标签：sts out over 定义 system getname 对象 set equal
讨论的问题：众所周知，set集合的特点是无重复的元素。一般Java类库中的类型比如String 类在添加到set中后，是不会出现重复现象的。那么自定义的类呢？
1.创建自定义类
public class Dog { private String name; private String kind; private String hobby; public Dog(String name,String kind,String hobby) { this.hobby=hobby; this.kind=kind; this.name=name; } public String getName() { return name; } public String getKind() { return kind; } public String getHobby() { return hobby; } @Override public String toString() { return "Dog [name=" + name + ", kind=" + kind + ", hobby=" + hobby + "]"; }
2.测试
public class TestSet { public static void main(String[] args) { Set<Dog> dogs=new HashSet<>(); Dog dog=new Dog("huahua","tianyuan","run"); Dog dog1=new Dog("tiantian","hashiqi","eat"); Dog dog2=new Dog("qiqi","jinmao","smell"); Dog dog3=new Dog("qiqi","jinmao","smell"); dogs.add(dog); dogs.add(dog1); dogs.add(dog2); dogs.add(dog3); System.out.println(dogs); } }
3.结果：
[Dog [name=qiqi, kind=jinmao, hobby=smell], Dog [name=huahua, kind=tianyuan, hobby=run], Dog [name=tiantian, kind=hashiqi, hobby=eat], Dog [name=qiqi, kind=jinmao, hobby=smell]]
4.分析：
可以看到在结构中有两个重复的元素。
自定义类的情况下，需要重写hashcode()和equals()方法，Object中的equals()仅能判断两个对象是否具有相同的引用，不能判断对象的内容细节，所以需要重写。
Note:hashcode是根据类的实例域生成的一个随机的整数，可为负。只要两个对象相同，他们的hashcode码就相同。
5.改进：在Dog类中重写如下方法：
@Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + ((hobby == null) ? 0 : hobby.hashCode()); result = prime * result + ((kind == null) ? 0 : kind.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; Dog other = (Dog) obj; if (hobby == null) { if (other.hobby != null) return false; } else if (!hobby.equals(other.hobby)) return false; if (kind == null) { if (other.kind != null) return false; } else if (!kind.equals(other.kind)) return false; if (name == null) { if (other.name != null) return false; } else if (!name.equals(other.name)) return false; return true; }
6.结果：
[Dog [name=tiantian, kind=hashiqi, hobby=eat], Dog [name=qiqi, kind=jinmao, hobby=smell], Dog [name=huahua, kind=tianyuan, hobby=run]]
不再有重复的元素！
标签：sts out over 定义 system getname 对象 set equal
原文地址：https://www.cnblogs.com/liu-chen/p/11877759.html
