java中的超级构造函数

发布于 2024-10-11 11:43:09 字数 690 浏览 4 评论 0原文

请解释一下

public class Contact {
    private String contactId;
   private String firstName;
    private String lastName;
    private String email;
    private String phoneNumber;

public Contact(String contactId,String firstName, String lastName,   String email,        String phoneNumber) {
    super();  //what does standalone super() define? With no args here?
    this.firstName = firstName;  
    this.lastName = lastName;     //when is this used?, when more than one args to be entered?
    this.email = email;
    this.phoneNumber = phoneNumber;
}

Super() 里面没有参数意味着有多个参数需要定义？这是在“this.xxx”的帮助下完成的吗？

为什么我们在“公共类联系人”本身中定义。为什么我们在这里再次定义并调用它的参数？

原文

Please explain

public class Contact {
    private String contactId;
   private String firstName;
    private String lastName;
    private String email;
    private String phoneNumber;

public Contact(String contactId,String firstName, String lastName,   String email,        String phoneNumber) {
    super();  //what does standalone super() define? With no args here?
    this.firstName = firstName;  
    this.lastName = lastName;     //when is this used?, when more than one args to be entered?
    this.email = email;
    this.phoneNumber = phoneNumber;
}

Super() with no arguments inside mean there are more than one arguments to be defined? And is this done with the help of "this.xxx" ?

Why dint we define in the "public class Contact" itself. Why we defined again and called its arguments here?

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岁月如刀 2024-10-18 11:43:09

Super() 里面没有参数意味着有多个参数需要定义？

不，super() 只是调用基类的无参数构造函数，在您的案例对象。

它实际上什么也没做。它只是在代码中明确表明您正在使用无参数构造函数构造基类。事实上，如果您将 super() 排除在外，编译器会隐式地将其添加回来。

那么如果 super() 是隐式添加的，那么它的作用是什么？嗯，在某些情况下，类没有无参数构造函数。此类的子类必须显式调用某些超级构造函数，例如使用super("hello")。

this.lastName = 姓氏； //什么时候使用这个？什么时候需要输入多个参数？

this.lastName = lastName;与super()<无关/代码>。它只是指出构造函数参数 lastName 的值应分配给成员变量 lastName。这相当于

public Contact(String contactId, String firstName, String lastNameArg,
               String email, String phoneNumber) {
    // ...
    lastName = lastNameArg;
    // ...

Super() with no arguments inside mean there are more than one arguments to be defined?

No, super() just calls the no-arg constructor of the base class, in your case Object.

It does nothing really. It just makes it explicit in the code, that you're constructing the base class using the no-arg constructor. In fact, if you left super() out, it would be added back implicitly by the compiler.

So what is the super() for if it's added implicitly anyway? Well, in some cases, a class does not have a no-arg constructor. Subclasses of this class must explicitly call some super-constructor, using for instance super("hello").

this.lastName = lastName; //when is this used?, when more than one args to be entered?

this.lastName = lastName; has nothing to do with super(). It merely states that the value of the constructor argument lastName should be assigned to the member variable lastName. This is equivalent to

public Contact(String contactId, String firstName, String lastNameArg,
               String email, String phoneNumber) {
    // ...
    lastName = lastNameArg;
    // ...

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