JDK源码研究之-----------HashMap AND HashCode

package com.bin.jdk.util;

public class Person {

    private String id;

    private String name;

    private int age;

    public Person() {
        
    }

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

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public void info() {
        System.out.println("name:" + getName() + " age:" + getAge());
    }

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((id == null) ? 0 : id.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        Person person = (Person) obj;
        if (id == null) {
            if (person.getId() != null)
                return false;
        } else if (!id.equals(person.getId())) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "Person [id=" + id + ", name=" + name + ", age = " + age + "]";
    }
}

package com.bin.jdk.util;

import java.util.HashMap;
import java.util.Map;

import com.bin.util.map.MapUtil;

public class Test {

    public static void main(String[] args) {
        
        /**
         * 判断相同元素条件
         * 1、首先判断hashcode返回值是否一样
         * 2、如果hashcode值一样，再调用equals方法判断 
         */
        Map<Object,Object> map = new HashMap<Object,Object>();
        
        Person p = new Person("1","xx",10);
        Person p2 = new Person("1","mm",11);
        map.put(p, "person1");
        map.put(p2, "person2");
    
        String result = MapUtil.foreachByKey(map);
        System.out.println(map.size()+"  "+result); //1  Person [id=1, name=xx, age = 10]:person2;    id相同，因此p、p2会被认为是同一对象，值覆盖
        
        
        Person p3 = new Person("2","nn",22);
        map.put(p3, "person3");
        String result2 = MapUtil.foreachByKey(map);
        System.out.println(map.size()+"  "+result2); //2  Person [id=2, name=nn, age = 22]:person3;Person [id=1, name=xx, age = 10]:person2;   正常加入p3
        
        p3.setId("1");
        String result3 = MapUtil.foreachByKey(map);
        System.out.println(map.size()+"  "+result3); //2  Person [id=1, name=nn, age = 22]:person2;Person [id=1, name=xx, age = 10]:person2;   更改了引用，并没有影响原先的对象引用
        
        map.put(p3, "person4");
        String result4 = MapUtil.foreachByKey(map);
        System.out.println(map.size()+"  "+result4); //2  Person [id=1, name=nn, age = 22]:person4;Person [id=1, name=xx, age = 10]:person4;   
        
        p3.setId("2");
        
        Person p5 = new Person("2","xxc",23);
        Object o = map.get(p5);
        System.out.println(o.toString());//person3   原先id为2的引用对象任然存在
    }
}

 具体Hash算法可参考：http://www.ibm.com/developerworks/cn/java/j-lo-hash/