java - 如果一个字符的猜测是错误的，打印出一条消息说错过。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");
            } 

所以我得到了这么多，它要求用户猜一个字母，但我不知道是如何做到的，所以如果他们猜的字母不正确，打印一个“错过”消息。

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;

上面的代码是在猜到时显示字母的代码。

Answer 1

你有你的secretWord.revealLetter(guess)which 返回count多少次guess是 in secretWord，当它是一个未命中时为 0。所以你可以把它放在if条件下：

if (secretWord.revealLetter(guess) == 0)  {
    System.out.println("Miss");
}

您可以在secretWord.revealLetter(guess)之前删除对 的调用if，因为您只需要调用一次。