对Java中的参数和构造函数感到困惑

所以基本上，我有一个任务，我有很多不同的事情要做。我被困在这一部分，我必须创建三个不同的构造函数，它们采用不同的参数

参数：

获取所有4个字段的参数-我想我在下面的构造函数中做了

地址行1-3和邮政编码的一个参数

地址行1和2以及邮政编码的参数

我尝试过用不同的参数创建不同的构造函数，但它不起作用，因此对此有任何见解都将不胜感激

class Address{

    private String addressLine1;
    private String addressLine2;
    private String addressLine3;
    private String addressLine4;
    private String postcode;

    public Address(String allFields){
        this.addressLine1 = allFields;
        this.addressLine2 = allFields;
        this.addressLine3 = allFields;
        this.addressLine4 = allFields;
        
        
     }
}

编辑：这是关于这个练习阶段的原始问题

为类提供三个不同的构造函数：一个为所有四个字段获取参数，一个仅为地址行1-3和邮政编码获取参数，另一个仅为地址行1和2以及邮政编码获取参数。省略某些字段的构造函数应将这些字段设置为包含空字符串值“”

多谢各位

获取所有4个字段的参数-我想我在下面的构造函数中做了

地址行1-3和邮政编码的一个参数

地址行1和2以及邮政编码的参数

---

根据你原来的问题答案应该如下：在你最初的问题中，是说要实现“一个为所有四个字段都取参数”，所以“强”>我将把四个字段看作ADRESSLIN字段>，因为已经有5个字段，包括PASCODE字段。 额外：在命名变量时，请遵循Java中的命名转换。例如：邮政编码应为邮政编码，尽量避免变量名称中的数字

public Address(String addressLine1, String addressLine2, String addressLine3, String addressLine4){//Here I considered the four fields as addressLine attributes
       this.addressLine1 = addressLine1;
       this.addressLine2 = addressLine2;
       this.addressLine3 = addressLine3;
       this.addressLine4 = addressLine4;
       this.postcode ="";
    }

---

从您的问题中，您将

设置所有构造函数中所有变量的值。对于这种调用内部构造函数的场景，Java提供了一个方法
this（）
。因此，以下应该是正确的组合

class Address{
    private String addressLine1;
    private String addressLine2;
    private String addressLine3;
    private String addressLine4;
    private String postcode;

    //1. one which takes parameters for only address lines 1 and 2 and the postcode
    public Address(String addressLine1, String addressLine2, String postCode){
        this(addressLine1, addressLine2, "", postCode);
    }

    //2. one which takes parameters for only address lines 1-3 and the postcode
    public Address(String addressLine1, String addressLine2, String addressLine3, String postCode){
        this(addressLine1, addressLine2, addressLine3, "");
        this.postcode = postCode;
    }
        
    //3. one which takes parameters for all four fields, 
    public Address(String addressLine1, String addressLine2, String addressLine3, String addressLine4){
       super();
       this.addressLine1 = addressLine1;
       this.addressLine2 = addressLine2;
       this.addressLine3 = addressLine3;
       this.addressLine4 = addressLine4;
    }
}

您可以使用不同的参数定义多个构造函数，但是如果（2）和（3）都需要两个字符串参数，那么它们就无法区分。是正确的，毕竟它说构造函数应该为所有4个字段使用“参数”（复数），而不是为所有字段分配一个参数。你所做的没有任何意义。因为你正在为所有addressLine属性初始化相同的值。我想你需要一个具有多个参数的构造函数。（关于已实施的构造函数）你是否照搬了建设者的要求？或者该列表是否部分反映了你对作业的理解？因为你问题中给出的措辞会导致相当高级的作业，这可能不是你的讲师的意图。如果我是对的，请随意将原始措辞添加到问题中。“\u..所有4个字段考虑到
Address
对象实际上有五个字段（四个地址行和邮政编码），这有点令人困惑。

public Address(String addressLine1, String addressLine2, String postCode){
           this.addressLine1 = addressLine1;
           this.addressLine2 = addressLine2;    
           this.postCode = postCode;
        }


public Address(String addressLine1, String addressLine2, String addressLine3, String addressLine4){//Here I considered the four fields as addressLine attributes
       this.addressLine1 = addressLine1;
       this.addressLine2 = addressLine2;
       this.addressLine3 = addressLine3;
       this.addressLine4 = addressLine4;
       this.postcode ="";
    }

public Address(String addressLine1, String addressLine2, String addressLine3, String postcode){
       this.addressLine1 = addressLine1;
       this.addressLine2 = addressLine2;
       this.addressLine3 = addressLine3;
       this.addressLine4 = "";
       this.postcode = postcode;
    }


public Address(String addressLine1, String addressLine2, String postcode){
           this.addressLine1 = addressLine1;
           this.addressLine2 = addressLine2; 
           this.addressLine3 = "";
           this.addressLine4 = "";
           this.postcode = postcode;
        }


class Address{
    private String addressLine1;
    private String addressLine2;
    private String addressLine3;
    private String addressLine4;
    private String postcode;

    //1. one which takes parameters for only address lines 1 and 2 and the postcode
    public Address(String addressLine1, String addressLine2, String postCode){
        this(addressLine1, addressLine2, "", postCode);
    }

    //2. one which takes parameters for only address lines 1-3 and the postcode
    public Address(String addressLine1, String addressLine2, String addressLine3, String postCode){
        this(addressLine1, addressLine2, addressLine3, "");
        this.postcode = postCode;
    }
        
    //3. one which takes parameters for all four fields, 
    public Address(String addressLine1, String addressLine2, String addressLine3, String addressLine4){
       super();
       this.addressLine1 = addressLine1;
       this.addressLine2 = addressLine2;
       this.addressLine3 = addressLine3;
       this.addressLine4 = addressLine4;
    }
}