编译器没有认识到我已经从它实现的接口实现了compareTo（）方法

Question

我正在通过抽象类AbstractAffiliate实现一个接口'Comparable'，它由抽象类Abstract Faculty扩展，它由常规类Assistant扩展。

在助手类中实现compareTo方法（已在上面的所有类/接口中声明）后，编译器会给出此错误。

Assistant.java:1：错误：助手不是抽象的，并且不会覆盖抽象方法中的抽象方法compareTo（）公共类助手扩展AbstractFaculty {^ 1错误

我已经尝试添加作为实现Comparable的泛化器。

摘要会员

public abstract class AbstractAffiliate implements Printable, Comparable<AbstractAffiliate>{

protected String name;
protected int age;
protected String address;
protected int phoneNumber;
protected int yearTheyCameToChapman;

    /**
    * Default empty AbstractAffiliate constructor
    */
public AbstractAffiliate() {
    super();
  age = 0;
  address = " ";
  phoneNumber = 0;
  yearTheyCameToChapman = 0;
}

public AbstractAffiliate(String name, int age, String address, int phoneNumber, int yearTheyCameToChapman){
    this.name = name;
    this.age = age;
    this.address = address;
    this.phoneNumber = phoneNumber;
    this.yearTheyCameToChapman = yearTheyCameToChapman;
}

public abstract String print();

public abstract int compareTo();

}

抽象的教师

public abstract class AbstractFaculty extends AbstractAffiliate{

protected int facultyId;
protected String department;
protected int yearlySalary;
protected int numberOfPapers;

    /**
    * Default empty AbstractFaculty constructor
    */
public AbstractFaculty() {
    super();
  facultyId = 0;
  department = " ";
  yearlySalary = 0;
  numberOfPapers = 0;
}

    /**
    * Default AbstractFaculty constructor
    */
public AbstractFaculty(String name, int age, String address, int phoneNumber, int yearTheyCameToChapman, int facultyId, String department, int yearlySalary, int numberOfPapers) {
    super(name, age, address, phoneNumber, yearTheyCameToChapman);
    this.facultyId = facultyId;
    this.department = department;
    this.yearlySalary = yearlySalary;
    this.numberOfPapers = numberOfPapers;
    }

public abstract String print();

public abstract int compareTo();



}

助理

public class Assistant extends AbstractFaculty{

private int yearsUntilTenure;

public Assistant(){
  super();
  yearsUntilTenure = 0;
}

public Assistant(String name, int age, String address, int phoneNumber, int yearTheyCameToChapman, int facultyId, String department, int yearlySalary, int numberOfPapers, int yearsUntilTenure){
  super(name, age, address, phoneNumber, yearTheyCameToChapman, facultyId, department, yearlySalary, numberOfPapers);
  this.yearsUntilTenure = yearsUntilTenure;
}

public String print(){
  return "yup";
}


public int compareTo(AbstractAffiliate affiliate){
  if (this.yearTheyCameToChapman < affiliate.yearTheyCameToChapman){
    return 1;
  }
  if (this.yearTheyCameToChapman > affiliate.yearTheyCameToChapman){
    return -1;
  }
  else{
    return 0;
  }
}



}
```[enter image description here][1]


  [1]: https://i.stack.imgur.com/Xdz2F.png

Answer 1

您尚未实现抽象方法。抽象的.compareTo()方法不带参数。相比之下，您在Assistant类中实现的版本将AbstractAffiliate作为参数。由于它们具有不同的参数，因此它们使它们完全不同。

乍一看，看起来像参数的版本是你想要的那个，并且应该由你的基类实现Comparable<AbstractAffiliate>的事实来处理，所以只需完全删除你的抽象compareTo()方法，你应该没问题。

Answer 2

Comparable＃compareTo的方法签名如下：

public int compareTo(T o);

但是在Assistant和它的父类中，你添加了一个抽象方法

public abstract int compareTo();

这与Comparable#compareTo的方法不同。只需从其父类中删除public abstract int compareTo();方法就可以解决问题，不需要一次又一次地声明它。

Answer 3

这是因为你声明要实现Comparable<AbstractAffiliate>并且你没有实现该方法

int compareTo(AbstractAffiliate o)

你有一个不同的签名（没有参数）由于AbstractAffiliate和AbstractFaculty被声明为摘要，他们不必从Comparable接口实现该方法。助理班需要。

请参阅Comparable