如果某个字符的猜测错误，则打印出一条错过消息。刽子手游戏

Question

   do {            
            System.out.println("Word: " + secretWord.getWordMask());
            //System.out.print("Guesses: " + guesses);
            System.out.print("Enter your guess: ");
            Scanner keyboard = new Scanner(System.in);
            String guess = keyboard.next();              
            WordHider revealChar = new WordHider();                
            revealChar.revealLetter(guess);
            System.out.println(revealChar.getWordMask());
            secretWord.revealLetter(guess);
            if (guess != ???)  {
                System.out.println("Miss");
            } 

所以我有那么多，它做了什么它要求用户猜一个字母，但我不知道如何使它如此，如果他们猜的字母是正确的，打印一个“未命中”的消息。 / p>

public int revealLetter(String letter) {
    int count = 0;
    String newFoundWord = "";
    if (letter.length() == 1) {
        for (int i = 0; i < secretWord.length(); i++) {
            if ((secretWord.charAt(i) == letter.charAt(0))
                    && (wordMask.charAt(i) == HIDE_CHAR.charAt(0))) {
                count++;
                newFoundWord += letter;
            }
            else {
                newFoundWord += wordMask.charAt(i);
            }
        }
    }
    wordMask = newFoundWord;
    return count;

上面的代码是显示该字母的代码，如果猜到的话。

Answer 1

您的secretWord.revealLetter(guess)已经count返回guess secretWord在if中的次数，当它是未命中时为0。所以你可以把它放在if (secretWord.revealLetter(guess) == 0) { System.out.println("Miss"); } 条件：

secretWord.revealLetter(guess)

您可以在if之前删除对{{1}}的来电，因为您只需要拨打一次。