在Android Studio上与Jack和Java1.8一起使用时，sort（list，comparator）不反映更新的列表

Person.java

package com.example.android.cardview.HackerRank;

/**
* Created by Chhiring on 10/5/2017.
*/

public class Person {
   private String mFirstName;
   private String mLastNme;

   public Person() {
   }

   public Person(String mFirstName, String mLastNme) {
       this.mFirstName = mFirstName;
       this.mLastNme = mLastNme;
   }

   public String getmFirstName() {
       return mFirstName;
   }

   public void setmFirstName(String mFirstName) {
       this.mFirstName = mFirstName;
   }

   public String getmLastNme() {
       return mLastNme;
   }

   public void setmLastNme(String mLastNme) {
       this.mLastNme = mLastNme;
   }
}

package com.example.android.cardview.HackerRank;

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

/**
* Created by Chhiring on 10/9/2017.
*/
public class LambdaExpressionExample {
   public static void main(String[] args) {

       List<Person> myList = new ArrayList<>();

       myList.add(new Person("Chhiring", "Gurung"));
       myList.add(new Person("Jon", "Skeet"));
       myList.add(new Person("Joseph", "Rooney"));
       myList.add(new Person("Jack","Johnson"));

       System.out.println("Sort Based on first name");
       Collections.sort(myList, (o1, o2) -> o1.getmFirstName().compareTo(o2.getmFirstName()));
       for(Person p:myList){
           System.out.println(p.getmFirstName()+" "+p.getmLastNme());
       }

       System.out.println("\nSort Based on last name");
       Collections.sort(myList, (o1, o2) -> o1.getmLastNme().compareTo(o2.getmLastNme()));
       for(Person p:myList){
           System.out.println(p.getmFirstName()+" "+p.getmLastNme());
       }
   }
}

LambdaExpressionExample.java

package com.example.android.cardview.HackerRank;

/**
* Created by Chhiring on 10/5/2017.
*/

public class Person {
   private String mFirstName;
   private String mLastNme;

   public Person() {
   }

   public Person(String mFirstName, String mLastNme) {
       this.mFirstName = mFirstName;
       this.mLastNme = mLastNme;
   }

   public String getmFirstName() {
       return mFirstName;
   }

   public void setmFirstName(String mFirstName) {
       this.mFirstName = mFirstName;
   }

   public String getmLastNme() {
       return mLastNme;
   }

   public void setmLastNme(String mLastNme) {
       this.mLastNme = mLastNme;
   }
}

package com.example.android.cardview.HackerRank;

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

/**
* Created by Chhiring on 10/9/2017.
*/
public class LambdaExpressionExample {
   public static void main(String[] args) {

       List<Person> myList = new ArrayList<>();

       myList.add(new Person("Chhiring", "Gurung"));
       myList.add(new Person("Jon", "Skeet"));
       myList.add(new Person("Joseph", "Rooney"));
       myList.add(new Person("Jack","Johnson"));

       System.out.println("Sort Based on first name");
       Collections.sort(myList, (o1, o2) -> o1.getmFirstName().compareTo(o2.getmFirstName()));
       for(Person p:myList){
           System.out.println(p.getmFirstName()+" "+p.getmLastNme());
       }

       System.out.println("\nSort Based on last name");
       Collections.sort(myList, (o1, o2) -> o1.getmLastNme().compareTo(o2.getmLastNme()));
       for(Person p:myList){
           System.out.println(p.getmFirstName()+" "+p.getmLastNme());
       }
   }
}

---

但是，当我重建并运行时，我会看到更新的结果。为什么会这样

我不理解这个问题。您看到了什么以及希望看到什么？运行不会更新我更改的列表。它显示列表中以前的值。重新生成并运行后，它将更新列表值。为什么？我只看到一套插页。以前的值是什么？更新后的值是什么？@chhring它可能看起来很迂腐，但即使它看起来很琐碎，您仍然需要在您的问题中包括观察到的输出和预期的。。。后者并不像我们很容易看到的那样重要，因为它需要基于代码，但请包括前者（观察到的）。您可以在顶部@ValentinRuano上看到Person.java类