浅析重不重写hashcode和equals对于HashSet添加元素的影响

一，重写hashcode和equals的数据结构

通俗来说，hashcode决定了新天街的元素挂在数组那个位置，equals比较当前索引下两个内容是否相等，相等新添加的不成功，决定了能不能加。

二，重不重写hashcode和equals对于HashSet添加元素的影响（可否添加重复元素）

实验一：不重写hashcode的equals

import java.util.HashSet;

public class DemoHashSet {

    public static void main(String[] args) {
        HashSet<Student> students = new HashSet<>();
        Student s1 = new Student("张明",12);
        Student s2 = new Student("张杰",14);
        Student s3 = new Student("张杰",14);
        students.add(s1);
        students.add(s2);
        students.add(s3);
        for (Student student : students) {

            System.out.println(student.getName()+student.getAge()+"岁");
        }
    }

}
import java.util.Objects;

public class Student {
private String name;
private int age;

    public Student(String name, int age) {
        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;
    }
/*

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Student)) return false;
        Student student = (Student) o;
        return getAge() == student.getAge() &&
                Objects.equals(getName(), student.getName());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getName(), getAge());
    }
*/
}

结果：

张杰14岁
张明12岁
张杰14岁

Process finished with exit code 0

实验二：重写hashcode和equals

import java.util.HashSet;

public class DemoHashSet {

    public static void main(String[] args) {
        HashSet<Student> students = new HashSet<>();
        Student s1 = new Student("张明",12);
        Student s2 = new Student("张杰",14);
        Student s3 = new Student("张杰",14);
        students.add(s1);
        students.add(s2);
        students.add(s3);
        for (Student student : students) {

            System.out.println(student.getName()+student.getAge()+"岁");
        }
    }

}
import java.util.Objects;

public class Student {
private String name;
private int age;

    public Student(String name, int age) {
        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;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Student)) return false;
        Student student = (Student) o;
        return getAge() == student.getAge() &&
                Objects.equals(getName(), student.getName());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getName(), getAge());
    }
}

结果：

张明12岁
张杰14岁

没有出现重复元素。

三。原因分析：

HashMap底层

add-》put-》hashmap

    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;
    }

之前的总结

重写前：2号是==比较地址值不同false，不重写equals比较地址值3号false，结果false，向下执行，添加元素。

重写了hashcode，比较hash值，地址值肯定不同2号false，equal比较的是成员信息3号为true，结果true，执行括号中的，不添加。

 hashmap也是同理

import java.util.HashMap;
import java.util.HashSet;

public class DemoHashMap {
    public static void main(String[] args) {
        HashMap<String,Integer> students = new HashMap<>();
        Student s1 = new Student("张明",12);
        Student s2 = new Student("张杰",14);
        Student s3 = new Student("张杰",14);
        students.put("张明",12);
        students.put("王明",12);
        students.put("张明",12);
        String s = students.toString();

        System.out.println(s);}
}

结果：

{张明=12, 王明=12}

 hashmap的键如果是自定义类型不要我忘记重写hashcode和equals。

hashmap是不同步，可以有null键null值

hashtable是同步的，不可以有null键null值。