import java.util.Set;

import java.util.HashSet;

public class SetTest {

	public static void main(String[] args) {

		/*

		 *对于用户自己定义类型的数据放在容器（Set）中

		 *务必重写equals和hashCode方法

		 *要不然stu1和stu2放在容器中，和觉得是两个不同的元素

		 **/

		//set中存放的元素是无序的

		//set中存储的元素是不能够反复的（依据equals方法和hashCode方法推断）

		Set set = new HashSet();

		Student stu1 = new Student(1, "aaa");

		Student stu2 = new Student(1, "aaa");

		Student stu3 = new Student(2, "ccc");

		Student stu4 = new Student(8, "fff");

		set.add(stu1);

		set.add(stu2);

		set.add(stu3);

		set.add(stu4);

		System.out.println(set);

	}

}

class Student {

	private int id;

	private String name;

	public Student(int id, String name) {

		this.id = id;

		this.name = name;

	}

	@Override

	public String toString() {

		return this.id + " " + this.name;

	}

	@Override

	public boolean equals(Object obj) {

		Student stu = (Student) obj;

		return this.id == stu.id && this.name.equals(stu.name);

	}

	@Override

	public int hashCode() {

		return this.id*this.name.hashCode();

	}

}

输出结果：

[8 fff, 1 aaa, 2 ccc]

假设不重写hashCode和equals方法

import java.util.Set;

import java.util.HashSet;

public class SetTest {

	public static void main(String[] args) {

		/*

		 *对于用户自己定义类型的数据放在容器（Set）中

		 *务必重写equals和hashCode方法

		 *要不然stu1和stu2放在容器中，和觉得是两个不同的元素

		 **/

		//set中存放的元素是无序的

		//set中存储的元素是不能够反复的（依据equals方法和hashCode方法推断）

		Set set = new HashSet();

		Student stu1 = new Student(1, "aaa");

		Student stu2 = new Student(1, "aaa");

		Student stu3 = new Student(2, "ccc");

		Student stu4 = new Student(8, "fff");

		set.add(stu1);

		set.add(stu2);

		set.add(stu3);

		set.add(stu4);

		System.out.println(set);

	}

}

class Student {

	private int id;

	private String name;

	public Student(int id, String name) {

		this.id = id;

		this.name = name;

	}

	@Override

	public String toString() {

		return this.id + " " + this.name;

	}

}

输出结果：

[1 aaa, 1 aaa, 8 fff, 2 ccc]