以这种方式使用多个构造函数是否正确？

Question

我不太确定这是如何工作的，但是如果我想给一个类的对象提供更多或更少的变量的选项，这是否适用于这样的多个构造函数？

假设我想创建一个多项选择问卷，但是我不知道我的用户想要输入多少个答案，可能是 2、3、4、5、6？所以为此：

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

也许我可以用某种枚举或开关做 1 个构造函数？归根结底，在尝试了这种方法之后，出于某种原因，将它放入一个哈希映射中，然后将其序列化到一个文件中是行不通的，而与 1 个构造函数一样，它可以工作，但不会在其中写入所有内容。我对问题所在有点困惑，也许这与我的 toString 覆盖有关，但无论如何，请告诉我这个问题，这样我就不必担心一个令人困惑的问题。

Answer 1

对于您发布的代码，这将是一种简单的方法：

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;
    }
}

对于改进的方法，我建议您查看问题的“可变参数”。由于您有可变数量的问题，您可以将其String ... questions作为最后一个构造函数参数（因此rightAnswer必须先行）。

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。

super()另一条评论：在构造函数中调用 no-args通常是多余的（您不需要它们）。