equals()：用来判断两个对象是否相等；注意不是**==**，==一般是比较内存地址
hashCode():获取哈希码，也称为散列码，返回一个int整数。这个哈希码的作用是通过该对象在哈希表中的索引位置。
直接看如下的几个例子
例子1:
class A{
public String name;
public A(String name) {
this.name = name;
}
// @Override
// public boolean equals(Object o) {
// if (this == o) return true;
// if (o == null || getClass() != o.getClass()) return false;
// A a = (A) o;
// return Objects.equals(name, a.name);
// }
// @Override
// public int hashCode() {
// return Objects.hash(name);
// }
}
public class Test {
public static void main(String[] args) {
Map<A, Integer> map = new HashMap<>(16);
A a1 = new A("A");
map.put(a1, 1);
A a2 = new A("A");
map.put(a2, 2);
System.out.println("a1 hashcode:" + a1.hashCode());
System.out.println("a2 hashcode:" + a2.hashCode());
System.out.println("equals:" + a1.equals(a2));
System.out.println("v1:" + map.get(a1));
System.out.println("v2:" + map.get(a2));
}
}
输出
a1 hashcode:1456208737
a2 hashcode:288665596
equals:false
v1:1
v2:2
A对象在业务上看来是一个对象，相等的；但是equals方法和hashcode方法都没有重写，实际是不等的，hashcode也不同；在HashMap中是两个key存储；
例子2:
class A{
public String name;
public A(String name) {
this.name = name;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
A a = (A) o;
return Objects.equals(name, a.name);
}
// @Override
// public int hashCode() {
// return Objects.hash(name);
// }
}
public class Test {
public static void main(String[] args) {
Map<A, Integer> map = new HashMap<>(16);
A a1 = new A("A");
map.put(a1, 1);
A a2 = new A("A");
map.put(a2, 2);
System.out.println("a1 hashcode:" + a1.hashCode());
System.out.println("a2 hashcode:" + a2.hashCode());
System.out.println("equals:" + a1.equals(a2));
System.out.println("v1:" + map.get(a1));
System.out.println("v2:" + map.get(a2));
}
}
输出：
a1 hashcode:1456208737
a2 hashcode:288665596
equals:true
v1:1
v2:2
只重写了equals方法，两个对象相等，但是hashcode不同，在HashMap中是两个key存储, 取出来不同；
那么只重写equals方法，两个相等对象会出现一样的hashcode产生误解吗；例子3
class A{
public String name;
public A(String name) {
this.name = name;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
A a = (A) o;
return Objects.equals(name, a.name);
}
// @Override
// public int hashCode() {
// return Objects.hash(name);
// }
}
public class Test {
public static void main(String[] args) {
Map<A, Integer> map = new HashMap<>(16);
A a1 = new A("A");
map.put(a1, 1);
A a2 = new A("A");
map.put(a2, 2);
System.out.println("a1 hashcode:" + a1.hashCode());
System.out.println("a2 hashcode:" + a2.hashCode());
System.out.println("equals:" + a1.equals(a2));
System.out.println("v1:" + map.get(a1));
System.out.println("v2:" + map.get(a2));
Integer v3 = map.get(new A("A"));
int times = 0;
while(v3 == null){
v3 = map.get(new A("A"));
times ++;
}
System.out.println("v3:" + v3 + " times:" + times);
}
}
输出：
a1 hashcode:1456208737
a2 hashcode:288665596
equals:true
v1:1
v2:2
v3:1 times:33872685
这就和例子2出现不同的结果了，都是new A("a")这个对象，在33872685次后的某个new A("a")和a1对象的hashcode值一样了；那么可以想象他也可以取到和a2一样的hashcode值，那么结果就是2了。逻辑上也是完全无法接受new A("a')会出现不同值的
例子4： 重写equals和hashcode
class A{
public String name;
public A(String name) {
this.name = name;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
A a = (A) o;
return Objects.equals(name, a.name);
}
@Override
public int hashCode() {
return Objects.hash(name);
}
}
public class Test {
public static void main(String[] args) {
Map<A, Integer> map = new HashMap<>(16);
A a1 = new A("A");
map.put(a1, 1);
A a2 = new A("A");
map.put(a2, 2);
System.out.println("a1 hashcode:" + a1.hashCode());
System.out.println("a2 hashcode:" + a2.hashCode());
System.out.println("equals:" + a1.equals(a2));
System.out.println("v1:" + map.get(a1));
System.out.println("v2:" + map.get(a2));
Integer v3 = map.get(new A("A"));
int times = 0;
while(v3 == null){
v3 = map.get(new A("A"));
times ++;
}
System.out.println("v3:" + v3 + " times:" + times);
}
}
输出
a1 hashcode:96
a2 hashcode:96
equals:true
v1:2
v2:2
v3:2 times:0
可以看到，new A("a')总是相等且有相同的hashcode值的，那么Map中只有这么一个key；即每次业务都是合理的，某个对象只有一个存储到map中，以最后的更新值为准，不会出现什么意外。
综合上述的例子，也就说明了为什么重写equals方法必须重写hashcode方法。而且也是强烈建议业务对象务必按照实际意义对这两个方法进行重写。
另外回顾一下HashMap的put方法: 添加元素重复性校验的标准是先检查hashCode值，后判断equals()
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
if (e != null) {
// existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}