Question

你好，英语不是我的主要语言我不确定如何解释我的问题或者标题是否应该标题

所以......请跳到House课堂！
提前谢谢

主要课程

public class Main {

    public static void main(String[] args) {

        Furniture f1 = new Furniture("chair", 2);
        Furniture f2 = new Furniture("bed", 20);
        Furniture f3 = new Furniture("workbench", 7);
        Furniture f4 = new Furniture("drawer", 15);

        Furniture [] foo = {f1,f2,f3}; 

        // public House(Furniture[] fur)
        House h1 = new House(foo);

        //  public House(Furniture f, House p) 
        House h2 = new House(f4, h1);

    }

}

家具类

public class Furniture {

    String name;
    int weight;

    public Furniture(String name, int weight) {
        this.name = name;
        this.weight = weight;
    }

    public static void display(Furniture f) {
    System.out.println("This furniture is a " + f.name + " and weights " + f.weight);
    }

}

众议院班级

// My House class contains one attribut which is an array of Furniture
public class House {

    Furniture[] fur;

    // Should build an empty house
    public House() {

    }

    // Should define the house with the array content
    public House(Furniture[] fur) {
        this.fur = fur;
    }

    // Should build an house containing the Furniture f and the furniture of the House p
    public House(Furniture f, House p) {

    // I'm so confused here, I'm not sure how to start 



    }

}

我已经考虑制作一个新的数组，其长度是房子中包含的数组并添加+1（对于家具f）然后做一个循环以获得房子p的所有家具并将它们添加到新的阵列与新家具f

我试着做p.length（p为房子），但它不起作用。我有点理解为什么但另一方面我没有，我怎么能访问房子的阵列？这是错误的做法，我找不到另一种方式

Answer 1

变量p不是数组（它是一个类（House类）），因此您不能直接使用p.length调用毛发的长度。而是使用p.fur.length;来获取fur数组的长度。因为这样你首先引用类：p.fur然后访问该类（fur）中的数组。

Answer 2

你不能这样访问数组，因为你的House不是数组，你必须在其他地方说houseInstance.fur才能访问数组，其中houseInstance是House类的一个实例。

考虑如何用旧家具和另一件新家具制作新房子：

这是你如何使用你的方法来做到的：

    public House(Furniture f, House p) {

        // I'm so confused here, I'm not sure how to start 
        this.fur = new Furniture[p.fur.length+1];

        this.fur[0] = f; // Add new furniture

        // Make new instances of furniture from another house(They don't share reference but have same values)
        // Suggested way of doing it
        for(int i =0; i < p.fur.length;i++){
            this.fur[i+1] = new Furniture(p.fur[i].name, p.fur[i].weight);
        }
        /*
           This will share furniture reference between those 2 houses
            so changing one will change another
            This should be avoided
        for(int i =0; i < p.fur.length;i++){
            this.fur[i+1] = p.fur[i];
        }*/
    }

操纵对象，构造函数？

2 个答案: