JavaHangman类方法错误

Question

我几乎完成了一个基本的刽子手游戏，但我在课堂上遇到了“makeGuess”方法的困难。我试图测试用户输入（他们的一个字符猜测）与秘密单词，并更新伪装的单词，如果他们的猜测是正确的或不正确的。任何正确方向的建议或指点都将不胜感激。每次运行时都会出现以下错误：

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 11
at java.lang.String.charAt(String.java:686)
at hangmandemo.Hangman.makeGuess(Hangman.java:42)
at hangmandemo.HangmanDemo.main(HangmanDemo.java:24)
Java Result: 1

这是刽子手课：

package hangmandemo;

public class Hangman {

private String secretWord;
private String disguisedWord;
private int guesses;
private int wrongGuesses;
boolean found=false;

public String getDisguisedWord() {
    return disguisedWord;
}

public String getSecretWord() {
    return secretWord;
}

/**
 * returns number of guesses
 */
public int getGuesses() {
    return guesses;
}

/**
 * continues game until secretWord is found
 */
public boolean isFound() {
    return found;
}

/**
 * brings in a char to test against String
 */
public void makeGuess(char c) {
    for (int i = 0; i <= disguisedWord.length(); i++) {

        if (c == secretWord.charAt(i)) {
            disguisedWord = disguisedWord.substring(0, i-1) + c + disguisedWord.substring(i+1);

        }
        if (secretWord.equals(disguisedWord)) {
            found = true;
        }
    }
    wrongGuesses++;
}

/**
 * assigns secret String to secret word
 */
public void createSecretWord(String secret) {
    secretWord = secret;
}

/**
 * disguises secret word with "???"
 */
public void createDisguisedWord() {

    for (int i = 0; i < secretWord.length(); i++) {
        disguisedWord += "*";
    }
}

/**
 * returns number of wrong guesses
 */
public int getWrongGuesses() {
    return wrongGuesses;
}

}

这是我的演示课

package hangmandemo;
import java.util.Scanner;
public class HangmanDemo {
public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    Hangman game = new Hangman();
    game.createSecretWord("assessments");
    game.createDisguisedWord();
    System.out.println("This is the hangmandemo. Press the number '3' to quit.");
    System.out.println("The disguised word is..."+ game.getDisguisedWord());
    while (!game.isFound()) {
        System.out.println("Enter a one character guess");
        char c = keyboard.next().charAt(0);
        if (c == 3) {
            System.out.println("game over, the word was... " + game.getSecretWord());
            System.exit(0);
        }
        game.makeGuess(c);
        System.out.println(game.getDisguisedWord());
    }

    System.out.println("you guessed it, the word was... " + game.getSecretWord());
    System.out.println("you had " + game.getGuesses() + " guesses.");
    System.out.println("you had " + game.getWrongGuesses() + " wrong guesses.");
}

}

Answer 1

将替换为（int i=0； i

带for（int i=0；i

在JAVA中，所有东西都是基于0的索引：“abc”在索引0处有字符'a'，'b'在1处和'c'在2处，其长度是3。

Answer 2

int i = 0; i <= disguisedWord.length(); i++

这一行，您必须更改

Answer 3

这里有3个问题。

在你的圈子里

for (int i = 0; i <= disguisedWord.length(); i++) 

convergedword的最后一个元素是'convergedword[convergedword.length-1]，因为数组的索引从0开始。

按照惯例，循环的格式如下

for (int i = 0; i < disguisedWord.length(); i++) 

但这还不够，因为

disguisedWord.substring(i+1)

会在那一点上出界。

还有这个表达

disguisedWord.substring(0, i-1)

将在i=0时引发异常；

你需要更仔细地关注你这里的边缘条件。