正确重写equals和hashcode方法

1. 重写的原因

如有个User对象如下：

public class User {private String name;private Integer age;
}

如果不重写equals方法和hashcode方法，则：

public static void main(String[] args) {User user1 = new User("userA", 30);User user2 = new User("userB", 30);System.out.println(user1.equals(user2)); // falseSystem.out.println(user1.hashCode() == user2.hashCode()); // falseSystem.out.println(user1.hashCode());System.out.println(user2.hashCode());}

2. 重写方法

equals:

@Override
public boolean equals(Object obj) {if (null == obj) {return false;}if (obj == this) {return true;}if (!(obj instanceof User)) {return false;}User anoUser = (User) obj;if (Objects.equals(this.name, anoUser.name) && Objects.equals(this.age, anoUser.age)) {return true;}return false;
}

hashcode:

@Override
public int hashCode() {return Objects.hash(this.name, this.age);
}// 或
@Override
public int hashCode() {int result = 1;if (this.name == null && this.age == null) {return result;}result = 31 * result + (this.name == null ? 0 : this.name.hashCode());result = 31 * result + (this.age == null ? 0 : this.age.hashCode());return result;
}

如果hashcode返回一个固定值，则HashMap每次都要找同一个位置，导致链表很长，效率很低。