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。