Object 裡面有個hashcode 方法

如以下程式碼：

package com.collectiondo;

class Person {

   private Integer age;
   private String  name;

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

}


public class TestHash {
    public static void main(String[] args) {
        Person person1 
 
= new Person(19,"路飛");
        Person person2 = new Person(19,"路飛");

        System.out.println("person1:"+person1.hashCode());//1173230247
        System.out.println("person2:"+person2.hashCode());//1173230247

        //String 重寫了hashcode的方法
        String m=new String("路飛");
        String n="路飛";
        System.out.println(
 
"m:"+m.hashCode());//1165519
        System.out.println("n:"+n.hashCode());//1165519


        //這兩個是特殊的
        String str1 = "通話";
        String str2 = "重地";
        System.out.println("通話："+"通話".hashCode());//雜湊值1179395
        System.out.println("重地："+"重地".hashCode());//雜湊值1179395
    }
}

hashset 儲存資料的結構如下圖

hashset 不能存重複元素的原理，如下圖

用hashset 儲存自定義型別的資料（比如自己定義的物件，person ,student）

以下是示例：

package com.collectiondo;
import java.util.HashSet;
import java.util.Set;

class Person {

   private Integer age;
   private String  name;

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

    @Override
    public String toString() {
        return "Person{" +
                "age=" + age +
                ", name='" + name + '\'' +
                '}';
    }
}

public class TestHash {
    public static void main(String[] args) {
        Set pset=new HashSet<Person>();
        Person person1 = new Person(19,"路飛");
        Person person2 = new Person(19,"路飛");

        pset.add(person1);
        pset.add(person2);

        System.out.println("person1:"+person1.hashCode());//1173230247
        System.out.println("person2:"+person2.hashCode());//1173230247
        System.out.println("pset:"+pset);
}

輸出結果是：

person1:1173230247
person2:856419764
pset:[Person{age=19, name='路飛'}, Person{age=19, name='路飛'}]

從輸出的結果可以看出來，兩個person 雖然裡面的name 和age 屬性的值都是一樣的，但是由於hashcode 是不一樣的，所以被認為是不同的兩個例項。

顯然這樣是不合適的。所以我們要來改造一下person 類，在person 類裡面重寫hashcode 方法和equals方法

class Person {

   private Integer age;
   private String  name;

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

    @Override
    public String toString() {
        return "Person{" +
                "age=" + age +
                ", name='" + name + '\'' +
                '}';
    }
    //重寫equals方法
    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        Person person = (Person) o;
        return age.equals(person.age) &&
                name.equals(person.name);
    }
    //重寫hashCode方法
    @Override
    public int hashCode() {
        return Objects.hash(age, name);
    }
}

測試程式碼：

public class TestHash {
    public static void main(String[] args) {
        Set pset=new HashSet<Person>();

        Person person1 = new Person(19,"路飛");
        Person person2 = new Person(19,"路飛");
        Person person3= new Person(18,"娜美");
        Person person4 = new Person(20,"索隆");
        Person person5 = new Person(21,"山治");


        pset.add(person1);
        pset.add(person2);
        pset.add(person3);
        pset.add(person4);
        pset.add(person5);

        System.out.println("person1:"+person1.hashCode());//1173230247
        System.out.println("person2:"+person2.hashCode());//1173230247
        System.out.println("pset:"+pset);

    }
}

輸出的結果是：

person1:1167069
person2:1167069
pset:[Person{age=18, name='娜美'}, Person{age=19, name='路飛'}, Person{age=21, name='山治'}, Person{age=20, name='索隆'}]