从多种方法中选择随机方法

Question

我试图从类中创建的方法中选择一个随机方法。有没有办法创建ArrayList并传递方法？我试图这样做，但是当我尝试将该方法添加到数组时，我收到错误。

public class Monkey{

    private int energy;

    String[] food = {"Meat", "Fish", "Bugs", "Grain"};
    ArrayList<Activity> monkeyActivity = new ArrayList<>();

    public Monkey(int energy) {
        this.energy = energy;
    }

    public int getEnergy() {
        System.out.println("Monkey energy level: " + energy);
        return energy;
    }

    public void sound() {
        System.out.println("Monkey: Oooo Oooo~!");
        energy -= 3;
        monkeyActivity.add(sound()); //I get an error message here when trying 
                                 //to add the method to the array
    }

    public void play(){
        if (energy >= 8){
            System.out.println("Monkey begins to play.");
            energy -= 8;
        }else {
            System.out.println("Monkey does not have enough energy to play");
        }
        System.out.println("Energy remaining: " + energy);
    }

    public void eat(){
        Random random = new Random();
        int index = random.nextInt(food.length);
        System.out.println("Monkey beings to eat " + food[index]);
        energy += 5;
        System.out.println("Energy remaining: " + energy);
    }

    public void sleep(){
        System.out.println("Monkey is sleeping: Zzz...");
        energy += 10;
        System.out.println("Energy remaining: " + energy);
    }
}

这是我为通用Activity创建的单独类。

public class Activity {

    private String sleep;
    private String eat;
    private String sound;
    private String play;

    public Activity(String sleep, String eat, String sound, String play) {
        this.sleep = sleep;
        this.eat = eat;
        this.sound = sound;
        this.play = play;
    }

    public String getSleep() {
        return sleep;
    }

    public String getEat() {
        return eat;
    }

    public String getSound() {
        return sound;
    }

    public String getPlay() {
        return play;
    }

    public void setSleep(String sleep) {
        this.sleep = sleep;
    }

    public void setEat(String eat) {
        this.eat = eat;
    }

    public void setSound(String sound) {
        this.sound = sound;
    }

    public void setPlay(String play) {
        this.play = play;
    }
}

Answer 1

你正在混淆概念。

技术问题：

返回值冲突

public void sound() {
    // ...
    monkeyActivity.add(sound());

方法sound()的返回值是 void （这意味着没有返回值），但是您尝试添加其（不存在的）返回值作为List的元素。这是你的编译器抱怨的。

非预期的递归

public void sound() {
    System.out.println("Monkey: Oooo Oooo~!");
    energy -= 3;
    monkeyActivity.add(sound());

在最后一行中，您执行递归调用，这意味着您调用与此代码完全相同的方法。如果这种情况无意中发生，则几乎会产生StackOverflowError。

在没有适当分析的情况下编写课程

你有一个班级Activity。 但是如果你仔细看看这不是单个活动（正如类名所暗示的那样），但它是所有可能的活动。

因此，您的收藏集monkeyActivity无法将单项活动作为元素。

做一个狂野的猜测我认为你想要的更像是这样：

interface Activity{
  void do();
}

public class Monkey{
    private int energy;
    String[] food = {"Meat", "Fish", "Bugs", "Grain"};
    List<Activity> monkeyActivity = new ArrayList<>();
    // ...        
    public void sound() {
        monkeyActivity.add(new Activity(){
              public void do(){                
                System.out.println("Monkey: Oooo Oooo~!");
                energy -= 3;
              }
           }); 
    }

Answer 2

您可以将每个方法存储为Runnable，与任何＆＃34;操作＆＃34;是no-arg void方法，满足Runnable功能接口：

List<Runnable> actions = Arrays.asList(this::sound, this::play, this::eat, this::sleep);

执行随机方法，只需：

Random rnd = new Random();
actions.get(rnd.nextInt(actions.size())).run();