如何在不破坏封装的情况下使它们对彼此“可见”？

Question

Manager 和 Employee 类都是 EnterpriseMember 的子类。如何为 Employee 类编写一个“getManager”方法（返回在其报表列表中包含此 Employee 的 Manager 实例）？提前致谢！public class Manager extends EnterpriseMember {    /*Fields */    private List reports = new ArrayList();    /*Constructor */    public Manager(String name){        super(name);    }    /*Methods */    public void addReport(Employee employee){                    reports.add(employee);    }// How can "employee" know it is in this List?}public class Employee extends EnterpriseMember {    /*Constructor */    public Manager(String name){        super(name);    }    /*Methods */    public Manager getManager(){                    return ???;    }}

Answer 1

像这样的东西：

public class Manager {

    private List<Employee> reports = new ArrayList<Employee>();

    public void addReport(Employee e) {

        if (e != null) {

            this.reports.add(e);

            e.setManager(this);

        }

    }

}

public class Employee {

    private Manager manager;

    public void setManager(Manager m) {

        if (m != null) {

            this.manager = m;

        }

    }

}

以防万一不清楚，您应该添加您需要的所有其他方法。我只说明了如何在将参考添加到直接报告时更新Manager参考。EmployeeList

您还应该有一个从removeReport中删除 an并将其设置为的方法。EmployeeListManagernull

你打算如何Employee在这个中找到一个List？按名字？员工ID？提示：考虑覆盖equals并hashCode适当地为您的课程。

不Managers也是Employees吗？老板没有老板吗？这是一个层次结构，一棵树。

Answer 2

通常具有不同属性的对象如下所示：

public class Employee extends EnterpriseMember {

    private Manager manager;

    private String name; // You probably don't need this because you defined it in the Superclass.

    .

    .

    .

    /*Constructor */

    public Employee(String name){

        super(name);

    }

    /*Methods */

    public Manager getManager(){            

        return manager;

    }

    public void setManager(Manager manager){            

        this.manager = manager

    }

    // Other getters and setters for the attributes.

}