在HashMap與HashSet中，是不允許元素重復(fù)的。

在判定時比較元素的hashCode()是否相等，equals()是否返回true，那么問題來了

?

這是一個Person類

public class Person {
?? ?private String id;
?? ?private String name;
?? ?private Integer age;
?? ?public String getId() {
?? ??? ?return id;
?? ?}
?? ?public void setId(String id) {
?? ??? ?this.id = id;
?? ?}
?? ?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;
?? ?}
?? ?@Override
?? ?public String toString() {
?? ??? ?return "Person [id=" + id + ", name=" + name + ", age=" + age + "]";
?? ?}
?? ?@Override
?? ?public int hashCode() {
?? ??? ?final int prime = 31;
?? ??? ?int result = 1;
?? ??? ?result = prime * result + ((age == null) ? 0 : age.hashCode());
?? ??? ?result = prime * result + ((id == null) ? 0 : id.hashCode());
?? ??? ?result = prime * result + ((name == null) ? 0 : name.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 other = (Person) obj;
?? ??? ?if (age == null) {
?? ??? ??? ?if (other.age != null)
?? ??? ??? ??? ?return false;
?? ??? ?} else if (!age.equals(other.age))
?? ??? ??? ?return false;
?? ??? ?if (id == null) {
?? ??? ??? ?if (other.id != null)
?? ??? ??? ??? ?return false;
?? ??? ?} else if (!id.equals(other.id))
?? ??? ??? ?return false;
?? ??? ?if (name == null) {
?? ??? ??? ?if (other.name != null)
?? ??? ??? ??? ?return false;
?? ??? ?} else if (!name.equals(other.name))
?? ??? ??? ?return false;
?? ??? ?return true;
?? ?}
?? ?/*@Override
?? ?public int hashCode() {
?? ??? ?System.out.println("id = " + id + ";id.hashCode = " + id.hashCode());
?? ??? ?return id.hashCode();
?? ?}
?? ?@Override
?? ?public boolean equals(Object obj) {
?? ??? ?System.out.println("equals");
?? ??? ?Person p = (Person)obj;
?? ??? ?if(this.id == p.id) {
?? ??? ??? ?return true;
?? ??? ?}else{
?? ??? ??? ?return false;
?? ??? ?}
?? ?}*/
}

?

這是一個Test類

public class TestHashCode {
?? ?public static void main(String[] args) {
?? ??? ?Set<Person> set = new HashSet<Person>();
?? ??? ?Person p1 = new Person();
?? ??? ?p1.setId("11");
?? ??? ?p1.setName("張三");
?? ??? ?p1.setAge(20);
?? ??? ?Person p2 = new Person();
?? ??? ?p2.setId("11");
?? ??? ?p2.setName("李四");
?? ??? ?p2.setAge(30);
?? ??? ?set.add(p1);
?? ??? ?set.add(p2);
?? ??? ?System.out.println(p1.hashCode() == p2.hashCode());
?? ??? ?System.out.println(p1.equals(p2));
?? ??? ?for (Person person : set) {
?? ??? ??? ?System.out.println(person.toString());
?? ??? ?}
?? ?}
}

?

如果Person的id屬性作為主鍵，且只重寫equals方法，我們認(rèn)為id相等的兩個對象為同一個對象

此時p1.equals(p2)返回true,但是hashCode()卻不相等，而hashSet判斷唯一的依據(jù)是hashCode()值相等且equals()返回true

所以我們的預(yù)期是p1和p2為同一個對象，但是hashSet認(rèn)為他們是兩個對象。

所以我們需要同時重寫hashCode()和equals()方法。

轉(zhuǎn)載于:https://www.cnblogs.com/cz305679760/p/4560137.html

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