c＃要包含在参数化构造函数中的字段，该类是从派生自抽象类的另一个类派生的类

Question

我有一个课程Accommodation如下：

class Accommodation {
    int accommodationNo
    String address1
    String address2
    String postcode
    double valuation
    double rent
    int noRentPaymentsPerYe
}

班级Housing扩展了Accommodation：

class Housing : Accomodation {
    String type
    int noBedrooms
    int noCarParkingSites
}

课程Flat扩展Housing：

class Flat : Housing {
    double maintenanceCharge
}

创建参数为Flat class的构造函数时，是否要将每个字段都包含在Accommodation之内？参数类Flat的构造函数的示例将受到高度赞赏。

Answer 1

是的，您必须包含每个字段，并将其传递给构造函数链（这是字段的缩写，因为每个类只需要演示几个 - 您需要添加其他字段）：< / p>

public abstract class Accommodation {
    private int accommodationNo;
    private String address1;
    protected Accomodation(int acomNo, string addr1) {
        this.accomodationNo = acomNo;
        this.address1 = addr1;
    }
}
public class Housing {
    private String type;
    private int noBedrooms;
    public Housing(int acomNo, string addr1, string typ, int noBeds) 
            : base(acomNo, addr1) {
        this.type = typ;
        this.noBedrooms = noBeds;
    }
}
public class Flat {
    double maintenanceCharge;
    public Flat(int acomNo, string addr1, string typ, int noBeds, double maintCharge) 
            : base(acomNo, addr1, typ, noBeds) {
        this.maintenanceCharge = maintCharge;
    }
}