1

我有这个代码

public class Student extends Person {
    //id represents the student's ID
    private int id;
    //grade represents the student's grade in the course
    private Grade grade;

    //constructor allows user to define first and last names, id, and grade of student in demo
    public Student(String fName, String lName, int id, Grade grade) {
        super(fName, lName);
        this.id=id;
        this.grade=grade;
    }
    //get methods for fields
    public int getId() {
        return id;
    }
    public Grade getGrade() {
        return grade;
    }
    //toString prints out the string from person class along with id and grade fields in formatted string
    public String toString() {
        return super.toString()+"'s id is " + id + "." +getGrade();
    }
}

还有这段代码。问题是toString 方法使用 passFailGrade getGrade() 返回值而不是位于类中的方法

public class Grade {

    private double score;

    public Grade(double score) {
        this.score=score;
    }
    public void setScore(double score) {
        this.score=score;
    }
    public double getScore() {
        return score;
    }
    public char getGrade() {
        if (getScore()>=90)
            return 'A';
        else if (getScore()>=80)
            return 'B';
        else if (getScore()>=70)
            return 'C';
        else if (getScore()>=60)
            return 'D';
        else
            return 'F';
    }
    public String toString() {
        return "\nThe student recieved a " + getGrade() + 
                " and had a mark of " + getScore() + ".";
    }
}

不确定是否有问题PassFailGrade：

public class PassFailGrade extends Grade {

    public PassFailGrade(double score) {
        super(score);
    }
    public char getGrade() {
        if (getScore()>=50)
            return 'Y';
        else
            return 'N';
    }
    public String toString() {
        return "(Y for yes/N for no) The student passed their course (" 
                + getGrade()+ ")." + super.toString();
    }
}

然后演示类只是在构造函数中定义并打印

public class StudentDemo {
    public static void main(String[] args) {
        PassFailGrade bo= new PassFailGrade(98);
        Student s1 = new Student("bob", "blake", 123, bo);
        System.out.println(s1); 
    } 
}

输出：

bob blake 的 id 是 123。（Y 代表是/N 代表否）学生通过了他们的课程 (Y)。该学生得了一个 Y 并且得分为 98.0。
4

1 回答 1

3

您实际上是在覆盖该getGrade()方法。这是您打算做的事情，但现在您在调用时遇到了问题super.toString()- 它仍然使用您的覆盖方法。

您可以通过以下方式更改您toString的 in来解决此问题PassFailGrade：

@Override
public String toString() {
    return "(Y for yes/N for no) The student passed their course ("
            + getGrade()+ ")." + "\nThe student recieved a " + super.getGrade() +
            " and had a mark of " + getScore() + ".";
}

请注意，我只调用super.getGrade(). 这将产生正确的结果。

于 2013-07-08T05:42:02.403 回答