我正在为我的O.O.P做一些修改。考试即将在下周进行,我会遇到一个问题。问题基本上是一个狗和跳蚤之间的双向关联的例子。到目前为止,我有一只带跳蚤的狗。我坚持的部分是,“修改狗类,以便狗对象最多只能容纳5个跳蚤对象(打印”你的狗有太多的跳蚤!“如果有超过5个跳蚤)。这是我的代码到目前为止:
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:
弗雷迪狗(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)。答案 0 :(得分:1)
在此处添加检查您的情况:
public void hostFlea(Flea flea) {
if(fleas.size() >= 5) {
System.out.println("This dog has too many fleas!");
} else {
fleas.add(flea);
}
}
不是你的列表变量的定义(正如你所做的那样),因为你刚刚添加了an instance initialization block。
答案 1 :(得分:1)
我猜你被迫使用ArrayList。所以你必须确保没有人修改其他作为你的狗类的列表。因此,将列表设为私有
private ArrayList<Flea> fleas = new ArrayList<Flea>();
返回副本,不要在getDogFlea
方法
public ArrayList<Flea> getDogFlea() {
return new ArrayList(fleas); //use copy constructor to not expose original list
}
并在hostFlea
方法中添加支票。
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);
}
对于你的考试而言,也许最后一次是足够的,但是在真实的程序中,getter会出现问题,确保不会有更多的5个元素。
答案 2 :(得分:1)
狗假设添加跳蚤的公共方法是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()
会返回内部ArrayList
,因此无法阻止确定的用户调用dog.getDogFlea().add(flea6)
。为了防止此类行为,您可以复制数据:
public ArrayList<Flea> getDogFlea() {
return new ArrayList<>(fleas);
}
或者,在放宽API以返回List
而不是ArrayList
的“成本”时,教科书解决方案将使用Collections.unmodifiableList
:
public List<Flea> getDogFlea() {
return Collections.unmodifiableList(fleas);
}
答案 3 :(得分:0)
在你的hostFlea方法中: -
在将Flea对象添加到Dog类中的跳蚤ArrayList之前,只需检查此arraylist的大小。
如: -
public void hostFlea(Flea flea){
if(fleas.size() > 5) {
System.out.println("This dog has too many fleas!");
}
else{
fleas.add(flea);
}
}