目录
2、为什么重写equals方法必须重写hashCode方法？
1、从Object类源码说起
equals和hashcode都是Object对象的方法。因此所有的Java类都会默认继承这两个方法。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™ 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，源码给出了一些常规协定：
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。也就是说，这两个方法，具有以下特性：
（1）如果两个对象相同（即用equals比较返回true），那么它们的hashCode值一定要相同；
（2）如果两个对象不同（即用equals比较返回false），那么它们的hashCode值可能相同也可能不同；
（3）如果两个对象的hashCode相同（存在哈希冲突），那么它们可能相同也可能不同(即equals比较可能是false也可能是true)；
（4）如果两个对象的hashCode不同，那么他们肯定不同(即用equals比较返回false)；
2、为什么重写equals方法必须重写hashCode方法？
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);
});
}（1）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;
}
}输出如下：
p的hashCode:356573597
p1的hashCode:1735600054
false
false
com.blueskyli.练习.Person@677327b6
888
com.blueskyli.练习.Person@1540e19d
888（2）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);
}
}输出如下：
p的hashCode:356573597
p1的hashCode:1735600054
true
false
com.blueskyli.练习.Person@677327b6
888
com.blueskyli.练习.Person@1540e19d
888（3）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();
}
}输出如下：
p的hashCode:3254239
p1的hashCode:3254239
true
false
com.blueskyli.练习.Person@31a7df
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，会使得在使用map的时候出现与预期不一样的结果。原因如下：
public V put(K key, V value) {
if (table == EMPTY_TABLE) {
inflateTable(threshold);
}
if (key == null)
return putForNullKey(value);
int hash = hash(key);
//这里通过哈希值定位到对象的大概存储位置
int i = indexFor(hash, table.length);
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
Object k;
//if语句中，先比较hashcode，再调用equals()比较
//由于“&&”具有短路的功能，只要hashcode不同，也无需再调用equals方法
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(hash, key, value, i);
return null;
}3、总结
重写equals()方法就重写hashCode()方法。重写hashcode方法是为了将数据存入HashSet / HashMap / Hashtable 类时进行比较，以免出现与预期不一样的结果。