我的代碼中發生StringIndexOutOfBoundsException的原因是什麼

Question

在我的學校任務中​​，我必須製作遞歸方法，並且只能使用.charAt，.indexOf，.substring，.toLowerCase，.concat和其他一些我擁有的方法在代碼中使用

這是代碼

/* 
* Lab to perform different functions on Strings 
* all methods are static 
* only two methods should be public 
* all other methods are internal (only used in the class) 
*/ 

package stringutil; 

/** 
* @author [REDACTED] 
*/ 

public class StringUtil { 

    public static boolean inOut(String input){//the argument is in main 
     int len = input.length(); 
     boolean test; 

     input = input.toLowerCase(); 

     //call the cleaners 
     input = StringUtil.cleanse(input, len); 

     //this is le final product 
     String reverse = StringUtil.flip(input); 
     test = input.equals(reverse); 

     return test; 
    } 

    private static String cleanse(String raw, int count){ 
     if (count < 0) 
      return (""); 
     //this means that there was invalid punctuation 
     else{ 
      char ch; 
      ch = raw.charAt(count); 

      if (ch >= 97 && ch <= 122 || ch >= 48 && ch<= 57){ 
       //call method again with count-1 | string is same 
       return cleanse(raw, count-1); 
      } 
      else{ //character ain't ok yo 
       if (raw.indexOf(count) == -1){ 
        raw = raw.substring(raw.length()-count, count-1); 
       } 
       else 
        raw = raw.substring(0,count-1).concat(raw.substring(count+1)); 
       return cleanse(raw, count); 
      } 
     } 
    } 

    public static String flip(String input){ 
     String newer; 
     // base case 
     if (input.length() == 1){ 
      return input; 
     } 
     else{ 
     //take the last letter and make it the new start 
      newer = input.substring(input.length()-1); 
      input = input.substring(0, input.length()-1); 
      return newer + flip(input); 
     } 
     //input = newer + 
     // flip(input.substring(0, input.length()-1)); 
    } 

/** 

* @param args the command line arguments 

*/ 

    public static void main(String[] args) { 
     // TODO code application logic here 
     System.out.println(StringUtil.flip("aashf")); 
     System.out.println(StringUtil.inOut("what, t;haw")); 
    } 
} 

所以我得到這個作爲運行

fhsaa 
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 11 
    at java.lang.String.charAt(String.java:646) 
    at stringutil.StringUtil.cleanse(StringUtil.java:36) 
    at stringutil.StringUtil.inOut(StringUtil.java:21) 
    at stringutil.StringUtil.main(StringUtil.java:78) 
Java Result: 1 
BUILD SUCCESSFUL (total time: 0 seconds) 

我已經與方法擺弄周圍用於清潔字符to後帽子不是字母或數字，但系統似乎喜歡從字符串到字符的轉移。

我翻轉方法的工作，但我的清洗經常遇到錯誤的地方是超出範圍。我曾嘗試添加很多東西以確保它在範圍內，但這隻會增加問題。

Answer 1

有幾個邏輯錯誤的;我想這是你試圖做：

/* 
* Lab to perform different functions on Strings 
* all methods are static 
* only two methods should be public 
* all other methods are internal (only used in the class) 
*/ 

package stringutil; 

/** 
* @author [REDACTED] 
*/ 

public class StringUtil { 

    public static boolean inOut(String input) {// the argument is in main 
     int len = input.length(); 
     boolean test; 

     input = input.toLowerCase(); 

     // call the cleaners 
     input = StringUtil.cleanse(input, len - 1); 

     // this is le final product 
     String reverse = StringUtil.flip(input); 
     test = input.equals(reverse); 

     return test; 
    } 

    private static String cleanse(String raw, int count) { 
     if (count < 0) 
      return raw; 
     // this means that there was invalid punctuation 
     else { 
      char ch; 
      ch = raw.charAt(count); 

      if (ch >= 97 && ch <= 122 || ch >= 48 && ch <= 57) { 
       // call method again with count-1 | string is same 
       return cleanse(raw, count - 1); 
      } else { // character ain't ok yo 
       raw = raw.substring(0, count).concat(raw.substring(count + 1)); 
       return cleanse(raw, count - 2); 
      } 
     } 
    } 

    public static String flip(String input) { 
     String newer; 
     // base case 
     if (input.length() == 1) { 
      return input; 
     } else { 
      // take the last letter and make it the new start 
      newer = input.substring(input.length() - 1); 
      input = input.substring(0, input.length() - 1); 
      return newer + flip(input); 
     } 
     // input = newer + 
     // flip(input.substring(0, input.length()-1)); 
    } 

    /** 
    * 
    * @param args 
    *   the command line arguments 
    * 
    */ 

    public static void main(String[] args) { 
     // TODO code application logic here 
     System.out.println(StringUtil.flip("aashf")); 
     System.out.println(StringUtil.inOut("what, t;ahw")); 
    } 
} 

---

基本上，你的// character ain't ok yo代碼段是有缺陷的，因爲是你給cleanse電話。我還沒有檢查出你的代碼的其餘部分，但main中調用的代碼部分似乎現在正在工作。此外，作爲前TA，請考慮在您的代碼中添加註釋。

P.S：我也改變了輸入字符串爲呼叫inOut輸出正確的。

Answer 2

是這樣的.length()爲一個字符串返回字符串內的數量或字符，所以你總是檢查越界，因爲你需要做一個減號後得到的長度爲了不超出界限，因爲.charAt()返回字母在上述特定地點

Answer 3

問題是與此否則代碼塊，如果檢查

raw = raw.substring(0,count-1).concat(raw.substring(count+1)); 

您正在嘗試與子的位置，這是出界後Concat的。 count是String的長度，並且您正在嘗試使count + 1位置的子字符串。