java - 以这种方式使用多个构造函数是否正确？

Question

这个问题已经有答案了:

Constructor overloading in Java - best practice (5 个回答)

已关闭 6 年前。

我不太确定这是如何工作的，但是如果我想提供为类的对象提供更多或更少变量的选项，这可以与像这样的多个构造函数一起使用吗？

假设我想创建一个多项选择问卷，但我不知道我的用户想要输入多少个答案，也许是 2,3,4,5,6？因此:

public class Quiz {
    private int counter;
    private String question;
    private String answer1;
    private String answer2;
    private String answer3;
    private String answer4;
    private String answer5;
    private String answer6;
    private String rightAnswer;

    public Quiz(int counter,String question, String answer1, String answer2, String rightAnswer){
        super();
        this.counter = counter;
        this.question = question;
        this.answer1 = answer1;
        this.answer2 = answer2;
        this.rightAnswer = rightAnswer;
    }
    public Quiz(int counter, String question, String answer1, String answer2, String answer3, String rightAnswer) {
        super();
        this.counter = counter;
        this.question = question;
        this.answer1 = answer1;
        this.answer2 = answer2;
        this.answer3 = answer3;
        this.rightAnswer = rightAnswer;
    }
    public Quiz(int counter, String question, String answer1, String answer2, String answer3, String answer4,
                String rightAnswer) {
        super();
        this.counter = counter;
        this.question = question;
        this.answer1 = answer1;
        this.answer2 = answer2;
        this.answer3 = answer3;
        this.answer4 = answer4;
        this.rightAnswer = rightAnswer;
    }
    //...more options

也许我可以只用某种枚举或开关做 1 个构造函数？归根结底，在尝试此方法后，由于某种原因，将其放入 HashMap 中，然后将其序列化到文件不起作用，与 1 个构造函数一样，它可以工作，但不会写入所有内容在那里。我对问题所在感到有点困惑，也许这与我的 toString 覆盖有关，但无论如何，请告诉我这个问题，这样我就不用再担心一个令人困惑的问题了。

Answer 1

对于您发布的代码，这将是一个简单的方法:

package com.steve.research;

public class Quiz {

    private int counter;
    private String question;
    private String answer1;
    private String answer2;
    private String answer3;
    private String answer4;
    private String answer5;
    private String answer6;
    private String rightAnswer;

    public Quiz(int counter, String question, String answer1, String answer2, String rightAnswer) {
        this(counter, question, answer1, answer2, null, null, rightAnswer);
    }

    public Quiz(int counter, String question, String answer1, String answer2, String answer3, String rightAnswer) {
        this(counter, question, answer1, answer2, answer3, null, rightAnswer);
    }

    public Quiz(int counter, String question, String answer1, String answer2, String answer3, String answer4, String rightAnswer) {
        this.counter = counter;
        this.question = question;
        this.answer1 = answer1;
        this.answer2 = answer2;
        this.answer3 = answer3;
        this.answer4 = answer4;
        this.rightAnswer = rightAnswer;
    }
}

要获得改进的方法，我建议您查看“varargs”来查找问题。由于您的问题数量不定，因此您可以将 String ... questions 作为最后一个构造函数参数(因此 rightAnswer 必须放在前面)。

public class Quiz {

    private int counter;
    private String question;
    private String rightAnswer;
    private String[] answers;

    public Quiz(int counter, String question, String rightAnswer, String... answers) {
        this.counter = counter;
        this.question = question;
        this.rightAnswer = rightAnswer;
        this.answers = answers;
    }

    public static void main(String[] args) {
        new Quiz(1, "one plus one", "two", "one", "two", "three");
        new Quiz(1, "one plus one", "two", "one", "two", "three", "four");
        new Quiz(1, "one plus one", "two", "one", "two", "three", "four", "five");
    }
}

请注意，answers 现在是一个字符串数组 String[]，您可以引用 answers.length、answers[0] 等等。

还有一点评论:在构造函数中调用无参数 super() 通常是多余的(您不需要它们)。