所以我有一个动物园程序,我想要一个动物园中的房间列表,以及每个房间中的猫列表。
我有 3 个类:Felid
、Housecat
和 Wildcat
- Housecat
和 Wildcat
扩展Felid
。根据动物的字面类别(现在我有老虎、波斯猫、暹罗猫和猎豹 - 老虎和猎豹延伸野猫,波斯猫和暹罗猫延伸家猫)某些属性将自动分配。
类图 - /image/oDydn.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;
}
如果猫是野猫,当用户使用猫名创建 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)
问题是它把所有猫都算作野猫,我认为这是因为我用于猫列表的列表是:
ArrayList<Felid> catList = new ArrayList<Felid>();
所以我猜 catName 将始终为 null,因为添加到列表中的家猫将仅算作“felid”类型。
如何创建一个列表,可以将所有猫放入其中,但仍将它们视为各自的类别?
编辑:感谢您指出赋值运算符错误,但它仍然只打印野性
最终编辑:非常感谢“DoubleDouble”向我指出如何使用“Super()”——这不是我预期的问题。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;
}
}
最佳答案
if(feral = true){
这是一个赋值语句,而不是比较。使用 ==
关于java - 如何在ArrayList中使用多态性?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30173183/