泛型:java.lang.String不能应用于Student

时间:2013-09-26 19:47:12

标签: java generics

我在课堂上,我的教授说她的代码有效,并且没有任何问题,而且一定是我。我已经查看了她的代码,并按照她的说法逐字复制,但我仍然收到错误:

  

Pair.java:28:set中的set(java.lang.String,java.lang.Double)无法应用于(Student,java.lang.Double)

我加粗的部分是我收到错误的部分。设定方法不正确吗?因为带有错误的行返回到set方法

这是她的代码:

Student.java

import java.io.*;

public class Student implements Person {
  String id;
  String name;
  int age;

  //constructor
  public Student(String i, String n, int a) {
    id = i;
    name = n;
    age = a;
  }

  public String getID() {
    return id;
  }

  public String getName() {
    return name; //from Person interface
  } 

  public int getAge() {
    return age; //from Person interface
  }

  public void setid(String i) {
    this.id = i;
  }

  public void setName(String n) {
    this.name = n;

  }

  public void setAge(int a) {
   this.age = a;
  }

  public boolean equalTo(Person other) {
    Student otherStudent = (Student) other;
    //cast Person to Student
    return (id.equals(otherStudent.getID()));
  }

  public String toString() {
    return "Student(ID: " + id + 
      ",Name: " + name +
      ", Age: " + age +")";
  }
}

Person.java

import java.io.*;

public interface Person {
  //is this the same person?
  public boolean equalTo (Person other);
  //get this persons name
  public String getName();
  //get this persons age
  public int getAge();
}

Pair.java

import java.io.*;

public class Pair<K, V> {

  K key;
  V value;
  public void set (K k, V v) {
    key = k;
    value = v;
  }

  public K getKey() {return key;}
  public V getValue() {return value;}
  public String toString() {
    return "[" + getKey() + "," + getValue() + "]";
  }

  public static void main (String [] args) {
    Pair<String,Integer> pair1 = 
      new Pair<String,Integer>();
    pair1.set(new String("height"),new
                Integer(36));
    System.out.println(pair1);
    Pair<String,Double> pair2 = 
      new Pair<String,Double>();

    //class Student defined earlier
    **pair2.set(new Student("s0111","Ann",19),**
              new Double(8.5));
    System.out.println(pair2);
  }
}

4 个答案:

答案 0 :(得分:3)

对于实例化:

Pair<String,Double> pair2 = new Pair<String,Double>();

您的set()方法签名相当于:set(String, Double)。并且您在下面的调用中传递了Student引用,这不起作用,因为Student不是String

pair2.set(new Student("s0111","Ann",19), new Double(8.5));

要避免此问题,请将pair2的声明更改为:

Pair<Student,Double> pair2 = new Pair<Student,Double>();

答案 1 :(得分:2)

错误非常自我解释。 pair2定义为Pair<String, Double>。您正在尝试设置Student, Double。那不行。

答案 2 :(得分:2)

Pair<String,Double> pair2 = new Pair<String,Double>();

应该是:

Pair<Student,Double> pair2 = new Pair<Student,Double>();

答案 3 :(得分:1)

您的代码pair2中的

被定义为Pair<String,Double>类型,即pair2 set()方法期望StringDouble作为参数,但是您正在通过StudentDouble 所以

Pair<String,Double> pair2 = new Pair<String,Double>();  

应该是

Pair<Student,Double> pair2 = new Pair<Student,Double>();