我目前正在通过java处理一个版本的hangman。我最近被一个我似乎无法解决的问题所困扰。
我有一个数组easyDifficulty和一个字符串hiddenWord。我希望这样做，以便每次游戏开始时，我的方法randomWord()将从数组easyDifficulty中选择一个随机单词来启动程序。我的很多方法都使用hiddenWord所以我粘贴了所有代码，因为你需要检查一下。否则，我认为导致问题的主要部分是在我声明我的字段变量，randomWord方法和main方法的区域。
import java.util.Arrays;
import java.util.*;
import java.util.Scanner;
public class HangmanGame {
static String[] easyDifficulty = new String[]{"orange", "jacket","shirt"
,"rocket","airplane","circle","balloon","swing","truck","caterpillar"};
static Random rand = new Random();
static String hiddenWord = new String("");
static char[] hiddenWordToChar = hiddenWord.toCharArray();
static int triesLeft = 6;
static boolean done = false;
static int length = hiddenWord.length();
static Scanner input = new Scanner(System.in);
final static int maxLength = 30;
static char[] repeatChecker = new char[30]; //this is to help prevent the user from putting the same char multiple times
static char[] fillWord = new char[length];
static int var;
static int var2;
static char underscore = '_';
static int chooseDifficultyInt;
public static String randomWord(int n) {
int randomNum = rand.nextInt(9);
int difficultyInt = n;
if (difficultyInt == 1){
//System.out.println(easyDifficulty[randomNum]);
return easyDifficulty[randomNum];
}
return null;
}
public static boolean contains(char[] arr, char i) {
for (char n : arr) {
if (i == n) {
return true;
}
}
return false;
}
public static boolean multiLetter(char[] arr, char i){
int multiple = 0;
for (char n : arr){
if (i == n){
multiple++;
//continue;
}
}
System.out.println(multiple);
if (multiple > 1){
return true;
}
else {
return false;
}
}
public static void createSpaces(int n){
for (int i = 0; i <= n-1; i++){
fillWord[i] = underscore;
}
System.out.println(fillWord);
}
public static void tryAgain(){
System.out.println("Would you like to try again? Enter Y/N to go again or quit!");
char goAgain = input.next().charAt(0);
goAgain = Character.toLowerCase(goAgain);
if (goAgain == 'y') {
triesLeft = 6;
repeatChecker = new char[20];
main(null);
}
else if (goAgain == 'n') {
System.out.println("Bye!");
System.exit(0);
}
else {
System.out.println("Invalid input");
tryAgain();
}
}
public static void arrayLetters(){
System.out.println("This is a " + length + " letter word. Please enter a letter to guess: ");
char charInput = input.next().charAt(0);
char[] letters = new char[20];
//This code only runs if the user inputs a letter that repeats
//throughout hiddenWord
if (multiLetter(hiddenWordToChar, charInput)){
for (int n = 0; n < length; n++){
char multiWordLetter = hiddenWord.charAt(n);
letters[n] = multiWordLetter;
if (letters[n] == charInput){
fillWord[n] = charInput;
}
if (contains(repeatChecker, charInput)){
System.out.println("You already did that word, try again!");
arrayLetters();
}
if (Arrays.equals(fillWord, hiddenWordToChar)){
System.out.println("Congratulations, you win! The word was '" + hiddenWord + "'!");
System.out.println("You completed the challenge in " + triesLeft + " tries! \n\n");
tryAgain();
}
}
System.out.println(fillWord);
System.out.println("Nice! There is a(n) " + charInput + " in this word!");
System.out.println("You have " + triesLeft + " tries left!\n");
arrayLetters();
}
//This block of code runs when the user input a letter that only occurs once
//in hiddenWord
for (int i = 0; i < length; i++){
char wordLetter = hiddenWord.charAt(i);
letters[i] = wordLetter;
if (contains(letters, charInput)){
/*
if (multiLetter(letters, charInput)){
System.out.println("aylmao");
}
*/
//System.out.println(multiLetter(hiddenWordArray, charInput));
if (contains(repeatChecker, charInput)){
System.out.println("You already did that word, try again!");
arrayLetters();
}
repeatChecker[var] = charInput;
var++;
fillWord[i] = charInput;
if (Arrays.equals(fillWord, hiddenWordToChar)){
System.out.println("Congratulations, you win! The word was '" + hiddenWord + "'!");
System.out.println("You completed the challenge in " + triesLeft + " tries! \n\n");
tryAgain();
}
System.out.println(fillWord);
System.out.println("Nice! There is a(n) " + charInput + " in this word!");
System.out.println("You have " + triesLeft + " tries left!\n");
arrayLetters();
}
if (i == length-1){
System.out.println("There is no " + charInput + " in this word!");
triesLeft--;
System.out.println("You have " + triesLeft + " tries left!\n");
if (triesLeft <= 0){
System.out.println("You failed!\n\n");
tryAgain();
}
arrayLetters();
}
}
}
public static void main(String[] args) {
System.out.println("Welcome to my hangman game!");
System.out.println("Please input your prefered difficulty level.");
System.out.println("1. Easy");
System.out.println("2. Medium");
System.out.println("3. Hard");
chooseDifficultyInt = input.nextInt();
randomWord(chooseDifficultyInt);
hiddenWord = randomWord(chooseDifficultyInt);
createSpaces(length);
arrayLetters();
}
}
我是java编程的中间人，所以我已经知道字符串是不可变的。因此，我认为将hiddenWord设置为等于randomWord(chooseDifficultyInt);的原因是因为这个原因？
答案 0 :(得分：1)
我是java编程的中间人，所以我已经知道字符串是不可变的
这经常会引起新程序员的混淆。这是正确的，字符串是不可变的。但是，指向String 的变量可以重新分配 （假设它不是final）。
您可以根据需要多次修改隐藏的字词：
HangmanGame.hiddenWord = "MyNewWord";
HangmanGame.hiddenWord = "AnotherWord";
HangmanGame.hiddenWord = "ThisWorks";
如果字符串是不可变的，你可能想知道hiddenWord是如何变化的。
您的变量实际上正在发生变化。每次重新分配它时，都会创建一个新的String。这意味着字符串本身永远不会被修改。
是的，你可以打电话给
HangmanGame.hiddenWord = HangmanGame.randomWord(n);
它会完美运作。
答案 1 :(得分：0)
将hiddenWord初始化为new String("");是没用的。你可以安全地删除它。
当你这样做时：
hiddenWord = randomWord(chooseDifficultyInt);
您将hiddenWord设置为预定义列表的随机字符串的引用，而不是分配任何内容，只是重用现有的字符串引用。