects must have equal hash codes.
     *
     * @param   obj   the reference object with which to compare.
     * @return  {@code true} if this object is the same as the obj
     *          argument; {@code false} otherwise.
     * @see     #hashCode()
     * @see     java.util.HashMap
     */
    public boolean equals(Object obj) {
        return (this == obj);
    }

hashCode：是一个native方法，返回的是对象的内存地址，

equals：对于基本数据类型，==比较的是两个变量的值。对于引用对象，==比较的是两个对象的地址。

接下来我们看下hashCode的注释

1.在 Java 应用程序执行期间，在对同一对象多次调用 hashCode 方法时，必须一致地返回相同的整数，前提是将对象进行 equals 比较时所用的信息没有被修改。　从某一应用程序的一次执行到同一应用程序的另一次执行，该整数无需保持一致。
2.如果根据 equals(Object) 方法，两个对象是相等的，那么对这两个对象中的每个对象调用 hashCode 方法都必须生成相同的整数结果。
3.如果根据 equals(java.lang.Object) 方法，两个对象不相等，那么两个对象不一定必须产生不同的整数结果。　但是，程序员应该意识到，为不相等的对象生成不同整数结果可以提高哈希表的性能。

从hashCode的注释中我们看到，hashCode方法在定义时做出了一些常规协定，即

1，当obj1.equals(obj2) 为 true 时，obj1.hashCode() == obj2.hashCode()

2，当obj1.equals(obj2) 为 false 时，obj1.hashCode() != obj2.hashCode()

hashcode是用于散列数据的快速存取，如利用HashSet/HashMap/Hashtable类来存储数据时，都是根据存储对象的hashcode值来进行判断是否相同的。如果我们将对象的equals方法重写而不重写hashcode，当我们再次new一个新的对象的时候，equals方法返回的是true，但是hashCode方法返回的就不一样了，如果需要将这些对象存储到结合中（比如：Set，Map ...）的时候就违背了原有集合的原则，下面让我们通过一段代码看下。

/**
     * @see Person
     * @param args
     */
    public static void main(String[] args)
    {
        HashMap<Person, Integer> map = new HashMap<Person, Integer>();

        Person p = new Person("jack",22,"男");
        Person p1 = new Person("jack",22,"男");

        System.out.println("p的hashCode:"+p.hashCode());
        System.out.println("p1的hashCode:"+p1.hashCode());
        System.out.println(p.equals(p1));
        System.out.println(p == p1);

        map.put(p,888);
        map.put(p1,888);
        map.forEach((key,val)->{
            System.out.println(key);
            System.out.println(val);
        });
    }

equals和hashCode方法的都不重写

public class Person
{
    private String name;

    private int age;

    private String sex;

    Person(String name,int age,String sex){
        this.name = name;
        this.age = age;
        this.sex = sex;
    }
}

p的hashCode:356573597
p1的hashCode:1735600054
false
false
com.blueskyli.练习[email protected]
888
com.blueskyli.练习[email protected]1540e19d
888

只重写equals方法

public class Person
{
    private String name;

    private int age;

    private String sex;

    Person(String name,int age,String sex){
        this.name = name;
        this.age = age;
        this.sex = sex;
    }

    @Override public boolean equals(Object obj)
    {
        if(obj instanceof Person){
            Person person = (Person)obj;
            return name.equals(person.name);
        }
        return super.equals(obj);
    }
}

p的hashCode:356573597
p1的hashCode:1735600054
true
false
com.blueskyli.练习[email protected]
888
com.blueskyli.练习[email protected]1540e19d
888

equals和hashCode方法都重写

public class Person
{
    private String name;

    private int age;

    private String sex;

    Person(String name,int age,String sex){
        this.name = name;
        this.age = age;
        this.sex = sex;
    }

    @Override public boolean equals(Object obj)
    {
        if(obj instanceof Person){
            Person person = (Person)obj;
            return name.equals(person.name);
        }
        return super.equals(obj);
    }

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

p的hashCode:3254239
p1的hashCode:3254239
true
false
com.blueskyli.练习[email protected]
888

我们知道map是不允许存在相同的key的，由上面的代码可以知道，如果不重写equals和hashCode方法的话会使得你在使用map的时候出现与预期不一样的结果，具体equals和hashCode如何重写，里面的逻辑如何实现需要根据现实当中的业务来规定。

总结：

1，两个对象，用==比较比较的是地址，需采用equals方法（可根据需求重写）比较。

2，重写equals()方法就重写hashCode()方法。

3，一般相等的对象都规定有相同的hashCode。

4，String类重写了equals和hashCode方法，比较的是值。

5，重写hashcode方法为了将数据存入HashSet/HashMap/Hashtable（可以参考源码有助于理解）类时进行比较

原文地址：https://www.cnblogs.com/blueskyli/p/9936076.html