java.util.HashMap中的key为对象的问题，对象key获得相应的value值

java.util.HashMap中的key为对象的问题，对象key获得相应的value值
重写对象key的equals和hashCode的方法

package com;import java.util.HashMap;import java.util.Map;public class AMain {    public static void main(String[] args) {        Map<User, Integer> map = new HashMap<>();        String name = "aa";        // String name1 = "aa";        // System.out.println(name.hashCode());        // System.out.println(name1.hashCode());        Integer age = 123;        Integer result = null;        User u1 = new User();        u1.setName(name);        u1.setAge(age);        map.put(u1, 1000);        result = map.get(u1);        System.out.println(result);        User u2 = new User();        u2.setName(name);        u2.setAge(age);        result = map.get(u2);        System.out.println(result);        result = map.get(u1);        System.out.println(result);    }}

package com;public class User {    private String name;    private Integer age;    @Override    public boolean equals(Object obj) {        boolean result = false;        if (this == obj)            result = true;        if (obj == null || getClass() != obj.getClass())            result = false;        User user = (User) obj;        if (user.getName() == null || user.getAge() == null)            result = false;        if (user.getName().equals(name) && user.getAge().equals(age))            result = true;        System.out.println("equals：" + result);        return result;    }    @Override    public int hashCode() {        int a = 0;        if (name != null && age != null) {            a = name.hashCode() + age.hashCode();        }        if (name != null) {            a = name.hashCode();        }        if (age != null) {            a = age.hashCode();        }        System.out.println("hashCode：" + a);        return a;    }    public String getName() {        return name;    }    public void setName(String name) {        this.name = name;    }    public Integer getAge() {        return age;    }    public void setAge(Integer age) {        this.age = age;    }}

原因：

 /**     * Implements Map.get and related methods     *     * @param hash hash for key     * @param key the key     * @return the node, or null if none     */    final Node<K,V> getNode(int hash, Object key) {        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;        if ((tab = table) != null && (n = tab.length) > 0 &&            (first = tab[(n - 1) & hash]) != null) {            if (first.hash == hash && // always check first node                ((k = first.key) == key || (key != null && key.equals(k))))                return first;            if ((e = first.next) != null) {                if (first instanceof TreeNode)                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);                do {                    if (e.hash == hash &&                        ((k = e.key) == key || (key != null && key.equals(k))))                        return e;                } while ((e = e.next) != null);            }        }        return null;    }

HashMap在根据key获得相应的value的时候，用的hash值和equals来判断是否是同一个key