Question

这是我尝试使用Java重建战舰。我决定只用一艘船来测试更简单的游戏版本，并在游戏板上为游艇提供一个具体的位置。我发现我的代码有问题。无论我进入什么样的坐标，我最终“击中”了这艘船。

以下是我到目前为止编写的所有代码：

import java.util.Scanner;

class GameBoard {

    Scanner input = new Scanner(System.in);                                         // scanner object

    String[][] board = {                                                            // game board

    {"_", " 1", " 2", " 3", " 4", " 5", " 6", " 7", " 8", " 9", "10"},      
    {"A", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"B", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"C", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"D", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},            
    {"E", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"F", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"G", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"H", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"I", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"},
    {"J", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]", "[ ]"}

    };

    boolean frigateIsAlive = true;                                                  // the ship is still alive

    int numOfHitsOnFrigate = 0;                                                     // number of hits the player made on the frigate

    String [] frigate = {board[1][1], board[1][2]};                                 // ship 


    public void createBoard(){                                                      // draws the battleship game board
        for (int row = 0; row < board.length; row++) {
            for (int col = 0; col < board[row].length; col++) {
                System.out.print(board[row][col] + "\t");

            } // inner loop
            System.out.println();
            System.out.println();
            System.out.println();
        } // outer loop
}

    public String getUserGuess() {                                                  // takes the users guess

        System.out.println("Choose a coordinate on the board to fire at");
        int x = input.nextInt();                                                        
        int y = input.nextInt();

        String userGuess = board[x][y];
        return userGuess;
    }



    public void checkResult(String userGuess) {                                     // checks the user's guess

            if(userGuess.equalsIgnoreCase(frigate[0])){
                System.out.println("hit!");
                numOfHitsOnFrigate++;
                board[1][1] = " *";
                createBoard();
            }
            else if(userGuess.equalsIgnoreCase(frigate[1])) {
                System.out.println("hit!");
                numOfHitsOnFrigate++;
                board[1][2] = " *";
                createBoard();
            }
            else {
                System.out.println("miss!");
            }
            if (numOfHitsOnFrigate == 2) {
                System.out.println("Enemy frigate has been sunk!");
                frigateIsAlive = false;
            }

        } 



} // end class


public class Game {

public static void run() {
    GameBoard newGame = new GameBoard();

    newGame.createBoard();

    while(newGame.frigateIsAlive) {
    newGame.checkResult(newGame.getUserGuess());
    }






    }

}


public class App {

    public static void main(String[] args) {

        Game.run();

    }
}

Answer 1

由于frigate的声明是：

，船总是被击中

  frigate = {board[1][1], board[1][2]}

，最终将字符串'[ ]'分配给护卫舰的两个值。然后，当您正在寻找护卫舰并比较值时，将其与更多空字符串进行比较。

这可以通过在[1,2,3,4, n]和[A,B,C...,Letter_n]中创建位置x的板来修复。也就是说，护卫舰的坐标为Frigate.x = 1和Frigate.y = A。

我希望这有帮助！

我看到了你如何实现这一点的进一步问题。我会让Frigate成为一个有坐标列表的类：

this.x作为一个字母或数字
this.y作为一个点不是像你的例子中那样的类型
元组(this.x, this.y)在您的列表Frigate中运行良好
对Frigate列表中的任何其他点执行相同的操作。

在Frigate列表完成之后，还有两件事需要改变。

首先要改变的是如何检查用户是否正在调用您想要的范围内的内容。

必须改变的第二件事是如何确保不会一次又一次地调用同一点以“炸毁”一艘船。也就是说，当Frigate中的一个点被调用时，它应该从Frigate中移除。护卫舰中剩余的元组将成为护卫舰留下的“生命值”。要想起Frigate的原始大小，添加Frigate.initialSize()会非常方便，但这可能会在以后发生。

战舰代码不起作用

1 个答案: