从递归方法调用的对象数组上的 toString 方法答案

【问题标题】：toString method on array of objects called from recursion method从递归方法调用的对象数组上的 toString 方法
【发布时间】：2016-06-19 21:49:59
【问题描述】：

class Anadromi001
    {
        public void Anadromi01 (Foititis[] F, short ap01, short i)
            {
                if (ap01 == 4) System.exit(0);
                Anadromi002 obj02 = new Anadromi002();
                String onoma, epitheto;
                short MesOr;
                int AritMit, EtosEis;
                short ap02;
                boolean flag00, flag02, flag03;
                int k, f, l = 0;          
                k = F.length-1
[...]
case 3:
                        f = obj02.Anadromi02(F, l);
                        if (f == -1) System.out.println("O pinakas einai adeios.");
                        else 
                            {
                                for (int j = 0; j <= f; j++)
                                    {
                                        System.out.println(F[j].toString());
                                    }
                            }
                        //System.out.println(f);
                        break;
                    }
    }

这是我的代码的一部分，我想做的是如果我调用第二个递归（当前代码本身就是一个递归），然后使用第二个递归的值来打印我的 toString 方法（放置在对象类）。不用理会case 3，它只是一个“代码”，供用户在程序中选择要做什么。

class Foititis 
    {
        String onoma, epitheto;
        short MesOr;
        int AritMit, EtosEis;
        public Foititis (String on, String ep, int AM, int EE,  short MO)
            {
                onoma = on;
                epitheto = ep;
                AritMit = AM;
                EtosEis = EE;
                MesOr = MO;
            }

        public String toString()
            {
                String emf;
                emf = "--------------------" + "\n";
                emf = "Onoma: " + onoma + "\n";
                emf = "Epwnymo: " + epitheto + "\n";
                emf = "Arithmos Mitrwoy: " + AritMit + "\n";
                emf = "Etos Eisagwnis: " + EtosEis + "\n";
                emf = "Mesos Oros Mathimatwn: " + MesOr + "\n";
                emf = "--------------------";
                return emf;
            }
    }

当我执行程序时，我得到的输出是：--------------------

【问题讨论】：

标签： java arrays object recursion tostring

【解决方案1】：

您实际上是在每一行中分配新值，而不是追加。所以它只显示最后一行。从第二个作业开始，您可以使用

emf += <your string>

或者您可以使用字符串生成器并附加行......最后是 toString

【讨论】：

【解决方案2】：

使用+= 而不是= 来连接字符串，否则您将看到最后分配的内容。

public String toString()
    {
        String emf;
        emf = "--------------------" + "\n";
        emf += "Onoma: " + onoma + "\n";
        emf += "Epwnymo: " + epitheto + "\n";
        emf += "Arithmos Mitrwoy: " + AritMit + "\n";
        emf += "Etos Eisagwnis: " + EtosEis + "\n";
        emf += "Mesos Oros Mathimatwn: " + MesOr + "\n";
        emf += "--------------------";
        return emf;
    }

或者干脆

public String toString()
    {
        return "--------------------" + "\n" +
               "Onoma: " + onoma + "\n" +
               "Epwnymo: " + epitheto + "\n" +
               "Arithmos Mitrwoy: " + AritMit + "\n" +
               "Etos Eisagwnis: " + EtosEis + "\n" +
               "Mesos Oros Mathimatwn: " + MesOr + "\n" +
               "--------------------";
    }

【讨论】：

【解决方案3】：

使用

emf = emf + "Prevous line" + "\n";

在每个新行上。您唯一的最后一行是打印。

正确的格式是：

              emf = "--------------------" + "\n"+
              "Onoma: " + onoma + "\n"+
              "Epwnymo: " + epitheto + "\n"+
              "Arithmos Mitrwoy: " + AritMit + "\n"+
              "Etos Eisagwnis: " + EtosEis + "\n"+
              "Mesos Oros Mathimatwn: " + MesOr + "\n"+
               "--------------------";

或

        emf = "--------------------" + "\n";
        emf = emf +  "Onoma: " + onoma + "\n";
        emf = emf + "Epwnymo: " + epitheto + "\n";
        emf = emf + "Arithmos Mitrwoy: " + AritMit + "\n";
        emf = emf + "Etos Eisagwnis: " + EtosEis + "\n";
        emf = emf + "Mesos Oros Mathimatwn: " + MesOr + "\n";
        emf = emf + "--------------------";

【讨论】：