Question

所以我有这个动物园计划，我希望在动物园里有一个房间清单，以及每个房间里的猫咪清单。

我有3个班级：Felid，Housecat和Wildcat - Housecat和Wildcat扩展Felid。取决于动物的字面类（现在我有虎，波斯，暹罗和猎豹 - 虎和猎豹扩展野猫，波斯和暹罗延伸家猫）某些属性将被自动分配。

类图 - http://i.imgur.com/VTSNRVA.jpg

例如，这些是felid的字段：

String speciesName;
String furColour;
String gender;
int weightPounds;
boolean packAnimal;
String habitat;
int age;

这些是housecat的字段：

String ownerName;
String catName;
boolean feral;

这些是野猫的字段：

boolean manEater;

在我的housecat构造函数中，我有

    if(catName == null || catName.equals("")){
        feral = true;
    }

如果是猫的野性，当用户创建一个带有猫名的HouseCat并使用'printCatInfo（）'时：

@Override
public void printCatInfo(){
    if(feral){
        System.out.println("feral" + "\n" + speciesName + "\n" + furColour + "\n" + gender + 
            "\n" + weightPounds + "lbs\n" + "is not a pack animal" + "\n" + habitat + "\n" + age + " years old (human)");
    }
    else{
        System.out.println("owner name is: " + ownerName + "\n" + "cat name is: " + catName + "\n" + speciesName + "\n" + furColour + "\n" + gender + 
            "\n" + weightPounds + "lbs\n" + "is not a pack animal" + "\n" + habitat + "\n" + age + " years old (human)" + "\n");
    }     
}

它不会打印它的名字。

feral
siamese
White or grey abdomen with black legs, face and tail
male
8lbs
is not a pack animal
urban
7 years old (human)

麻烦的是，它把所有的猫都视为野性，我认为是因为我用于猫列表的列表是：

ArrayList<Felid> catList = new ArrayList<Felid>();

所以我猜catName将永远为null，因为添加到列表中的housecats只会算作'felid'类型。

如何创建一个列表，我可以将所有猫扔进去，但仍将它们视为各自的类？

编辑：感谢指出赋值操作符错误，它仍然只打印野性

最终编辑：非常感谢'DoubleDouble'指出如何使用'Super（）'给我 - 这不是我预期的问题所在。这就是问题所在：

public class Siamese extends HouseCat{
    public Siamese(int weightPounds, int age, String ownerName, String catName, String gender){
        this.speciesName = "siamese";
        this.furColour = "White or grey abdomen with black legs, face and tail";        
        this.ownerName = ownerName;
        this.catName = catName;
        this.weightPounds = weightPounds;
        this.age = age;
        this.gender = gender;
    }
}

新代码：

public class Siamese extends HouseCat{
    public Siamese(int weightPounds, int age, String ownerName, String catName, String gender){
        super(catName);
        this.speciesName = "siamese";
        this.furColour = "White or grey abdomen with black legs, face and tail";        
        this.ownerName = ownerName;
        this.catName = catName;
        this.weightPounds = weightPounds;
        this.age = age;
        this.gender = gender;
    }
}

Answer 1

if(feral = true){

这是一个赋值语句，而不是比较。使用==

Answer 2

在这一行：

if(feral = true){

您已将true分配给feral。您使用了赋值运算符=。比较运算符为==。但是，feral已经是boolean，因此您可以在条件中使用feral。

if (feral) {

Answer 3

你的构造函数似乎如下（如果我错了，请纠正我）：

public Siamese(int weightPounds, int age, String ownerName, String catName, String gender)
{
    this.weightPounds = weightPounds;
    this.age = age;
    this.ownerName = ownerName;
    this.catName = catName;
    this.gender = gender;
}

public HouseCat()
{
    if(catName == null || catName.equals(""))
    {
        feral = true;
    }
}

由于ownerName，catName和feral都是HouseCat类的一部分，因此最好让HouseCat构造函数处理这些字段

public HouseCat(String ownerName, String catName)
{
    this.ownerName = ownerName;
    this.catName = catName;
    if(catName == null || catName.equals(""))
    {
        feral = true;
    }
}

Siamese就是这样：

public Siamese(int weightPounds, int age, String ownerName, String catName, String gender)
{
    super(ownerName, catName);
    this.weightPounds = weightPounds;
    this.age = age;
    this.gender = gender;
}

如何在ArrayLists中使用多态？

3 个答案: