对于我的编程课程，我应该制作一个模拟tic tac toe游戏的程序。我的老师提供了所有的方法，并说我们不需要添加任何方法，也不需要任何方法，并告诉我们不要改变playGame方法，但不管我做什么，似乎在playGame之间存在某种脱节方法和processComputerMove / processHumanMove方法。每当我尝试运行程序时，似乎这些方法都被完全跳过，即使它们中没有任何东西，只是简单的打印语句。我想知道如何解决这个问题。到目前为止，这是我的（未完成和不工作）代码：
import java.util.*;
public class TicTacToe {
private static Scanner keyboard = new Scanner(System.in);
private static char[] board = new char[9];
// *******************************************************
public static void main(String[] args) {
String answer;
System.out.println("Welcome to Tic-Tac-Toe!!");
System.out.print("Do you want to play a game? (Y/N) ");
answer = keyboard.next();
if (answer.charAt(0) == 'Y' || answer.charAt(0) == 'y' ) {
playGame();
} else if (answer.charAt(0) == 'N' || answer.charAt(0) == 'n'){
System.out.println("Goodbye!");
}
}
// *******************************************************
private static void playGame() {
char currentPlayer = determineFirstPlayer();
System.out.println("The computer is X, the human is O.\n");
initializeGameBoard();
while (!isDone()) {
if (currentPlayer == 'c' || currentPlayer == 'C') { // the computer
processComputerMove();
currentPlayer = 'h'; // the human
} else {
displayGameBoard();
processHumanMove();
currentPlayer = 'c'; // the computer
}
}
displayGameBoard();
System.out.println("Game over!");
}
public static char determineFirstPlayer() {
System.out.print("Who would you like to make the first move? (h = human/c = computer) ");
String response = keyboard.next();
char player = response.charAt(0);
return player; // return 'c' or 'h' as appropriate
}
// *******************************************************
private static void initializeGameBoard() {
String playBoard = " | | \n---+---+---\n | | \n---+---+---\n | | ";
board[0] = playBoard.charAt(2);
board[1] = playBoard.charAt(6);
board[2] = playBoard.charAt(10);
board[3] = playBoard.charAt(13);
board[4] = playBoard.charAt(17);
board[5] = playBoard.charAt(21);
board[6] = playBoard.charAt(24);
board[7] = playBoard.charAt(28);
board[8] = playBoard.charAt(32);
}
// *******************************************************
private static void processComputerMove() {
Random rand = new Random();
int array = rand.nextInt(8);
int comp = rand.nextInt(8);
if (board[array] != 'X' || board[array] != 'O') {
board[array] = 'X';
System.out.println("The computer chooses cell "+board[array]+".");
} else {
board[array] = board[comp];
board[comp] = 'X';
}
}
// *******************************************************
private static void processHumanMove() {
System.out.println("Enter an empty position number (0-8) ");
int position = keyboard.nextInt();
if (board[position] != 'X' || board[position] != 'O') {
board[position] = 'O';
} else {
System.out.println("Error! Must enter an empty position from 0-9.");
}
}
// *******************************************************
private static void displayGameBoard() {
System.out.println(" "+board[0]+" | "+board[1]+" | "+board[2]+" \n---+---+---\n "+board[3]+" | "+board[4]+" | "+board[5]+" \n---+---+---\n "+board[6]+" | "+board[7]+" | "+board[8]+" ");
}
// *******************************************************
private static boolean isDone() {
isComputerWin();
isHumanWin();
isTie();
if (isComputerWin()) {
return true;
} else if(isHumanWin()) {
