【发布时间】:2014-12-12 15:54:18
【问题描述】:
嘿,伙计们,我一直在编写我的代码并且几乎可以正常工作,但是在方法 numGuess 和 hideWord 之间的某个地方我有一个逻辑错误,没有比较两者 我想要的方式。我正在尝试替换星号。所以如果用户输入是我想要的“a” 用“a”填充的星号进行更新。
public class Lab12 {
public static final int MAXWORD = 15000;
public static final int MAXCHAR = 20;
public static final int MAXGUESS = 8;
public static void main(String[] args) {//main
Scanner keyboard = new Scanner(System.in);
int cout = 0;
//method calling bulk of other methods
storage();
}//end main
public static int pickrandom(int count)//generates random number
{
Random generator = new Random();
return generator.nextInt(count);
}
public static int storage()//holds bulk of other methods also performs intro and reads in file
{
String wordList[] = new String[MAXWORD];
String wordLetter[] = new String[MAXCHAR];
String dictionary = "dictionary.txt";
Scanner readFileIn = null;
String dictionaryVal = " ";
int count = 0;
String cont = "";
try//Try to read in the file.
{
readFileIn = new Scanner(new File(dictionary));//creates object scanner and object file?
while (readFileIn.hasNextLine()) {
wordList[count] = readFileIn.nextLine();
count++;
}
System.out.println("");
System.out.println("H A N G M A N");
System.out.println("");
System.out.println("This is a word guessing game A word will be selected at random");
System.out.println("and kept hidden. You will try to figure out the secret word by");
System.out.println("guessing letters which you think are in the word. You will guess");
System.out.println("one letter at a time. If the letter you guess is correct, the");
System.out.println("position(s) of the letter in the secret word will be shown.");
System.out.println("You will be allowed 8 wrong guesses. If you guess incorrectly 8");
System.out.println("times, you lose the game. If you guess all of the letters in the");
System.out.println("word, you win.");
System.out.println("");
int rand = pickrandom(count);
String hiddenWord = wordList[rand];
char[] asterisks = new char[MAXCHAR];
clearScreen(cont);//clear screen method call
hideWord(hiddenWord);//hidden word method call
System.out.println((hideWord(hiddenWord)));//print out hidden word in asterisks
System.out.println("");
System.out.println("");
numGuess(hiddenWord, asterisks);//method call to number of guesses/comparisions
} catch (Exception e)//Catch error when trying to open/find the file dictionary.txt
{
System.out.println("Error can not open dictionary.txt");
}
return count;
}
public static void clearScreen(String cont)//clear screen method for user
{
Scanner keyboard = new Scanner(System.in);
System.out.println("Press enter to continue:");
cont = keyboard.nextLine();
if (cont.equals(""));
{
for (int i = 0; i < 100; i++) {
System.out.println("");
}
}
}
public static String hideWord(String hiddenWord)//puts hidden word in asterisks
{
int wordLength = hiddenWord.length();
char[] asterisks = new char[wordLength];
for (int i = 0; i < wordLength; i++) {
asterisks[i] = '*';//couldn't figure out how to double space asterisks
}
String hideAsterisks = String.valueOf(asterisks);
return hideAsterisks;//return the value of
}
public static void numGuess(String hiddenWord, char[] asterisks)//compares guesses and counts
//attempts, however I feel like
//like I'm so close but I have a
//logic error in comparisons.
{
Scanner keyboard = new Scanner(System.in);
String hiddenword = hideWord(hiddenWord);
int remAttempts = MAXGUESS;
char attempts;
do {
System.out.println("Enter a letter or 9 to quit");
attempts = keyboard.next().charAt(0);
remAttempts--;
System.out.println("Attempts remaining: " + remAttempts);
for (int i = 0; i < hiddenWord.length() - 1; i++) {
if (attempts == (hiddenword.charAt(asterisks[i])))//trying to compare user input to *
{
System.out.println("Nice job!");
}
}
} while (attempts != '9' && remAttempts > 0);
}
}
【问题讨论】:
-
为什么不直接问你想知道什么?写了这么长的问题,我真的不知道你想要什么。
-
如何将星号的值发送到我的 numGuess 方法,然后要求用户输入一个字符,如果他/她的字符与星号的值匹配,用他/她的字符替换它
-
Char array updating 的可能重复项
-
在您的
clearScreen方法中,您有一个不起作用的if:if (cont.equals("")); {...}。;就像一个空语句,告诉 java if 已经完成,所以大括号{...}中的代码将始终执行——它独立于if。 -
顺便说一句:你可以写
\n换一个新行——这样你就不必连续使用这么多System.out.println:)
标签: java arrays while-loop