所以我正在尝试使用返回随机单词的网站制作一个悬挂式游戏。我在刽子手游戏中使用那个随机词。
我坚持的是验证用户的猜测。这是代码,我只是将所有内容放在main中,然后在此工作之后为我做单独的方法来完成工作。
public static void main(String[] args) throws Exception {
randomWord = TestingStuff.sendGet();
int totalTries = 1;
char[] guesses = new char[26];
int length = randomWord.length();
Scanner console = new Scanner(System.in);
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);
System.out.print("\nWhat is your guess? ");
String guess = console.next();
char 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 (finalGuess != guesses[i]) {
for (int j = 0; i < length; j++) { //scans each letter of random word
if (finalGuess.equals(randomWord.charAt(j))) {
}
}
} else {
System.out.println("Letter already guessed, try again! ");
}
}
}
}
我坚持的是在while循环中,它说:
for (int i = 0; i < totalTries; i++) { //checks to see if the letter is already guessed
if (finalGuess != guesses[i]) {
for (int j = 0; i < length; j++) { //scans each letter of random word
if (finalGuess.equals(randomWord.charAt(j))) {
}
}
} else {
System.out.println("Letter already guessed, try again! ");
}
}
它给了我一个错误说&#34; char不能被解除引用&#34;。我在这里错过了什么吗?
答案 0 :(得分:3)
finalGuess
是原始char
- 您无法在其上使用equals
等方法。您可以使用char
运算符比较两个==
:
if (finalGuess == randomWord.charAt(j)) {