return true;
} else if (isTie()) {
return true;
} else {
return false;
}
}
// *******************************************************
private static boolean isComputerWin() {
if ((board[0] == 'X' && board[1] == 'X' && board[2] == 'X') || (board[0] == 'X' && board[3] == 'X' && board[6] == 'X') ||
(board[0] == 'X' && board[4] == 'X' && board[8] == 'X') || (board[3] == 'X' && board[4] == 'X' && board[5] == 'X') ||
(board[1] == 'X' && board[4] == 'X' && board[7] == 'X') || (board[2] == 'X' && board[5] == 'X' && board[8] == 'X') ||
(board[6] == 'X' && board[7] == 'X' && board[8] == 'X') || (board[6] == 'X' && board[4] == 'X' && board[2] == 'X')){
return true;
} else {
return false; // return true or false as appropriate
}
}
// *******************************************************
private static boolean isHumanWin() {
if ((board[0] == 'O' && board[1] == 'O' && board[2] == 'O') || (board[0] == 'O' && board[3] == 'O' && board[6] == 'O') ||
(board[0] == 'O' && board[4] == 'O' && board[8] == 'O') || (board[3] == 'O' && board[4] == 'O' && board[5] == 'O') ||
(board[1] == 'O' && board[4] == 'X' && board[7] == 'O') || (board[2] == 'O' && board[5] == 'O' && board[8] == 'O') ||
(board[6] == 'O' && board[7] == 'O' && board[8] == 'O') || (board[6] == 'O' && board[4] == 'O' && board[2] == 'O')){
return true;
} else {
return false; // return true or false as appropriate
}
}
// *******************************************************
private static boolean isTie() {
if (!((board[0] == 'O' && board[1] == 'O' && board[2] == 'O') || (board[0] == 'O' && board[3] == 'O' && board[6] == 'O') ||
(board[0] == 'O' && board[4] == 'O' && board[8] == 'O') || (board[3] == 'O' && board[4] == 'O' && board[5] == 'O') ||
(board[1] == 'O' && board[4] == 'X' && board[7] == 'O') || (board[2] == 'O' && board[5] == 'O' && board[8] == 'O') ||
(board[6] == 'O' && board[7] == 'O' && board[8] == 'O') || (board[6] == 'O' && board[4] == 'O' && board[2] == 'O')) ||
!((board[0] == 'X' && board[1] == 'X' && board[2] == 'X') || (board[0] == 'X' && board[3] == 'X' && board[6] == 'X') ||
(board[0] == 'X' && board[4] == 'X' && board[8] == 'X') || (board[3] == 'X' && board[4] == 'X' && board[5] == 'X') ||
(board[1] == 'X' && board[4] == 'X' && board[7] == 'X') || (board[2] == 'X' && board[5] == 'X' && board[8] == 'X') ||
(board[6] == 'X' && board[7] == 'X' && board[8] == 'X') || (board[6] == 'X' && board[4] == 'X' && board[2] == 'X'))) {
return true;
} else {
return false;
}
}
}
答案 0 :(得分：1)
您的isTie()方法是罪魁祸首之一。
使用调试器查看它，我发现每次检查“绑定”条件的布尔语句都是true的评估。
是的，我的意思是这个怪物声明：
(!((board[0] == 'O' && board[1] == 'O' && board[2] == 'O') || (board[0] == 'O' && board[3] == 'O' && board[6] == 'O') ||
(board[0] == 'O' && board[4] == 'O' && board[8] == 'O') || (board[3] == 'O' && board[4] == 'O' && board[5] == 'O') ||
(board[1] == 'O' && board[4] == 'X' && board[7] == 'O') || (board[2] == 'O' && board[5] == 'O' && board[8] == 'O') ||
(board[6] == 'O' && board[7] == 'O' && board[8] == 'O') || (board[6] == 'O' && board[4] == 'O' && board[2] == 'O'))
实际评估是错误的，但是由于你否定它，你将值翻转为true，从而使整个表达式短路，导致（最终）返回true。
我的直觉告诉我，你对领带的定义是偏斜的 - 不是一个领带更容易定义为既不是人类胜利，也不是计算机胜利？
如果我们减少isDone()方法只查看isHumanWin和isComputerWin（以及摆脱顶部无用的方法调用），并删除不必要的{ {1}}方法，我们得到（大多数）可玩的Tic-Tac-Toe游戏。它有点向第四步迈进，但这对读者来说是一种练习。
isTie