在运行hang子手（Java）的程序中出现逻辑错误

Question

所以这里还有另一个hang子手问题添加到库中。除了一个名为revealLetter（）的方法外，我的实体和边界类都是完整的，它用正确猜测的字母替换空白。它还会计算正确猜测字母的数量（如果有的话），并将该整数返回给驱动程序以确定它是否触发未命中或命中。如果用户输入错误的字母，revealLetter（）将返回零，否则返回正确字母的数量以确定正确的字母。我的问题是，尽管填写了正确的字母，revealLetter（）总是返回一个零。我已经投入几个sout来隔离发生了什么事情，并且在退出我的for循环后，似乎计数器被设置为零。我仍然在学习Java，所以很可能这很简单，但目前对我来说似乎很复杂。下面是驱动程序：

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!"); 
    } 

} 

}

这是从.txt到一个数组存储字，并产生随机字的类：

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;  
} 

}

而这是包含revealLetter（）的类，它也处理随机字：

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 += "*"; 
    } 

} 

}

还值得一提的是，我在一个CS大学课程，并有抄袭代码严格的法律，不属于我。因此，如果发生这种情况的任何一种灵魂，你能解释我在大部分英语中做错了吗？我仍然想把代码弄清楚，我只是逻辑上卡住了。在此先感谢

Answer 1

在你的驱动main()您有：

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

这就是为什么你会得到一个成功的呼叫，则第二次轮没有什么做的。还有我强调几个风格的问题，而是一个大的一个是：

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

为什么不letter.length() != 1，由于correctChars已经初始化为零，你不需要再这样做，所以整个“然后“部分可以被丢弃并且if变成letter.length() == 1。

另外：

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

和：

tempString += partiallyFoundWord.charAt(i);         

都做同样的事情，所以选择一种风格或其他。