java有多个超类的构造函数

Question

我有一个包含 3 个构造函数的超类，我想知道是否有更聪明的方法来编写子类构造函数

public class Person{

private String name;
private int age;
private String homeTown;


public Person(String name){
    this.name = name;
    this.age = 18;
    this.homeTown = "Atlanta";
}

public Person(String name, int age){
    this.name = name;
    this.age = age;
    this.homeTown = "Atlanta";
}

public Person(String name, int age, String homeTown){
    this.name = name;
    this.age = age;
    this.homeTown = homeTown;   
}

我还有一个继承超类的子类

public class Student extends Person{
private double avgGPA;
private int ID;
private String[] classes;

public Student(double avgGPA, int ID, String[] classes, String name){
    super(name);
    this.avgGPA = avgGPA;
    this.ID = ID;
    this.classes = classes;
}

public Student(double avgGPA, int ID, String[] classes, String name, int age){
    super(name, age);
    this.avgGPA = avgGPA;
    this.ID = ID;
    this.classes = classes;
}

public Student(double avgGPA, int ID, String[] classes, String name, int age, String homeTown){
    super(name, age, homeTown);
    this.avgGPA = avgGPA;
    this.ID = ID;
    this.classes = classes;
}

我的子类工作正常并且运行没有错误，但我想知道是否有另一种方法可以为子类编写构造函数而无需编写相同的构造函数 3 次，只是因为超类有 3 个不同的构造函数。

Answer 1

嗯，Java 中有一些东西可以简化您的超类。您可以使用this(); 调用同一类中的另一个构造函数。因此，不要为每个构造函数设置每个变量，而是使用一个变量设置构造函数并使用this(); 传递默认值。对于您的超类，您可以改用这些：

public Person(String name){
    this(name, 18, "Atlanta");
}

public Person(String name, int age){
    this(name, age, "Atlanta");
}

public Person(String name, int age, String homeTown){
    this.name = name;
    this.age = age;
    this.homeTown = homeTown;   
}

对于子类，我将创建一个名为 setVars 的私有方法，它接收您将使用的三个变量：double avgGPA、int ID 和 String[] classes。因此，您的类可以如下所示，而不是在每个构造函数中设置它们：

public Student(double avgGPA, int ID, String[] classes, String name){
    super(name);
    setVars(avgGPA, ID, classes);
}

public Student(double avgGPA, int ID, String[] classes, String name, int age){
    super(name, age);
    setVars(avgGPA, ID, classes);
}

public Student(double avgGPA, int ID, String[] classes, String name, int age, String homeTown){
    super(name, age, homeTown);
    setVars(avgGPA, ID, classes);
}

private void setVars(double avgGPA, int ID, String[] classes) {
    this.avgGPA = avgGPA;
    this.ID = ID;
    this.classes = classes;
}

我认为这与您获得的效率一样高，除非您想按照 QueenSvetlana 的回答所推荐的那样创建静态初始化方法。

Answer 2

类似这样的东西：

public final class Person{

    private final String name;
    private final int age;
    private final String homeTown;
    private double avgGPA;

    private Person(String name, int age, String homeTown, avgGPA){

        this.name = name;
        this.age = age;
        this.homeTown = homeTown;
        this.avgGPA = avgGPA;
    }

    public static Person createPerson(String name, age, homeTown, avgGPA){
        return new Person(name, age, homeTown, avgGPA);
    }

    public static Person createPersonwithoutHomeTown(String name, age,avgGPA){
        return new Person(name, age, "Atlanta", avgGPA);
    }

    public static Person createPersonwithoutAge(String name,avgGPA){
        return new Person(name, 18, "Atlanta", avgGPA);
    }

}

Immutable 对象是在创建后不改变其状态的对象，并且不允许子类化。从长远来看，不可变类是有利的。

Answer 3

我认为修改您的 Person 对象以使用构建器模式会对您有所帮助。