java - Java中的多重继承问题

Question

我想用Java建模以下类：

所以我提供了以下代码：

class Person
{
    private String name;
    private ing age;
    public Person(String name, int age){
        this.name=name;
        this.age=age;
    }
    //set and get methods
}

class Employee
{
    private String nameEmp;
    private double salary;
    public Employee(String nameEmp, double salary){
        this.nameEmp=nameEmp;
        this.salary=salary;
    }
    public double calcSalary(){}   //should this be an abstract method?
}

class Teacher extends Person implements Employee
{
    private String nameTeacher;
    private int ageTeacher;
    private String title;   //professor or lecturer
    public Teacher(String nameTeacher,int ageTeacher, String title){
        super(nameTeacher,ageTeacher);
        this.title=title;
    }
    public double calcSalary(){
        if (title.equals("Professor")) salary=salary*0,30;
        else if (title.equals("Lecturer")) salary=salary*0,10;
    }
}

我想使用接口对其进行建模，但我不太确定该怎么做。calcSalary 也应该是 Employee 中的抽象方法？如何使用 Java 中的接口来实现这一点？

谢谢

Answer 1

不，你不能这样做，你应该选择老师是 -> 员工是 -> 人。你不能在接口中实现任何东西！接口只能包含应该由类实现的方法。

Answer 2

你可以有这个：

public interface Person{
    // Only abstract methods here
}

public interface Employee extends Person {
    // Only abstract methods here specific to Employee
}

public class Teacher implements Employee {
    //Implements the methods 
}

Answer 3

Java 8 将允许您在接口中放置默认实现。在那之前，接口不能包含实现。

Answer 4

public interface Person {
    String getName();
    void setName(String name);
    int getAge();
    void setAge(int age);
}

public class PersonImpl implements Person {
    private String name;
    private int age;
    @override public String getName() { return this.name; }
    @override public void setName(String name) { this.name = name; }
    @override public int getAge() { return this.age; }
    @override public void setAge(int age) { this.age = age; }
}

public interface Employee extends Person {
    double getSalary();
    void setSalary(double salary);
    double calcSalary();
}

public abstract class EmployeeImpl extends PersonImpl implements Employee {
    private double salary;
    @override public double getSalary() { return this.salary; }
    @override public void setSalary(double salary) { this.salary = salary; }
}

public class Teacher extends EmployeeImpl
    @override public double calcSalary() { 
        if (title.equals("Professor")) salary=salary*0,30;
        else if (title.equals("Lecturer")) salary=salary*0,10;
    }
}

Answer 5

从人开始。你有一个错字：

public class Person
{
    private String name;
    private int age;
    public Person(String name, int age){
        this.name=name;
        this.age=age;
    }
    // set and get methods, equals, hashCode, toString,
    // perhaps an id for database storage.
}

现在，一些Persons 是Employees；其他人是学生、家长、准学生等等。如果任何方法Employee有一个实现，它就不能是一个接口。在任何情况下，拥有nameEmp都是错误的，因为它会在Person. 要么为 Employee 创建一个混入Teacher类的接口，要么从Employee.

任何一个：

public interface Employee {
    // Currency values should use BigDecimal or BigInteger, not double.
    BigDecimal salary();
    // Taxpayer Identification Number (SSN) in the USA, or the equivalent outside.
    String taxNumber();
}

或者：

public class Employee extends Person {

    private BigDecimal salary;
    private String taxNumber;
    public Employee(String name, int age, BigDecimal salary,
                    String taxNumber) {
        super(name, age);
        this.salary = salary;
        this.taxNumber = taxNumber;
        this.title = title;
    }
    // getters, setters, etc.
}

public class Teacher extends Employee {
    private Department department;
    private List<Course> courses = new ArrayList<>();
    // Constructors, getters, setters, etc.
}

如果需要完全不同的实现，您应该只在 Teacher 中使用薪水方法。您可能希望在 Employee 类中有某种薪酬计算器类（策略模式）。通过这种方式，您可以在适用的情况下处理应税扣除、保险、养老金计划和加班。我不理解您原始示例中的乘数。