如何限制 ArrayList 中的对象数量

How to put a limit on the amount of objects in an ArrayList

我正在为我的 O.O.P 做一些修改。下周就要考试了,我被一个问题困住了。这个问题基本上是举一个狗和跳蚤之间双向关联的例子。到目前为止,我有一只带跳蚤的狗。我坚持的部分是,"Modify the dog class so that a dog object can only hold up to 5 flea objects max (print "你的狗有太多跳蚤!”如果跳蚤超过 5 只。这是我目前的代码:

Dog.java

import java.util.ArrayList;

public class Dog {

    private String name;
    private int age;
    private String address;

    ArrayList<Flea> fleas = new ArrayList<Flea>(); {
        if(fleas.size() > 5) {
             System.out.println("This dog has too many fleas!");
        }
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }

    public void hostFlea(Flea flea) {
        fleas.add(flea);
    }

    public ArrayList<Flea> getDogFlea() {
         return fleas;
    }

    public String toString() {
        return name + " the Dog (aged " + age + ") has fleas. \nThey are: " + fleas + ".";
    }   

}

Flea.java

public class Flea {

    private String name;
    private int age; 

    public Flea (String name, int age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public String toString() {
        return name + " (aged " + age + ")";
    }

}

Test.java

public class Test {

    public static void main(String[] args) {

        Dog dog = new Dog();
            dog.setName("Freddy");
            dog.setAddress("Cork");
            dog.setAge(5);

        Flea flea1 = new Flea("John", 1);
        dog.hostFlea(flea1);

        Flea flea2 = new Flea("Patrick", 3);        
        dog.hostFlea(flea2);

        Flea flea3 = new Flea("Alan", 7);
        dog.hostFlea(flea3);

        Flea flea4 = new Flea("Steven", 2);
        dog.hostFlea(flea4);

        Flea flea5 = new Flea("Charles", 5);
        dog.hostFlea(flea5);

        Flea flea6 = new Flea("Derek", 1);
        dog.hostFlea(flea6);

        Flea flea7 = new Flea("Kevin", 8);
        dog.hostFlea(flea7);

        System.out.println(dog);

    }

}

控制台:

狗弗雷迪(5 岁)有跳蚤。 他们是:[John(1 岁)、Patrick(3 岁)、Alan(7 岁)、Steven(2 岁)、Charles(5 岁)、Derek(1 岁)、Kevin(8 岁)]。

在此处添加检查您的状况:

public void hostFlea(Flea flea) {
    if(fleas.size() >= 5) {
        System.out.println("This dog has too many fleas!");
    } else {
        fleas.add(flea);
    }
}

不在你的列表变量的定义中(就像你所做的那样),因为你刚刚添加了 an instance initialization block.

在你的 hostFlea 方法中:-

在将 Flea 对象添加到 Dog 中的跳蚤 ArrayList 之前 class,只需检查此 arraylist 的大小。

喜欢:-

public void hostFlea(Flea flea){
if(fleas.size() > 5) {
  System.out.println("This dog has too many fleas!");
        }
 else{
fleas.add(flea);
   }
}

我猜你被迫使用 ArrayList。所以你必须确保没有人像你的狗一样修改列表class。所以将列表设为私有

private ArrayList<Flea> fleas = new ArrayList<Flea>();

Return 一个副本,不要在 getDogFlea 方法中公开原始列表

public ArrayList<Flea> getDogFlea() {
     return new ArrayList(fleas); //use copy constructor to not expose original list
}

并在您的 hostFlea 方法中添加检查。

public void hostFlea(Flea flea) {
    if(fleas.size() == 5 ) {
        System.out.println("This dog has too many fleas!");
        return; //so the 6th element is not added
    }
    fleas.add(flea);
}

也许 last 对您的考试来说已经足够了,但是 getter 在实际程序中会出现问题,确保永远不会超过 5 个元素。

狗的public方法应该添加跳蚤是hostFlea,所以你需要改变它:

public void hostFlea(Flea flea) {
    // If the dog already has at least 5 fleas, you can't add another
    if (fleas.size() >= 5) {
        System.out.println("Your dog has too many fleas!");
    } else {
        fleas.add(flea);
    }
}

但是,getDogFlea() return 是内部 ArrayList,因此没有什么能阻止坚定的用户调用 dog.getDogFlea().add(flea6)。为了防止这种行为,您可以复制数据:

public ArrayList<Flea> getDogFlea() {
     return new ArrayList<>(fleas);
}

或者,在 "cost" 放宽 API 到 return 一个 List 而不是 ArrayList,教科书的解决方案是使用 Collections.unmodifiableList:

public List<Flea> getDogFlea() {
     return Collections.unmodifiableList(fleas);
}