我知道这是非常基本的但是我无法用一种不熟悉的语言拼凑起来。我正在将UML映射到Java代码,继承正在抛弃我。我有一个像这样的ERD:
Animal
----------------------------
-color: int
-weight: int
----------------------------
+ getColor() : int
+ getWeight(): int
----------------------------
^(Inheritance Triangle)
|
|
----------------------------
Dog
----------------------------
-breed: string
----------------------------
+ getBreed()
----------------------------
当然是狗IS-A动物,我可以从狗类等调用getColor。我的问题是关于变量,特别是构造函数。当我实现这个时,我有
public class Animal
{
private int color;
private int weight;
public Animal(int c, int w)
{
color = c;
weight = w;
}
...
}
public class Dog extends Animal
{
private string breed;
public Dog()
{
breed = "Shelty";
}
}
我的问题是,在狗类中使用颜色和重量的正确方法是什么?看看UML,我知道我可以在dog实体中添加颜色和重量,但我知道这是可选的。我还会在狗类中拥有私人颜色和重量属性吗?我会调用Animal构造函数(抛出错误)这里的表格是什么?
答案 0 :(得分:9)
Dog
类的职责应该是初始化它添加的 new 功能,并让超类初始化Dog
类继承的所有属性。
你可以这样做:
public class Dog extends Animal
{
private String breed;
public Dog(int color, int weight, String breed)
{
super(color,weight); //let superclass initialize these.
this.breed = breed;
}
}
答案 1 :(得分:3)
您使用关键字super。
public Dog(int c, int w){
super(c,w);
breed = "Shelty";
}
答案 2 :(得分:2)
除了manish的回答, 您还可以编写NoArg构造函数来提供默认初始化值。
public class Dog extends Animal
{
private String breed;
public Dog(int color, int weight, String breed)
{
super(color,weight); //let superclass initialize these.
this.breed = breed;
}
// Default initialization if required
public Dog() {
this ( 0, 0, "Shelty")
}
}
答案 3 :(得分:1)
public class Dog extends Animal
{
private String breed;
//Declaring default constructor is a good practice
public Dog()
{
//Some default values
int color=3,weight =8;
super(color,weight);
this.breed = "pug";
}
public Dog(int color, int weight, String breed)
{
super(color,weight);
this.breed = breed;
}
}