Question

我必须制作一个程序，从用户那里获取一个短语，然后向用户返回他们输入的短语的加密代码。我们使用数组或枚举列表来保存数据。我使用ROT13进行编码（将每个英文字母替换为沿着字母表向前或向后移动的13个字母）。我的程序运行但它只允许我输入一个短语，之后说：

java.lang.ArrayIndexOutOfBoundsException: 26
    at J4_1_EncryptionVer2.main(J4_1_EncryptionVer2.java:28)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
    at java.lang.reflect.Method.invoke(Unknown Source)
    at edu.rice.cs.drjava.model.compiler.JavacCompiler.runCommand(JavacCompiler.java:272)

我不确定我的程序有什么问题！请帮忙！感谢

        import java.io.*;

public class J4_1_EncryptionVer2
{
  public static void main (String [] args) throws IOException
  {
    BufferedReader myInput = new BufferedReader (new InputStreamReader (System.in));

   String letterA [] = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};
   String letterB [] = {"N","O","P","Q","R","S","T","U","V","W","X","Y","Z","A","B","C","D","E","F","G","H","I","J","K","L","M"};

    System.out.println ("Please enter a phrase: ");
    String message = myInput.readLine();

    int x = 0; 
    while (x < message.length()){

      String text = message;
      String letter = Character.toString(text.charAt(x));
        x++;

      int i = 0;

     if(letter.equals (letterA[i])){
        System.out.println (letterB[i]);
      }
      while (!(letter.equals (letterA[i]))){
        i++;
        if(letter.equals (letterA[i])){
        System.out.println (letterB[i]);
      }

    }
  }
}
}

Answer 1

三个问题：

while (x <= message.length())应为while (x < message.length())
x++应该在text.charAt(x)
使用equals来比较字符串值。

当然你只能用A-Z来测试

Answer 2

对于初学者来说，异常的原因是数组索引超出界限第28行。你不必费心去检查我是否在界限范围内。此外，您正在从用户输入的消息中取一封信，这封信可以是键盘上的任何内容，然后在letterA中查找，这只是26个大写字母！基本上你走完了字母的结尾，从未找到信。

Answer 3

您仅为大写字母字符提供 letterA [] 。你的比较循环是好的。但你必须检查 while（letter！= letterA [i]）循环。

确保将程序输入作为大写字母字符仅用于测试输出。

Answer 4

import java.io.*;

public class JavaApplication1 {

    public static void main(String[] args) throws IOException {
        BufferedReader myInput = new BufferedReader(new InputStreamReader(System.in));

        char letterA[] = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
        char letterB[] = {'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M'};

        System.out.println("Please enter a phrase: ");
        String message = myInput.readLine();
        message = message.toUpperCase();
        char letterC[] = new char[message.length()]; 

        int x = 0;
        for (int j = 0; j < message.length(); j++) {
            for (int i = 0; i < letterA.length; i++){
                if(message.charAt(j)==letterA[i]){
                    letterC[j]=letterB[i];
                }
            }

        }

        for (int j = 0; j < message.length(); j++) {
            System.out.print(letterC[j]);
        }
        System.out.println("  -- done");
    }
}

Answer 5

像上面提到的其他用户一样，你没有处理数组大小..并且没有什么能阻止你的while循环：

while (!(letter.equals (letterA[i])))

在处理数组时我不太喜欢。在我看来，这就是你的外表应该是什么样子。

      while (x < message.length())          
            {
                String text = message;
                String letter = Character.toString(text.charAt(x));
                x++;

                for(int i=0; i<letterA.length; i++)
                {
                    if(letter.equals(letterA[i]))if(letter.equals(letterA[i]))
                    {
                        System.out.print (letterB[i]);
                        break;
                    }
                    else if (letter.equals(" "))
                    {
                        System.out.print(" ");
                        break;
                    }
                }
            }

我在我的机器上进行了测试，效果很好。

输入：这是我的PLAINTEXT

输出：GUVF VF ZL CYNVAGRKG

Java程序中的字符串数组

5 个答案: