因此，这是在这里添加到库中的另一个子手问题。我的实体和边界类几乎全部完成，只不过有一个名为revealLetter()的方法，该方法用正确猜测的字母替换空格。它还会计算正确猜出的字母(如果有)的数量，并将该整数返回给驱动程序，以确定它是否未命中或命中。如果用户输入了错误的字母，revealLetter()将返回零，否则它将返回正确的字母数来确定正确的字母。我的问题是，即使填写正确的字母，revealLetter()始终返回零。我抛出了一些信号以隔离正在发生的事情，并且退出我的for循环后，计数器似乎设置为零。我仍在学习Java，因此很有可能它很简单，但目前对我来说似乎很复杂。这是驱动程序:
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!");
}
}
}
此类是将单词从.txt存储到数组并生成随机单词的类:
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;
}
}
这是包含revealLetter()的类，它还处理随机单词:
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 += "*";
}
}
}
还值得注意的是，我正在参加CS大学 class ，并且有严格的法律禁止复制不属于我的代码。 因此，如果发生这种情况，请您用英语解释我做错的事情。我仍然想弄清楚代码，在逻辑上我只是被卡住了。提前致谢
最佳答案
在驱动程序main()中，您具有:
hider.revealLetter(guessedLetter.toLowerCase());
if (hider.revealLetter(guessedLetter)== 0)
这就是为什么您接到一个成功的电话，然后第二回合无事可做的原因。我可以强调一些样式问题，但其中一个大问题是:
if (letter.length() < 1 || letter.length() > 1) {
correctChars = 0;
return correctChars;
} else {
为什么不只是
letter.length() != 1，又因为correctChars已经被初始化为零，所以您无需再次执行此操作，因此可以删除整个“then”部分，并且if变为letter.length() == 1。也:
tempString += Character.toString(hiddenWord.charAt(i));
和:
tempString += partiallyFoundWord.charAt(i);
两者都做相同的事情，因此选择一种样式或另一种样式。