【java】集合操作中的retainAll的坑

JDK1.8

首先，set1中有值， set2中无值

import com.alibaba.fastjson.JSON;
import com.google.common.collect.Sets;
import java.util.*;



public class ListTest {

    public static void main(String[] args) throws Exception {

        Set<Long> set1 = new HashSet<>();
        set1.add(1L);
        set1.add(2L);

        Set<Long> set2 = new HashSet<>();


        Sets.SetView<Long> intersection = Sets.intersection(set1, set2);
        System.out.println(JSON.toJSONString(intersection));

        System.out.println(JSON.toJSONString(set1));
        System.out.println(JSON.toJSONString(set2));


        set1.retainAll(set2);
        System.out.println(JSON.toJSONString(set1));
        System.out.println(JSON.toJSONString(set2));

    }
}

结果：

【java】集合操作中的retainAll的坑

再来一次，set2中放值

public static void main(String[] args) throws Exception {

        Set<Long> set1 = new HashSet<>();
        set1.add(1L);
        set1.add(2L);

        Set<Long> set2 = new HashSet<>();
        set2.add(1L);

        Sets.SetView<Long> intersection = Sets.intersection(set1, set2);
        System.out.println(JSON.toJSONString(intersection));

        System.out.println(JSON.toJSONString(set1));
        System.out.println(JSON.toJSONString(set2));


        set1.retainAll(set2);
        System.out.println(JSON.toJSONString(set1));
        System.out.println(JSON.toJSONString(set2));

    }

结果如下

【java】集合操作中的retainAll的坑