Question

我在尝试制作一个刽子手游戏时遇到了一些问题。我之前发过关于其他错误的帖子，但现在我遇到了一个我无法弄清楚的新错误。我正在尝试验证字母猜测是否尚未输入。但它正在跳过if / else语句的整个部分。当我运行此代码时：

public class TestingStuff {

static StringBuffer randomWord;
static Scanner console = new Scanner(System.in);
static int totalTries = 1;
static String guess;
static char finalGuess;

public static void main(String[] args) throws Exception {
    randomWord = TestingStuff.sendGet();
    char[] guesses = new char[26];
    int length = randomWord.length();

    System.out.print("* * * * * * * * * * * * * * *"
            + "\n*    Welcome to Hangman!    *"
            + "\n* * * * * * * * * * * * * * *");
    System.out.println("\nYou get 10 tries to guess the word by entering in letters!\n");
    System.out.println(randomWord);
    /*
     Cycles through the array based on tries to find letter
     */
    while (totalTries <= 10) {
        System.out.print("Try #" + totalTries + "\nWord: " + makeDashes(randomWord));

        //Right here: Search through the array of guesses, make it 26 characters to represent the alphabet
        //if the user guess equals an already guessed letter, add to try counter. If it's correct, then reveal the letter that is 
        //correct and do it again without adding to the try counter. 
        System.out.print("\nWhat is your guess? ");
        guess = console.nextLine();
        finalGuess = guess.charAt(0);
        guesses[totalTries - 1] = finalGuess; //Puts finalGuess into the array

            for (int i = 0; i < totalTries; i++) { //checks to see if the letter is already guessed
                if (guesses[i] != finalGuess) {
                    System.out.println(guesses[i]);
                    for (int j = 0; i < length; j++) { //scans each letter of random word
                        if (finalGuess == randomWord.charAt(j)) {
                            //put a method that swaps out dashes with the guessed letter
                            totalTries++;
                        }
                    }
                } else {
                    System.out.println("Letter already guessed, try again! ");
                }
            }
        }
    }

我得到了这个输出：

* * * * * * * * * * * * * * *
*    Welcome to Hangman!    *
* * * * * * * * * * * * * * *
You get 10 tries to guess the word by entering in letters!

ostracization
Try #1
Word: -------------
What is your guess? a
Letter already guessed, try again! 
Try #1
Word: -------------
What is your guess?

这只是说当数组中有一个空元素时，这个字母已经被猜到了。我在这里错过了什么吗？

Answer 1

让我们通过您的示例遍历代码（我强烈建议您自己使用调试器）：

guesses[totalTries - 1] = finalGuess; // guesses[0] = 'a'
if (guesses[i] != finalGuess) // i = 0, guesses[0] = 'a', finalGuess = 'a'
else System.out.println("Letter already guessed, try again! ");

你可以移动

guesses[totalTries - 1] = finalGuess; //Puts finalGuess into the array

在最外面的for循环的末尾。在处理之前，无需存储猜测。

Answer 2

是。变量totalTries最初为1.您可以阅读猜测，然后将guesses[totalTries - 1]设置为猜测的字符，这意味着guesses[0]等于finalGuess。然后将i从0循环到totalTries - 1，这也是0.循环执行一次，并检查第一个条目不是finalGuess。但事实是，我们只是设定它。

如果仅使用for循环来发现重复的猜测，您可以将第一个for循环中的条件更改为i < totalTries - 1，它应该可以工作，但您需要在下面移动标记刽子手的单词。为了对代码产生最小的影响，请使用m0skit0的解决方案。

Java - 制作刽子手游戏

2 个答案: