在運行hang子手（Java）的程序中出現邏輯錯誤

Question

所以這裏還有另一個hang子手問題添加到庫中。除了一個名爲revealLetter（）的方法外，我的實體和邊界類都是完整的，它用正確猜測的字母替換空白。它還會計算正確猜測字母的數量（如果有的話），並將該整數返回給驅動程序以確定它是否觸發未命中或命中。如果用戶輸入錯誤的字母，revealLetter（）將返回零，否則返回正確字母的數量以確定正確的字母。我的問題是，儘管填寫了正確的字母，revealLetter（）總是返回一個零。我已經投入幾個sout來隔離發生了什麼事情，並且在退出我的for循環後，似乎計數器被設置爲零。我仍然在學習Java，所以很可能這很簡單，但目前對我來說似乎很複雜。下面是驅動程序：

package hangman; 

import java.util.Scanner; 

public class Hangman { 

public static int NUMBER_MISSES = 5; 

public static void main(String[] args) { 

    String guessedLetter; 
    WordHider hider = new WordHider(); 
    Dictionary dictionary = new Dictionary(); 

    Scanner Keyboard = new Scanner(System.in); 
    hider.setHiddenWord(dictionary.getRandomWord()); 
    System.out.println(hider.getHiddenWord().length()); 
    System.out.println(hider.getHiddenWord()); 

    do { 
     hider.wordFound(); 
     System.out.printf(hider.getPartiallyFoundWord() + " Chances Remaing: %d \nMake a guess: ", NUMBER_MISSES); 
     guessedLetter = Keyboard.nextLine(); 
     hider.revealLetter(guessedLetter.toLowerCase()); 
     if (hider.revealLetter(guessedLetter)== 0) { 
      NUMBER_MISSES--; 
      if (NUMBER_MISSES == 4) { 
       System.out.println("Swing and a miss!"); 
      } 
      else if (NUMBER_MISSES == 3) { 
       System.out.println("Yup. That. Is. A. Miss."); 
      } 
      else if (NUMBER_MISSES == 2) { 
       System.out.println("MISS! They say third time is a charm."); 
      } 
      else if (NUMBER_MISSES == 1) { 
       System.out.println("Ouch. One guess left, think carefully."); 
      }    
     } else { 
      System.out.println("That's a hit!"); 
     } 
     if (hider.wordFound() == true) { 
      NUMBER_MISSES = 0; 
     } 
    } while (NUMBER_MISSES > 0); 

    if ((NUMBER_MISSES == 0) && (hider.wordFound() == false)) { 
     System.out.println("Critical Failure. The word was " + hider.getHiddenWord() + " try harder next time and you'll win."); 
    } else if ((NUMBER_MISSES == 0) && (hider.wordFound() == true)) { 
     System.out.println(hider.getHiddenWord() + "\nBingo! You win!"); 
    } 

} 

}

這是從.txt到一個數組存儲字，併產生隨機字的類：

package hangman; 

import java.util.Random; 
import java.io.File; 
import java.io.FileNotFoundException; 
import java.util.Scanner; 

public class Dictionary { 

//Random randomizer = new Random(); 
private static String randomWord; 
String[] dictionary = new String[81452]; 
private static String FILE_NAME = "dictionarycleaned.txt"; 

Dictionary() { 
    int words = 0; 
    Scanner infile = null; 
    try { 
     infile = new Scanner(new File(FILE_NAME)); 
     while (infile.hasNext()) { 
      dictionary[words] = infile.nextLine(); 
      words++; 

     } 
     //System.out.println(dictionary[81451]); 
    } catch (FileNotFoundException e) { 
     System.err.println("Error opening the file " + FILE_NAME); 
     System.exit(1); 
    } 

} 

public String getRandomWord(){ 
    //randomWord = (dictionary[randomizer.nextInt(dictionary.length)]); //Are either of these techniques better than the other? 
    randomWord = (dictionary[new Random().nextInt(dictionary.length)]); 
    return randomWord;  
} 

}

而這是包含revealLetter（）的類，它也處理隨機字：

package hangman; 

public class WordHider { 

private static String hiddenWord; 
private static String partiallyFoundWord; 

WordHider() { 

    hiddenWord = ""; 
    partiallyFoundWord = ""; 

} 

public String getHiddenWord() { 
    return hiddenWord; 
} 

public String getPartiallyFoundWord() { 

    return partiallyFoundWord; 

} 

public void setHiddenWord(String newHiddenWord) { 
    int charCount; 
    hiddenWord = newHiddenWord; 
    for (charCount = 0; charCount < hiddenWord.length(); charCount++) { 
     partiallyFoundWord += "*"; 
    } 

} 

public int revealLetter(String letter) { 
    int correctChars = 0; 

    if (letter.length() < 1 || letter.length() > 1) { 
     correctChars = 0; 
     return correctChars; 
    } else { 

     String tempString = ""; 

     for (int i = 0; i < hiddenWord.length(); i++) { 
      if ((letter.charAt(0) == hiddenWord.charAt(i)) && (partiallyFoundWord.charAt(i) == '*')) {     
       correctChars++; 
       tempString += Character.toString(hiddenWord.charAt(i)); 

      } else { 
       tempString += partiallyFoundWord.charAt(i); 



      } 

     } 
     partiallyFoundWord = tempString;   
    } 

    return correctChars; 
} 

public boolean wordFound() { 
    boolean won = false; 
    if (partiallyFoundWord.contains(hiddenWord)) { 
     won = true; 
    } 
    return won; 
} 

public void hideWord() { 
    for (int i = 0; i < hiddenWord.length(); i++) { 
     partiallyFoundWord += "*"; 
    } 

} 

}

還值得一提的是，我在一個CS大學課程，並有抄襲代碼嚴格的法律，不屬於我。因此，如果發生這種情況的任何一種靈魂，你能解釋我在大部分英語中做錯了嗎？我仍然想把代碼弄清楚，我只是邏輯上卡住了。在此先感謝

Answer 1

在你的驅動main()您有：

hider.revealLetter(guessedLetter.toLowerCase()); 
if (hider.revealLetter(guessedLetter)== 0) 

這就是爲什麼你會得到一個成功的呼叫，則第二次輪沒有什麼做的。還有我強調幾個風格的問題，而是一個大的一個是：

if (letter.length() < 1 || letter.length() > 1) { 
    correctChars = 0; 
    return correctChars; 
} else { 

爲什麼不letter.length() != 1，由於correctChars已經初始化爲零，你不需要再這樣做，所以整個「然後「部分可以被丟棄並且if變成letter.length() == 1。

另外：

tempString += Character.toString(hiddenWord.charAt(i)); 

和：

tempString += partiallyFoundWord.charAt(i);         

都做同樣的事情，所以選擇一種風格或其他。